Forces in Two Dimensions
| Hệ thống: | Young Education |
| Khoá học: | Forces |
| Book: | Forces in Two Dimensions |
| Được in bởi: | ゲストユーザ |
| Ngày: | Thứ Sáu, 25 tháng 9 2026, 1:01 AM |
1. Resolving Forces into Components
Learning outcomes
- I can resolve forces into horizontal and vertical components.
- I can use trigonometry to calculate force components.
- I can represent components on diagrams.
- I can reconstruct resultant forces from components.
- I can apply force components to physical situations.
What Does It Mean to Resolve a Force?
A force is a vector quantity, which means it has both:
- magnitude
- direction
When a force acts at an angle, it can often be easier to analyze if we separate it into two perpendicular forces called components.
This process is called resolving a force into components.
Usually, we resolve a force into:
Horizontal component: Fx
Vertical component: Fy
Together, these components have exactly the same effect as the original force.
One Force, Two Components
Imagine a person pulling a box using a rope that makes an angle with the ground.
The tension force acts diagonally.
However, that force has two effects:
- it pulls the box forward
- it pulls the box upward
We can therefore replace the diagonal force with:
Fx → horizontal component
and
Fy ↑ vertical component
The original force and its two components are equivalent.
Components Form a Right Triangle
When a force is resolved into horizontal and vertical components, the three vectors form a right-angled triangle.
The original force is the hypotenuse.
The horizontal and vertical components form the other two sides.
If the angle θ is measured from the horizontal, then:
- F is the hypotenuse
- Fx is adjacent to θ
- Fy is opposite θ
This means we can use trigonometry.
Using Sine and Cosine
Remember:
SOH CAH TOA
Sine:
sin θ = opposite ÷ hypotenuse
Cosine:
cos θ = adjacent ÷ hypotenuse
Tangent:
tan θ = opposite ÷ adjacent
For a force at an angle measured from the horizontal:
Fx = F cos θ
Fy = F sin θ
where:
- F = original force
- Fx = horizontal component
- Fy = vertical component
- θ = angle measured from the horizontal
Why Is Cosine Horizontal?
Suppose the angle is measured from the horizontal.
The horizontal component lies next to the angle, so it is the adjacent side.
Therefore:
cos θ = Fx / F
Rearranging:
Fx = F cos θ
The vertical component is opposite the angle:
sin θ = Fy / F
Therefore:
Fy = F sin θ
A useful rule is:
Angle from horizontal: cosine gives horizontal, sine gives vertical.
But do not rely only on memorization. Identify the adjacent and opposite sides of the triangle.
Worked Example 1: Resolving a Force
A person pulls a sled with a force of 100 N at an angle of 30° above the horizontal.
Calculate the horizontal and vertical components.
For the horizontal component:
Fx = F cos θ
Fx = 100 cos 30°
Fx ≈ 86.6 N
For the vertical component:
Fy = F sin θ
Fy = 100 sin 30°
Fy = 50.0 N
Therefore:
Horizontal component = 86.6 N
Vertical component = 50.0 N
The original 100 N force therefore has the same effect as:
86.6 N horizontally
and
50.0 N vertically
Understanding the Components
It is important to understand that the 100 N force has not been divided into 86.6 N + 50 N.
In fact:
86.6 + 50 ≠ 100
Vectors cannot generally be added using ordinary arithmetic because they act in different directions.
Instead, the components combine using vector mathematics.
The original force is:
F = √(Fx² + Fy²)
For our example:
F = √(86.6² + 50²)
F ≈ 100 N
What Happens as the Angle Changes?
Consider a force of constant magnitude.
If the angle above the horizontal increases:
horizontal component decreases
while:
vertical component increases
For example, consider a 100 N force:
| Angle | Horizontal Component | Vertical Component |
|---|---|---|
| 0° | 100 N | 0 N |
| 30° | 86.6 N | 50.0 N |
| 45° | 70.7 N | 70.7 N |
| 60° | 50.0 N | 86.6 N |
| 90° | 0 N | 100 N |
At 0°, the force is entirely horizontal.
At 90°, the force is entirely vertical.
At 45°, the horizontal and vertical components are equal.
Forces Measured from the Vertical
Be careful: sometimes the angle is measured from the vertical rather than the horizontal.
If θ is measured from the vertical:
The vertical component is adjacent to the angle.
Therefore:
Fy = F cos θ
The horizontal component is opposite the angle.
Therefore:
Fx = F sin θ
This is why simply memorizing "cosine is horizontal" can cause mistakes.
Always ask:
Which component is adjacent to the given angle?
Representing Components on a Diagram
Suppose a force acts upward and to the right.
The original force can be drawn as:
↗ F
Its components are:
↑ Fy
→ Fx
The horizontal and vertical arrows should begin from the same point as the original vector when illustrating the resolution.
Together they form a right triangle.
Correct diagrams are especially important because they help determine:
- which component uses sine
- which component uses cosine
- whether components are positive or negative
Positive and Negative Components
Components can act in different directions.
A common coordinate system uses:
Right = positive x
Left = negative x
Up = positive y
Down = negative y
Therefore:
A force acting upward and right has:
Fx positive
Fy positive
A force acting upward and left has:
Fx negative
Fy positive
A force acting downward and left has:
Fx negative
Fy negative
Signs become especially important when combining several forces.
Reconstructing a Force from Its Components
Sometimes we know the horizontal and vertical components and need to find the original force.
Suppose:
Fx = 40 N
Fy = 30 N
The components form a right triangle.
Use the Pythagorean theorem:
F² = Fx² + Fy²
Therefore:
F = √(Fx² + Fy²)
Substitute:
F = √(40² + 30²)
F = √(1600 + 900)
F = √2500
F = 50 N
Finding the Direction of the Resultant
We can use tangent to determine the angle.
Since:
tan θ = opposite ÷ adjacent
then:
tan θ = Fy / Fx
Therefore:
θ = tan⁻¹(Fy / Fx)
For the previous example:
θ = tan⁻¹(30 / 40)
θ ≈ 36.9°
Therefore, the resultant force is:
50 N at approximately 37° above the horizontal.
Components and Resultant Forces
Resolving forces becomes especially useful when several forces act on the same object.
Suppose an object experiences:
Force A:
60 N to the right
Force B:
40 N upward
These two forces are perpendicular.
The resultant is:
Fresultant = √(60² + 40²)
Fresultant = √5200
Fresultant ≈ 72.1 N
Direction:
θ = tan⁻¹(40 / 60)
θ ≈ 33.7°
Therefore:
Resultant ≈ 72 N at 34° above the horizontal
Multiple Forces
For systems containing several forces, it is usually easiest to resolve every force into x and y components.
Then calculate:
ΣFx = sum of horizontal components
and:
ΣFy = sum of vertical components
These totals represent the components of the resultant force.
Then:
Fresultant = √((ΣFx)² + (ΣFy)²)
and:
θ = tan⁻¹(ΣFy / ΣFx)
The signs of the components must be considered carefully.
Worked Example 2: Two Angled Forces
Suppose two forces act on an object.
Force A:
100 N at 30° above the horizontal
Force B:
40 N horizontally left
First resolve Force A.
Horizontal:
Fx = 100 cos 30°
Fx = 86.6 N right
Vertical:
Fy = 100 sin 30°
Fy = 50.0 N upward
Now include Force B:
Horizontal resultant:
ΣFx = 86.6 − 40
ΣFx = 46.6 N
Vertical resultant:
ΣFy = 50.0 N
Now find the resultant magnitude:
F = √(46.6² + 50²)
F ≈ 68.4 N
The system therefore has a resultant force of approximately 68 N.
Pulling an Object at an Angle
A very common physical situation involves pulling a box or sled using an angled rope.
Suppose the tension is:
T = 120 N
at:
θ = 25°
The horizontal component is:
Tx = T cos θ
Tx = 120 cos 25°
Tx ≈ 108.8 N
The vertical component is:
Ty = T sin θ
Ty = 120 sin 25°
Ty ≈ 50.7 N
The horizontal component pulls the object forward.
The vertical component pulls upward.
Angled Forces and the Normal Force
The vertical component of an angled pull can affect the normal force.
Suppose a box is on a horizontal floor.
Vertically, the forces are:
↑ N
↑ Ty
↓ mg
If there is no vertical acceleration:
N + Ty = mg
Therefore:
N = mg − Ty
This means an upward component of tension reduces the normal force.
Since friction can be calculated using:
Ffriction = μN
reducing the normal force can also reduce friction.
This explains why pulling a heavy object slightly upward can sometimes make it easier to move.
Worked Example 3: Pulling a Box
A 20 kg box is pulled with a force of 100 N at 30° above the horizontal.
Calculate the horizontal component of the pulling force.
Fx = F cos θ
Fx = 100 cos 30°
Fx = 86.6 N
Now calculate the upward component:
Fy = F sin θ
Fy = 100 sin 30°
Fy = 50 N
The weight of the box is:
Fg = mg
Fg = 20 × 9.8
Fg = 196 N
If there is no vertical acceleration:
N + Fy = Fg
N + 50 = 196
N = 146 N
Notice that the normal force is less than the object's weight because the pulling force has an upward component.
Pushing Down at an Angle
Now imagine pushing a box downward at an angle.
The vertical component now acts downward.
Therefore:
N = mg + Fy
The normal force becomes larger.
Because:
Ffriction = μN
the friction force can also become larger.
This is one reason why pulling an object upward at an angle can be easier than pushing it downward at the same angle and force.
Components on an Inclined Plane
Resolving forces is also extremely useful for objects on slopes.
For an object on a slope, it is often more convenient to choose axes:
- parallel to the slope
- perpendicular to the slope
The weight mg can then be resolved into components.
For a slope angle θ:
Component parallel to slope:
Fg∥ = mg sin θ
Component perpendicular to slope:
Fg⊥ = mg cos θ
The parallel component tends to make the object slide down the slope.
The perpendicular component pushes the object against the surface.
Worked Example 4: Object on a Slope
A 10 kg object rests on a 30° slope.
Its weight is:
Fg = mg
Fg = 10 × 9.8
Fg = 98 N
Component down the slope:
Fg∥ = mg sin θ
Fg∥ = 98 sin 30°
Fg∥ = 49 N
Component perpendicular to the slope:
Fg⊥ = mg cos θ
Fg⊥ = 98 cos 30°
Fg⊥ ≈ 84.9 N
If there are no other perpendicular forces:
N ≈ 84.9 N
Components and Newton's Second Law
Once forces have been resolved, Newton's Second Law can be applied separately in each direction.
Horizontal:
ΣFx = max
Vertical:
ΣFy = may
For many objects moving along a horizontal surface:
ay = 0
Therefore:
ΣFy = 0
But the object may accelerate horizontally:
ΣFx = max
This allows complicated two-dimensional force situations to be reduced to two simpler one-dimensional problems.
Worked Example 5: Components, Friction, and Acceleration
A 10 kg box is pulled with a force of 50 N at 30° above the horizontal. A friction force of 15 N acts opposite the motion.
First find the horizontal pulling component:
Fx = F cos θ
Fx = 50 cos 30°
Fx ≈ 43.3 N
Now calculate the horizontal net force:
Fnet,x = 43.3 − 15
Fnet,x = 28.3 N
Use Newton's Second Law:
Fnet = ma
28.3 = 10a
a = 2.83 m/s²
The box accelerates horizontally at approximately:
2.8 m/s²
Components in Engineering
Engineers regularly resolve forces into components when designing structures.
Examples include:
- suspension bridge cables
- crane cables
- roof supports
- towers
- guy wires
- trusses
- elevators
- aircraft structures
A cable pulling diagonally may create both horizontal and vertical forces on a structure.
Engineers must calculate these components to determine whether the structure can safely support the forces.
Components in Aircraft
Aircraft provide another important application.
During level flight, lift acts mainly upward.
When an aircraft banks during a turn, the lift force becomes tilted.
The lift can then be resolved into:
- a vertical component
- a horizontal component
The vertical component helps support the aircraft's weight.
The horizontal component contributes to the turning motion.
This is another example of how one angled force can produce effects in two directions.
Components in Everyday Life
Resolving forces helps explain many familiar situations.
Examples include:
- pulling a suitcase
- pulling a wagon
- towing a vehicle
- climbing with ropes
- pushing a lawn mower
- pulling a sled
- lifting objects with cranes
- supporting structures with cables
Whenever a force acts at an angle, resolving it into components can make the situation easier to analyze.
Did You Know?
Rock climbers and rescue teams must carefully consider force components when setting up ropes and anchors.
Two ropes supporting the same load do not necessarily experience half the load each.
The angles of the ropes matter.
As ropes become more horizontal, the tension required to support the same vertical load can become very large.
This is why rope angles are an important consideration in climbing, rescue systems, bridge cables, and engineering structures.
Common Mistakes
Mistake 1: Automatically using cosine for the horizontal component
Cosine gives the component adjacent to the angle.
If the angle is measured from the vertical, cosine gives the vertical component instead.
Mistake 2: Adding perpendicular components normally
You cannot simply calculate:
Fx + Fy = F
Perpendicular components must be combined using the Pythagorean theorem.
Mistake 3: Ignoring direction
Left and downward components are often represented as negative.
Mistake 4: Forgetting that components replace the original force
When you resolve a force into components for a calculation, do not then include both the original force and its components as separate forces. That would count the same force twice.
Mistake 5: Using degrees incorrectly on a calculator
For most introductory force problems, make sure the calculator is in degree mode, not radian mode.
A Strategy for Resolving Forces
When an angled force appears:
- Draw the force vector.
- Identify the angle and what direction it is measured from.
- Draw horizontal and vertical components.
- Form a right triangle.
- Identify the hypotenuse, adjacent side, and opposite side.
- Use sine or cosine to calculate the components.
- Assign positive or negative signs based on direction.
- Combine components from other forces.
- Use Fnet = ma if motion must be analyzed.
- Check whether the result makes physical sense.
Reconstructing a Resultant Force
When the horizontal and vertical components are known:
Magnitude:
F = √(Fx² + Fy²)
Direction:
θ = tan⁻¹(Fy / Fx)
Be careful with the direction and signs of the components when identifying the correct angle or quadrant.
Key Terms
Vector: A quantity with both magnitude and direction.
Component: One part of a vector acting in a particular direction.
Resolving a force: Separating a force into perpendicular components.
Horizontal component: The part of a force acting horizontally.
Vertical component: The part of a force acting vertically.
Resultant force: The single force that has the same overall effect as several combined forces.
Magnitude: The size of a vector quantity.
Direction: The orientation in which a vector acts.
Key Equations
For an angle measured from the horizontal:
Fx = F cos θ
Fy = F sin θ
Resultant magnitude:
F = √(Fx² + Fy²)
Direction:
θ = tan⁻¹(Fy / Fx)
Newton's Second Law in two dimensions:
ΣFx = max
ΣFy = may
For an object on a slope:
Fg∥ = mg sin θ
Fg⊥ = mg cos θ
Key Takeaways
- Forces are vectors and can be separated into perpendicular components.
- Resolving a force means replacing one angled force with equivalent components.
- Horizontal and vertical components form a right-angled triangle with the original force.
- If the angle is measured from the horizontal, Fx = F cos θ and Fy = F sin θ.
- Always determine which side is adjacent and which is opposite rather than relying only on memorized rules.
- Components have directions and may therefore be positive or negative.
- The original force can be reconstructed using the Pythagorean theorem.
- The direction of a resultant can be calculated using inverse tangent.
- Multiple forces can be analyzed by adding their horizontal components and vertical components separately.
- Force components can be combined with Newton's Second Law to analyze acceleration.
- Angled pulling forces can change the normal force and therefore affect friction.
- On inclined planes, gravity can be resolved into components parallel and perpendicular to the slope.
- Resolving forces is widely used in engineering, transportation, construction, aviation, and everyday mechanics.
2. Equilibrium in Two Dimensions
Learning outcomes
- I can identify conditions for equilibrium in two dimensions.
- I can calculate unknown forces in equilibrium systems.
- I can analyze force vectors acting in multiple directions.
- I can solve equilibrium problems using components.
- I can explain real-world examples of equilibrium.
What Is Equilibrium?
An object is in equilibrium when the resultant force acting on it is zero.
In one dimension, this might simply mean that forces to the left equal forces to the right.
In two dimensions, forces can act horizontally, vertically, and at angles. For equilibrium, the forces must balance in both dimensions.
Therefore:
Net horizontal force = 0
and:
Net vertical force = 0
We write these conditions as:
ΣFx = 0
ΣFy = 0
The symbol Σ means "the sum of."
So ΣFx means the sum of all horizontal force components.
Equilibrium Does Not Necessarily Mean Stationary
An object in equilibrium does not have to be at rest.
Newton's First Law tells us that if:
Fnet = 0
then:
a = 0
The object could therefore be:
- stationary
- moving at constant velocity
For example, a car traveling along a straight road at constant velocity may have:
Driving force = resistive forces
and:
Normal force = weight
The forces are balanced even though the car is moving.
Equilibrium in Two Dimensions
Consider an object with forces acting in several directions.
Some forces may act:
- horizontally
- vertically
- diagonally
For equilibrium, the vector sum of all these forces must equal zero.
Instead of trying to balance diagonal forces directly, we can resolve them into horizontal and vertical components.
Then we analyze the two directions separately.
Horizontal:
ΣFx = 0
Vertical:
ΣFy = 0
This turns a complicated two-dimensional problem into two simpler one-dimensional problems.
A Simple Equilibrium Example
Imagine a hanging sign.
Its weight acts downward.
Two cables pull upward at angles.
The forces might look like:
↖ T₁
●
T₂ ↗
↓
W
The sign remains stationary because all of the forces balance.
Horizontally:
leftward component of T₁ = rightward component of T₂
Vertically:
upward components of T₁ and T₂ = weight
Resolving Forces in Equilibrium Problems
Suppose a tension force T acts at an angle θ above the horizontal.
Its components are:
Horizontal:
Tx = T cos θ
Vertical:
Ty = T sin θ
If the force points upward and to the right:
Tx is positive
Ty is positive
If the force points upward and to the left:
Tx is negative
Ty is positive
Correct signs are essential when writing equilibrium equations.
Choosing Positive Directions
Before solving a problem, choose positive directions.
A common convention is:
Right = +x
Left = −x
Up = +y
Down = −y
Then equilibrium requires:
ΣFx = 0
ΣFy = 0
For example:
50 N right + 30 N left
becomes:
ΣFx = +50 − 30
Direction is part of the calculation.
Worked Example 1: Horizontal and Vertical Forces
An object experiences four forces:
- 80 N right
- 80 N left
- 120 N upward
- 120 N downward
Horizontally:
ΣFx = 80 − 80
ΣFx = 0 N
Vertically:
ΣFy = 120 − 120
ΣFy = 0 N
Therefore:
Fnet = 0
The object is in equilibrium.
Equilibrium with Angled Forces
Now consider a more realistic situation.
A 100 N object is supported by two identical cables. Each cable makes an angle of 30° above the horizontal.
Because the system is symmetrical:
T₁ = T₂ = T
The horizontal components point in opposite directions.
Therefore, they cancel:
T cos 30° − T cos 30° = 0
The vertical components must support the 100 N weight.
Each cable contributes:
T sin 30°
Therefore:
T sin 30° + T sin 30° = 100
or:
2T sin 30° = 100
Since:
sin 30° = 0.5
we get:
2T(0.5) = 100
T = 100 N
Each cable has a tension of 100 N.
Why Can Tension Be Larger Than Expected?
Students sometimes assume that if two cables support a 100 N object, each cable must provide 50 N of tension.
That is only true if both cables pull straight upward.
When cables act at angles, only the vertical component of each tension supports the object's weight.
A cable may have a tension of 100 N while providing only 50 N of upward force.
The rest of the tension acts horizontally.
This is an extremely important idea in engineering and structural design.
Cable Angle and Tension
Consider a load supported symmetrically by two cables.
If the cables are steep, a large proportion of each tension acts vertically.
If the cables become more horizontal, a smaller proportion acts vertically.
Therefore, a larger tension is required to support the same load.
For two identical cables making angle θ above the horizontal:
2T sin θ = W
Therefore:
T = W / (2 sin θ)
As θ becomes smaller, sin θ becomes smaller.
Therefore:
smaller angle above horizontal → larger tension
This principle is extremely important when lifting heavy loads.
Worked Example 2: Supporting a Load
A 600 N load is supported by two identical cables. Each cable makes an angle of 45° above the horizontal.
Calculate the tension in each cable.
Vertical equilibrium requires:
ΣFy = 0
The upward forces are:
T sin 45° + T sin 45°
Therefore:
2T sin 45° = 600
Using:
sin 45° ≈ 0.707
2T(0.707) = 600
1.414T = 600
T ≈ 424 N
Tension in each cable ≈ 424 N
Notice that each tension is greater than 300 N because only part of each tension acts vertically.
Free-Body Diagrams for Equilibrium
A free-body diagram is one of the most useful tools for solving equilibrium problems.
A good free-body diagram should:
- show the object as a simple point or shape
- show every external force
- show the direction of each force
- label forces clearly
- include relevant angles
- avoid showing forces that do not act on the object
Once the diagram is complete, angled forces can be resolved into components.
A Strategy for Solving Two-Dimensional Equilibrium Problems
A reliable approach is:
- Identify the object being analyzed.
- Draw a free-body diagram.
- Choose positive x and y directions.
- Resolve angled forces into components.
- Write the horizontal equilibrium equation:
ΣFx = 0
- Write the vertical equilibrium equation:
ΣFy = 0
- Substitute known values.
- Solve for the unknown force or forces.
- Check that the forces balance in both directions.
The free-body diagram should usually come before the equations.
Worked Example 3: Cable and Horizontal Force
A 200 N object is held stationary by a diagonal cable and a horizontal rope.
The cable makes an angle of 60° above the horizontal.
Let the diagonal cable tension be T.
Its vertical component supports the weight:
T sin 60° = 200
T(0.866) = 200
T ≈ 231 N
Now find the horizontal component:
Tx = T cos 60°
Tx = 231 × 0.5
Tx ≈ 116 N
The horizontal rope must balance this force.
Therefore:
Horizontal rope tension ≈ 116 N
The system is balanced in both dimensions.
Unknown Forces Using Components
Sometimes two unknown forces must be determined.
For example, a sign may be supported by two cables at different angles.
Suppose:
T₁ acts upward-left at angle α.
T₂ acts upward-right at angle β.
Weight W acts downward.
Horizontal equilibrium:
T₂ cos β − T₁ cos α = 0
Therefore:
T₂ cos β = T₁ cos α
Vertical equilibrium:
T₁ sin α + T₂ sin β − W = 0
Therefore:
T₁ sin α + T₂ sin β = W
These two equations can be solved simultaneously to determine T₁ and T₂.
Worked Example 4: Two Different Cable Angles
A 500 N sign is supported by two cables.
The left cable makes an angle of 30° above the horizontal.
The right cable makes an angle of 60° above the horizontal.
Let their tensions be T₁ and T₂.
Horizontal equilibrium:
T₂ cos 60° = T₁ cos 30°
0.5T₂ = 0.866T₁
Therefore:
T₂ = 1.732T₁
Now use vertical equilibrium:
T₁ sin 30° + T₂ sin 60° = 500
0.5T₁ + 0.866T₂ = 500
Substitute:
0.5T₁ + 0.866(1.732T₁) = 500
0.5T₁ + 1.5T₁ = 500
2T₁ = 500
T₁ = 250 N
Then:
T₂ = 1.732 × 250
T₂ ≈ 433 N
Therefore:
Left cable tension = 250 N
Right cable tension ≈ 433 N
Checking the Answer
We can check the vertical forces.
Left vertical component:
250 sin 30° = 125 N
Right vertical component:
433 sin 60° ≈ 375 N
Total upward force:
125 + 375 = 500 N
This balances the 500 N weight.
Horizontally:
250 cos 30° ≈ 216.5 N
433 cos 60° ≈ 216.5 N
The horizontal forces also balance.
Therefore:
ΣFx = 0
and:
ΣFy = 0
The answer is consistent with equilibrium.
Equilibrium on an Inclined Plane
Two-dimensional equilibrium can also occur on a slope.
For an object resting on an incline, it is often useful to choose axes:
- parallel to the slope
- perpendicular to the slope
Weight can then be resolved into:
Parallel component:
Fg∥ = mg sin θ
Perpendicular component:
Fg⊥ = mg cos θ
If the object remains stationary, another force such as static friction must balance the component acting down the slope.
Therefore:
ΣFparallel = 0
and:
ΣFperpendicular = 0
Worked Example 5: Object on a Slope
A 10 kg box rests on a 30° slope.
Its weight is:
Fg = mg
Fg = 10 × 9.8
Fg = 98 N
Component parallel to the slope:
Fg∥ = 98 sin 30°
Fg∥ = 49 N
Component perpendicular to the slope:
Fg⊥ = 98 cos 30°
Fg⊥ ≈ 84.9 N
If the box remains stationary, static friction must balance the force down the slope:
Ffriction = 49 N upward along the slope
The normal force balances the perpendicular component:
N = 84.9 N
Therefore, the box is in equilibrium.
Equilibrium and Newton's Laws
Equilibrium is closely connected to Newton's Laws of Motion.
Newton's Second Law states:
Fnet = ma
If an object is in equilibrium:
Fnet = 0
Therefore:
ma = 0
For an object with mass:
a = 0
This is why equilibrium means there is no acceleration.
The object may remain stationary or continue moving at constant velocity.
Equilibrium in Bridges
Bridges contain many structures that must remain in equilibrium.
Forces may include:
- weight
- tension
- compression
- support forces
- wind forces
- forces produced by vehicles
Engineers calculate the horizontal and vertical components of these forces to make sure the structure remains stable.
If the forces were not properly balanced, parts of the structure could accelerate, deform, or fail.
Equilibrium in Cranes
Cranes provide another important example.
When a load hangs motionless from cables:
ΣFx = 0
ΣFy = 0
The vertical components of the cable tensions must balance the weight of the load.
If several angled cables are used, their horizontal components must also cancel.
Rigging systems therefore depend heavily on two-dimensional force calculations.
Equilibrium in Guy Wires
Tall towers and masts are often stabilized using guy wires.
The wires pull diagonally on the tower.
Each tension can be separated into horizontal and vertical components.
When designed correctly, the forces from different wires help balance one another and stabilize the structure.
Equilibrium in the Human Body
The human body also provides examples of force equilibrium.
When a person stands still:
- gravity pulls downward
- the ground exerts an upward normal force
When a climber remains stationary on a wall, forces from:
- gravity
- hands
- feet
- ropes
must combine to produce zero resultant force.
Biomechanics uses these principles to study movement, posture, muscles, joints, and sports performance.
Vector Polygons and Equilibrium
There is another way to recognize equilibrium.
If all force vectors are drawn head-to-tail, an equilibrium system produces a closed vector shape.
For three forces, this may form a triangle of forces.
If the final vector returns to the starting point:
Resultant = 0
Therefore, the forces are in equilibrium.
This graphical method provides a useful visual way of understanding why balanced forces have no resultant.
Translational and Rotational Equilibrium
So far, we have considered whether an object accelerates in a straight line. This is called translational equilibrium.
For translational equilibrium:
ΣFx = 0
ΣFy = 0
However, an object could have zero resultant force and still begin rotating if the forces produce an unbalanced turning effect.
For complete static equilibrium, we also need rotational equilibrium:
Στ = 0
where τ represents torque.
Therefore, complete static equilibrium requires:
ΣFx = 0
ΣFy = 0
Στ = 0
Torque is usually studied separately, but it is important to recognize that balancing forces alone does not guarantee that an extended object cannot rotate.
Did You Know?
When engineers lift heavy objects using two angled slings, making the slings nearly horizontal can create extremely large tensions.
This happens because only the vertical components of the tensions support the load.
As the cables become more horizontal, the vertical component becomes a smaller fraction of the total tension.
The cable tension must therefore increase dramatically to produce the required upward force.
This is why sling and cable angles are a major safety consideration in construction, climbing, rescue work, and engineering.
Common Mistakes
Mistake 1: Checking only one direction
For two-dimensional equilibrium, both conditions must be satisfied:
ΣFx = 0
ΣFy = 0
Mistake 2: Assuming two cables each carry half the weight
The cable angles determine how much vertical force each tension provides.
Mistake 3: Using the full angled force in both directions
An angled force must first be resolved into components.
Mistake 4: Ignoring negative directions
Leftward and downward components should be treated consistently as negative if right and up are chosen as positive.
Mistake 5: Thinking equilibrium means no forces
Many forces may act on an object in equilibrium. Their vector sum is zero.
Mistake 6: Thinking equilibrium always means stationary
An object moving at constant velocity can also be in translational equilibrium.
A Problem-Solving Checklist
For an equilibrium problem:
- Identify the object.
- Draw a free-body diagram.
- Label every force and angle.
- Choose positive directions.
- Resolve angled forces into components.
- Write ΣFx = 0.
- Write ΣFy = 0.
- Solve the equations.
- Check the horizontal forces.
- Check the vertical forces.
- Make sure the calculated directions and magnitudes make physical sense.
For extended objects, also consider whether rotational equilibrium must be checked.
Key Terms
Equilibrium: A condition in which the resultant force on an object is zero.
Translational equilibrium: A condition in which the net force is zero and there is no linear acceleration.
Static equilibrium: Equilibrium in which an object remains at rest.
Force component: The part of a force acting along a chosen direction.
Resultant force: The vector sum of all forces acting on an object.
Free-body diagram: A diagram showing all external forces acting on an object.
Tension: A pulling force transmitted through a rope, cable, or string.
Torque: The turning effect of a force.
Key Equations
Horizontal equilibrium:
ΣFx = 0
Vertical equilibrium:
ΣFy = 0
Angled force measured from the horizontal:
Fx = F cos θ
Fy = F sin θ
Weight:
Fg = mg
Newton's Second Law:
Fnet = ma
For complete static equilibrium:
ΣFx = 0
ΣFy = 0
Στ = 0
Key Takeaways
- An object is in translational equilibrium when the resultant force is zero.
- In two dimensions, horizontal and vertical forces must balance separately.
- The conditions are ΣFx = 0 and ΣFy = 0.
- Equilibrium means zero acceleration, not necessarily zero velocity.
- Angled forces can be resolved into horizontal and vertical components.
- Free-body diagrams are essential for analyzing equilibrium systems.
- Unknown forces can be calculated by applying the equilibrium equations separately in each direction.
- Cable angles strongly affect the tension required to support a load.
- More horizontal support cables generally require larger tensions.
- Objects on slopes can be analyzed using components parallel and perpendicular to the slope.
- Bridges, cranes, towers, rigging systems, and the human body provide real-world examples of two-dimensional equilibrium.
- Complete static equilibrium of an extended object also requires the net torque to be zero.
3. Inclined Planes
Learning outcomes
- I can identify forces acting on an object on an incline.
- I can resolve weight into parallel and perpendicular components.
- I can calculate net force on an inclined plane.
- I can analyze motion on slopes with and without friction.
- I can solve problems involving inclined planes.
What Is an Inclined Plane?
An inclined plane is a flat surface that is tilted at an angle to the horizontal.
Common examples include:
- ramps
- hills
- roads on slopes
- playground slides
- loading ramps
- roofs
- ski slopes
When an object is placed on an incline, gravity still acts vertically downward. However, part of the gravitational force tends to pull the object down the slope.
This makes inclined planes an important application of force components.
Forces Acting on an Inclined Plane
Consider a box resting on a slope.
Several forces may act on the box:
- weight (Fg) acting vertically downward
- normal force (N) acting perpendicular to the surface
- friction (Ff) acting along the surface when appropriate
- tension or applied forces if the object is being pulled or pushed
A common mistake is to draw the normal force vertically upward.
On an incline, the normal force is not vertical.
The normal force always acts:
perpendicular to the surface
Choosing Axes on an Incline
For horizontal surfaces, we normally use horizontal and vertical axes.
For inclined planes, a more useful coordinate system is:
x-axis: parallel to the slope
y-axis: perpendicular to the slope
This makes calculations much easier because the normal force and friction already lie along these axes.
Gravity is then the force that must be resolved into components.
Resolving Weight on an Incline
The gravitational force is:
Fg = mg
and always acts vertically downward.
On an incline, we resolve weight into two components:
Fg∥ = component parallel to the slope
Fg⊥ = component perpendicular to the slope
For a slope at angle θ:
Parallel component:
Fg∥ = mg sin θ
Perpendicular component:
Fg⊥ = mg cos θ
The parallel component pulls the object down the slope.
The perpendicular component pushes the object into the surface.
Why Is It mg sin θ Down the Slope?
The geometry of the force triangle shows that the component parallel to the slope is opposite the angle θ.
Therefore:
sin θ = Fg∥ / mg
Rearranging:
Fg∥ = mg sin θ
The perpendicular component is adjacent to θ:
cos θ = Fg⊥ / mg
Therefore:
Fg⊥ = mg cos θ
These two equations are central to solving inclined-plane problems.
The Normal Force on an Incline
If there are no other forces acting perpendicular to the slope and the object does not accelerate away from the surface:
N = Fg⊥
Therefore:
N = mg cos θ
Notice:
N ≠ mg
except when the surface is horizontal.
As the incline becomes steeper, the normal force becomes smaller.
At:
θ = 0°
cos 0° = 1
so:
N = mg
On a very steep incline, the normal force becomes much smaller.
Worked Example 1: Resolving Weight
A 10 kg box rests on a 30° incline.
Calculate the components of its weight.
First calculate weight:
Fg = mg
Fg = 10 × 9.8
Fg = 98 N
Parallel component:
Fg∥ = mg sin θ
Fg∥ = 98 sin 30°
Fg∥ = 49 N
Perpendicular component:
Fg⊥ = mg cos θ
Fg⊥ = 98 cos 30°
Fg⊥ ≈ 84.9 N
Therefore:
Force down the slope = 49 N
Force into the slope ≈ 84.9 N
If there are no other perpendicular forces:
N ≈ 84.9 N
A Frictionless Inclined Plane
First consider a perfectly smooth incline with no friction.
The forces are:
- weight
- normal force
The normal force balances the perpendicular component of weight.
Therefore:
N = mg cos θ
But there is nothing to balance the parallel component:
Fnet = mg sin θ
The object therefore accelerates down the slope.
This relationship is especially clear in the interactive inclined-plane model:
a = gsinθ
The mass cancels when Newton's Second Law is applied. Therefore, on an ideal frictionless incline, the acceleration depends on the slope angle and gravitational field strength, not on the object's mass.
Worked Example 2: Frictionless Incline
A 5.0 kg block slides down a frictionless 20° incline.
Calculate its acceleration.
Parallel force:
Fnet = mg sin θ
Using:
Fnet = ma
we get:
mg sin θ = ma
Cancel m:
g sin θ = a
Therefore:
a = 9.8 sin 20°
a ≈ 3.35 m/s²
Acceleration ≈ 3.4 m/s² down the slope
Notice that the mass did not affect the final acceleration.
Why Doesn't Mass Affect the Acceleration?
A heavier object has a larger gravitational force.
However, it also has greater inertia.
For a frictionless incline:
F = mg sin θ
and:
F = ma
Therefore:
mg sin θ = ma
The mass appears on both sides and cancels.
So:
a = g sin θ
This is similar to free fall, where objects of different masses experience the same gravitational acceleration when air resistance is ignored.
Inclined Planes with Friction
Real surfaces usually produce friction.
If an object slides down a slope, kinetic friction acts up the slope, opposing the motion.
The friction force can be modeled as:
Ff = μN
Since:
N = mg cos θ
then:
Ff = μmg cos θ
The gravitational component down the slope is:
Fg∥ = mg sin θ
Therefore, if the object slides downward:
Fnet = mg sin θ − Ff
Worked Example 3: Incline with Friction
A 10 kg box slides down a 30° slope. The coefficient of kinetic friction is 0.20.
Calculate its acceleration.
First calculate weight:
mg = 10 × 9.8 = 98 N
Parallel component:
Fg∥ = 98 sin 30°
Fg∥ = 49 N
Normal force:
N = 98 cos 30°
N ≈ 84.9 N
Calculate friction:
Ff = μN
Ff = 0.20 × 84.9
Ff ≈ 17.0 N
Now calculate the net force down the slope:
Fnet = 49 − 17
Fnet = 32 N
Use:
Fnet = ma
32 = 10a
a = 3.2 m/s²
The box accelerates at approximately:
3.2 m/s² down the slope
Friction Always Opposes Relative Motion
It is important not to automatically draw friction pointing up the slope.
Friction opposes the relative motion or tendency for relative motion between the surfaces.
If an object is sliding downward:
friction acts upward
If an object is being pulled upward:
friction acts downward
Always determine the direction of motion—or attempted motion—before choosing the direction of friction.
Static Friction on an Incline
An object placed on a slope does not always slide.
Static friction may prevent motion.
The force pulling the object down the slope is:
mg sin θ
Static friction acts up the slope to oppose this tendency.
For equilibrium:
Fstatic = mg sin θ
However, static friction has a maximum value:
Fs(max) = μsN
and:
N = mg cos θ
Therefore:
Fs(max) = μsmg cos θ
The object remains stationary only if the required static friction does not exceed this maximum.
Will the Object Slide?
To determine whether an object will slide:
Calculate the force pulling it down the slope:
Fg∥ = mg sin θ
Then calculate the maximum static friction:
Fs(max) = μsN
If:
mg sin θ ≤ Fs(max)
the object can remain stationary.
If:
mg sin θ > Fs(max)
static friction is insufficient and the object begins to slide.
Worked Example 4: Will It Slide?
A 20 kg box rests on a 25° slope. The coefficient of static friction is 0.50.
First calculate weight:
mg = 20 × 9.8
mg = 196 N
Force down the slope:
Fg∥ = 196 sin 25°
Fg∥ ≈ 82.8 N
Normal force:
N = 196 cos 25°
N ≈ 177.6 N
Maximum static friction:
Fs(max) = μsN
Fs(max) = 0.50 × 177.6
Fs(max) ≈ 88.8 N
Compare:
Required friction = 82.8 N
Maximum available friction = 88.8 N
Since:
82.8 < 88.8
static friction is strong enough to prevent sliding.
The box remains stationary.
The actual static friction is 82.8 N, not 88.8 N.
The Critical Angle
As the angle of a slope increases:
mg sin θ increases
while:
mg cos θ decreases
Therefore:
- the force pulling the object down the slope increases
- the normal force decreases
- the maximum possible friction decreases
Eventually the object may begin to slide.
At the point where sliding is just about to begin:
mg sin θ = μsmg cos θ
Cancel mg:
sin θ = μs cos θ
Divide by cos θ:
tan θ = μs
Therefore:
μs = tan θcritical
This relationship can be used experimentally to determine the coefficient of static friction.
Pulling an Object Up an Incline
Suppose a rope pulls an object up the slope.
Forces along the slope may include:
- tension upward
- gravity component downward
- friction downward
If upward along the slope is positive:
Fnet = T − mg sin θ − Ff
Using Newton's Second Law:
T − mg sin θ − Ff = ma
If the object moves at constant velocity:
a = 0
Therefore:
T = mg sin θ + Ff
Worked Example 5: Pulling a Box Up a Ramp
A 15 kg box is pulled up a 20° ramp at constant velocity. Friction acts with a force of 25 N.
Calculate the required tension.
Because the velocity is constant:
a = 0
Therefore:
Fnet = 0
Calculate the component of weight down the slope:
Fg∥ = mg sin θ
Fg∥ = 15 × 9.8 × sin 20°
Fg∥ ≈ 50.3 N
The tension must balance both gravity and friction:
T = Fg∥ + Ff
T = 50.3 + 25
T ≈ 75.3 N
The required tension is approximately:
75 N
Applied Forces at an Angle to the Incline
Sometimes an applied force is itself at an angle to the slope.
In that case, the applied force must also be resolved into components.
One component acts parallel to the incline.
The other acts perpendicular to the incline.
The perpendicular component can change the normal force.
Because:
Ff = μN
changing the normal force can also change the friction force.
This is why carefully drawing the free-body diagram is essential.
Net Force on an Inclined Plane
For any inclined-plane problem, identify all forces acting parallel to the slope.
Then calculate:
Fnet,parallel = sum of forces up slope − sum of forces down slope
or choose the opposite sign convention.
The important thing is to remain consistent.
Then use:
Fnet = ma
to calculate acceleration.
For the perpendicular direction, an object that remains in contact with the surface usually has:
aperpendicular = 0
Therefore:
ΣFperpendicular = 0
Inclined Planes and Energy
Inclined planes can also be analyzed using energy.
When an object moves down a slope, its height decreases.
Therefore, its gravitational potential energy decreases.
Without friction:
gravitational potential energy → kinetic energy
With friction:
some energy is also transferred into thermal energy.
This provides another way to analyze motion on ramps and slopes.
Why Use an Inclined Plane?
An inclined plane is one of the simplest machines.
A ramp allows an object to be raised vertically using a smaller force over a longer distance.
Instead of lifting an object straight upward against its entire weight, it can be pushed or pulled along a slope.
For an ideal frictionless ramp, the force required along the slope is:
F = mg sin θ
A shallow ramp requires a smaller force but a longer distance.
A steep ramp requires a larger force but a shorter distance.
Inclined Roads
Roads through mountainous areas often use gradual slopes and switchbacks instead of traveling directly up steep hills.
Reducing the slope angle reduces the component of gravity pulling a vehicle downhill:
Fg∥ = mg sin θ
This makes it easier for vehicles to climb and reduces the forces required from engines and brakes.
Skiing and Snowboarding
Inclined-plane physics is particularly important in skiing and snowboarding.
Gravity provides a component down the slope.
Friction and drag oppose the motion.
On a steeper slope:
mg sin θ becomes larger
so the skier can experience a larger downhill force.
At higher speeds, air resistance also becomes increasingly important.
Vehicles on Hills
A parked car on a hill is an example of static equilibrium on an inclined plane.
Gravity has a component pulling the car downhill.
Static friction between the tires and road, along with the braking system, prevents the vehicle from moving.
If the available friction becomes insufficient—for example, on ice—the vehicle may begin sliding downhill.
Avalanches and Landslides
Inclined-plane physics also helps explain natural hazards.
Material on a slope experiences a gravitational component pulling it downhill.
Friction and other forces resist this motion.
Changes in:
- slope angle
- water content
- snow structure
- surface conditions
- friction
can affect whether the material remains stable or begins moving.
Did You Know?
The angle at which loose material naturally forms a stable slope is called the angle of repose.
Different materials have different angles of repose because their particles interact differently.
Dry sand, gravel, grains, and other granular materials therefore form piles with characteristic slopes.
This concept is important in geology, construction, mining, agriculture, and material storage.
Common Mistakes
Mistake 1: Drawing weight perpendicular to the slope
Weight always acts vertically downward.
Mistake 2: Drawing the normal force vertically upward
The normal force acts perpendicular to the surface.
Mistake 3: Using mg as the force down the slope
Only the parallel component acts down the incline:
mg sin θ
Mistake 4: Assuming N = mg
On a simple incline:
N = mg cos θ
Mistake 5: Automatically drawing friction up the slope
Friction opposes relative motion or attempted relative motion. Its direction depends on the situation.
Mistake 6: Assuming static friction always equals μsN
The equation:
Fs(max) = μsN
gives the maximum static friction. Actual static friction may be smaller.
A Strategy for Solving Inclined-Plane Problems
When solving an inclined-plane problem:
- Draw the object and slope.
- Draw a free-body diagram.
- Draw weight vertically downward.
- Draw the normal force perpendicular to the slope.
- Identify friction, tension, or applied forces.
- Choose axes parallel and perpendicular to the slope.
- Resolve weight:
Fg∥ = mg sin θ
Fg⊥ = mg cos θ
- Calculate the normal force.
- Calculate friction if required.
- Find the net force parallel to the slope.
- Use Fnet = ma.
- Check that the direction and magnitude of the answer make sense.
Key Terms
Inclined plane: A flat surface tilted at an angle to the horizontal.
Parallel component: The component of a force acting along the slope.
Perpendicular component: The component of a force acting at right angles to the slope.
Normal force: The contact force exerted perpendicular to a surface.
Friction: A force opposing relative motion or attempted relative motion between surfaces.
Coefficient of friction: A value describing the frictional interaction between two surfaces.
Angle of repose: The steepest stable angle at which loose material can remain without sliding.
Net force: The vector sum of all forces acting on an object.
Key Equations
Weight:
Fg = mg
Weight parallel to an incline:
Fg∥ = mg sin θ
Weight perpendicular to an incline:
Fg⊥ = mg cos θ
For a simple incline:
N = mg cos θ
Friction:
Ff = μN
Maximum static friction:
Fs(max) = μsN
Newton's Second Law:
Fnet = ma
Frictionless acceleration:
a = g sin θ
Critical angle:
μs = tan θcritical
Key Takeaways
- An inclined plane is a surface tilted relative to the horizontal.
- Weight always acts vertically downward, even when an object is on a slope.
- The normal force acts perpendicular to the inclined surface.
- It is usually easiest to choose axes parallel and perpendicular to the slope.
- Weight can be resolved into mg sin θ parallel to the slope and mg cos θ perpendicular to it.
- The parallel component of gravity pulls an object down the slope.
- On a simple incline, the perpendicular component determines the normal force.
- On a frictionless incline, a = g sin θ.
- The acceleration on an ideal frictionless incline is independent of mass.
- Friction opposes relative motion or attempted relative motion along the surface.
- Static friction may prevent an object from sliding.
- Kinetic friction reduces the net force when an object slides.
- Steeper slopes produce a larger component of gravity down the incline.
- Inclined-plane physics helps explain ramps, roads, skiing, vehicles on hills, landslides, and many engineering systems.
4. Atwood Machines
Learning outcomes
- I can draw free-body diagrams for Atwood machines.
- I can identify tension and weight forces in pulley systems.
- I can apply Newton's Second Law to connected objects.
- I can calculate acceleration in Atwood systems.
- I can determine tension forces in pulley problems.
What Is an Atwood Machine?
An Atwood machine is a simple pulley system consisting of two masses connected by a string or rope that passes over a pulley.
The basic system contains:
- two masses, usually called m₁ and m₂
- a light string connecting the masses
- a pulley that allows the string to change direction
If the two masses are different, the heavier mass tends to move downward while the lighter mass moves upward.
Atwood machines are useful because they allow us to study the relationship between force, mass, tension, and acceleration.
The Basic Atwood Machine
Suppose:
m₂ > m₁
Then:
m₂ moves downward
and:
m₁ moves upward
Because the masses are connected by the same taut string, they move together.
In an ideal Atwood machine, both masses have the same magnitude of acceleration.
If m₂ accelerates downward at 2.0 m/s², then m₁ accelerates upward at 2.0 m/s².
Their directions are opposite, but the magnitudes are equal.
Assumptions for an Ideal Atwood Machine
Introductory Atwood-machine problems usually assume an ideal system.
This means:
- the string has negligible mass
- the string does not stretch
- the pulley has negligible mass
- the pulley has negligible friction
- the string does not slip on the pulley
Under these assumptions, the tension is the same throughout the string.
Therefore:
T₁ = T₂ = T
Real pulley systems may behave differently, but the ideal model allows us to understand the basic physics clearly.
Forces Acting on Each Mass
Each hanging mass experiences two main forces.
Weight acts downward:
Fg = mg
Tension acts upward:
T
For m₁:
↑ T
● m₁
↓ m₁g
For m₂:
↑ T
● m₂
↓ m₂g
The forces may look similar, but if the masses are different, their weights are different.
Drawing Free-Body Diagrams
It is usually best to draw a separate free-body diagram for each mass.
Suppose:
m₂ > m₁
For m₁, which accelerates upward:
↑ T
●
↓ m₁g
Because the acceleration is upward:
T > m₁g
For m₂, which accelerates downward:
↑ T
●
↓ m₂g
Because the acceleration is downward:
m₂g > T
o7I?purpose=inline" alt="This gives us an important relationship:
m₂g > T > m₁g
when m₂ is the heavier accelerating mass.
Applying Newton's Second Law
Newton's Second Law states:
Fnet = ma
We apply this law separately to each mass.
Because the masses accelerate in opposite directions, it is helpful to choose the direction of motion as positive for each mass.
Equation for the Lighter Mass
Suppose m₁ moves upward.
For m₁:
↑ T
↓ m₁g
The net force upward is:
Fnet = T − m₁g
Using Newton's Second Law:
T − m₁g = m₁a
This is our first equation.
Equation for the Heavier Mass
Now consider m₂, which moves downward.
The forces are:
↓ m₂g
↑ T
The net force downward is:
Fnet = m₂g − T
Therefore:
m₂g − T = m₂a
This is our second equation.
So the two equations are:
T − m₁g = m₁a
m₂g − T = m₂a
These equations describe the motion of the entire ideal Atwood machine.
Finding the Acceleration
We can add the two equations:
T − m₁g = m₁a
m₂g − T = m₂a
Adding gives:
m₂g − m₁g = m₁a + m₂a
The tension forces cancel.
Factor:
(m₂ − m₁)g = (m₁ + m₂)a
Therefore:
a = ((m₂ − m₁)g) / (m₁ + m₂)
This is the acceleration equation for an ideal Atwood machine when:
m₂ > m₁
Understanding the Acceleration Equation
The numerator:
(m₂ − m₁)g
represents the difference between the two weights.
This difference provides the driving force for the system.
The denominator:
m₁ + m₂
represents the total mass that must be accelerated.
Therefore:
Acceleration = driving force ÷ total mass
This is simply another application of:
Fnet = ma
Treating the Two Masses as One System
There is another useful way to understand the Atwood machine.
Instead of analyzing each mass separately, imagine treating both masses as a single connected system.
The tension is an internal force within the combined system.
Therefore, it does not appear in the net external force for the entire system.
The driving force is:
Fnet = m₂g − m₁g
or:
Fnet = (m₂ − m₁)g
The total mass is:
mtotal = m₁ + m₂
Using:
Fnet = mtotal a
gives:
(m₂ − m₁)g = (m₁ + m₂)a
which produces the same acceleration equation.
Worked Example 1: Finding Acceleration
An Atwood machine has:
m₁ = 3.0 kg
m₂ = 5.0 kg
Calculate the acceleration.
Because:
m₂ > m₁
the 5.0 kg mass moves downward.
Use:
a = ((m₂ − m₁)g) / (m₁ + m₂)
Substitute:
a = ((5.0 − 3.0) × 9.8) / (3.0 + 5.0)
a = 19.6 / 8.0
a = 2.45 m/s²
Therefore:
Acceleration = 2.45 m/s²
The 5.0 kg mass accelerates downward and the 3.0 kg mass accelerates upward.
Finding the Tension
Once acceleration is known, tension can be calculated using the equation for either mass.
For the lighter mass:
T − m₁g = m₁a
Therefore:
T = m₁g + m₁a
or:
T = m₁(g + a)
For the heavier mass:
m₂g − T = m₂a
Therefore:
T = m₂g − m₂a
or:
T = m₂(g − a)
Both methods should give the same result.
Worked Example 2: Finding Tension
Using the previous system:
m₁ = 3.0 kg
m₂ = 5.0 kg
a = 2.45 m/s²
Use the lighter mass:
T = m₁(g + a)
T = 3.0(9.8 + 2.45)
T = 3.0(12.25)
T = 36.75 N
Therefore:
T ≈ 36.8 N
We can check using the heavier mass:
T = m₂(g − a)
T = 5.0(9.8 − 2.45)
T = 5.0(7.35)
T = 36.75 N
The answers agree.
Why Isn't Tension Equal to Weight?
For a stationary hanging object:
T = mg
But the masses in an Atwood machine are usually accelerating.
Therefore, tension does not equal either object's weight.
For the lighter mass moving upward:
T > m₁g
For the heavier mass moving downward:
T < m₂g
In our example:
Weight of m₁:
m₁g = 3 × 9.8
m₁g = 29.4 N
Tension:
T = 36.8 N
Weight of m₂:
m₂g = 5 × 9.8
m₂g = 49.0 N
Therefore:
29.4 N < 36.8 N < 49.0 N
which is exactly what we expect.
What Happens When the Masses Are Equal?
Suppose:
m₁ = m₂
Then the acceleration equation becomes:
a = ((m₂ − m₁)g) / (m₁ + m₂)
The numerator becomes zero.
Therefore:
a = 0
If the system begins at rest, it remains at rest.
If it is already moving ideally, it can continue at constant velocity.
The system is in translational equilibrium because there is no resultant force causing acceleration.
What Happens When the Difference in Mass Increases?
Suppose the total mass remains similar but the difference between the masses becomes larger.
The driving force:
(m₂ − m₁)g
becomes larger.
Therefore, the acceleration increases.
For example:
A system with 5 kg and 4 kg has a small mass difference.
A system with 8 kg and 1 kg has a much larger mass difference.
The second system experiences a much larger acceleration.
What Happens When Both Masses Become Large?
Suppose the difference between the masses stays the same, but both masses become larger.
Compare:
2 kg and 3 kg
with:
20 kg and 21 kg
Both systems have a mass difference of 1 kg.
However, the second system has much more total mass to accelerate.
Therefore, its acceleration is smaller.
This illustrates the basic idea:
Acceleration depends on both the driving force and the total inertia of the system.
Worked Example 3: Larger Masses
An Atwood machine has:
m₁ = 8 kg
m₂ = 12 kg
Calculate the acceleration.
Use:
a = ((m₂ − m₁)g) / (m₁ + m₂)
a = ((12 − 8) × 9.8) / (8 + 12)
a = 39.2 / 20
a = 1.96 m/s²
Now calculate tension using m₁:
T = m₁(g + a)
T = 8(9.8 + 1.96)
T = 8(11.76)
T = 94.08 N
Therefore:
Acceleration ≈ 1.96 m/s²
Tension ≈ 94.1 N
Acceleration Is Less Than g
For a normal Atwood machine with two positive masses:
a < g
Why?
The heavier mass does not fall freely.
The lighter mass and the tension in the string resist its downward motion.
A freely falling object has:
a = g
But an Atwood machine has:
a = ((m₂ − m₁)/(m₁ + m₂))g
The fraction:
(m₂ − m₁)/(m₁ + m₂)
is less than 1.
Therefore:
a < g
Atwood Machines and Kinematics
Once the acceleration has been calculated, the system can also be analyzed using constant-acceleration equations.
For example:
vf = vi + at
and:
Δx = vit + ½at²
This allows us to determine:
- velocity after a certain time
- distance traveled
- time required to move a certain distance
The Atwood-machine force calculation provides a, and the kinematics equations describe the resulting motion.
Worked Example 4: Velocity After a Time
An Atwood machine accelerates at:
a = 2.0 m/s²
The system starts from rest.
Calculate its speed after 3.0 s.
Use:
vf = vi + at
vf = 0 + (2.0)(3.0)
vf = 6.0 m/s
Both masses have a speed of:
6.0 m/s
but they move in opposite directions.
Worked Example 5: Distance Traveled
The same Atwood machine starts from rest and accelerates at:
2.0 m/s²
How far does each mass move in 3.0 s?
Use:
Δx = vit + ½at²
Since:
vi = 0
Δx = ½(2.0)(3.0²)
Δx = 1 × 9
Δx = 9.0 m
Each mass moves 9.0 m in opposite directions, provided the physical system allows that much movement.
A Mass on a Table Connected to a Hanging Mass
A related pulley system has one mass on a horizontal surface and another hanging over the edge.
Suppose:
m₁ is on the table.
m₂ hangs vertically.
For m₁ on a frictionless table:
T = m₁a
For m₂:
m₂g − T = m₂a
Adding the equations gives:
m₂g = (m₁ + m₂)a
Therefore:
a = m₂g / (m₁ + m₂)
This is not technically the basic two-hanging-mass Atwood machine, but it uses the same connected-object reasoning.
Adding Friction to the Table
If the horizontal surface has friction:
Ff = μN
For the mass on the table:
N = m₁g
so:
Ff = μm₁g
If m₂ moves downward, the net driving force for the system becomes:
Fnet = m₂g − Ff
Therefore:
m₂g − Ff = (m₁ + m₂)a
and:
a = (m₂g − Ff)/(m₁ + m₂)
This combines ideas from:
- tension
- friction
- Newton's Second Law
- connected objects
Worked Example 6: Pulley with Friction
A 6 kg block rests on a horizontal table and is connected to a 4 kg hanging mass.
The friction force acting on the block is 10 N.
Calculate the acceleration.
Driving force from the hanging mass:
m₂g = 4 × 9.8
m₂g = 39.2 N
Subtract friction:
Fnet = 39.2 − 10
Fnet = 29.2 N
Total mass:
mtotal = 6 + 4
mtotal = 10 kg
Therefore:
a = Fnet / mtotal
a = 29.2 / 10
a = 2.92 m/s²
The hanging mass accelerates downward while the block accelerates horizontally toward the pulley.
Atwood Machines with Real Pulleys
Real pulleys are not perfectly ideal.
A real pulley may:
- have mass
- experience friction at its axle
- rotate with significant rotational inertia
- have a rope with non-negligible mass
In these situations, the tension on one side of the pulley may not equal the tension on the other side.
Some of the net force is needed to produce the angular acceleration of the pulley.
More advanced models therefore include rotational dynamics.
For introductory problems, however, the pulley and string are usually assumed to be ideal.
Experimental Atwood Machines
Atwood machines are commonly used in physics laboratories.
Students can vary:
- m₁
- m₂
- total mass
- difference in mass
and measure the resulting acceleration.
This allows Newton's Second Law to be tested experimentally.
For example, students can investigate whether:
greater net force → greater acceleration
and whether:
greater total mass → smaller acceleration for the same net force
Atwood Machines in Elevators and Lifting Systems
The basic principles of Atwood machines appear in real mechanical systems.
Elevators often use a counterweight.
The elevator car and counterweight are connected by cables passing over a pulley system.
The counterweight reduces the amount of force the motor must provide to move the elevator.
Although real elevators are much more complicated than ideal Atwood machines, the basic idea of connected masses and tension is similar.
Counterweights
Counterweights are also used in:
- cranes
- drawbridges
- stage equipment
- lifting systems
- industrial machinery
A counterweight helps balance another load.
This reduces the net force required from motors or people and can make lifting systems more efficient and controllable.
Did You Know?
The Atwood machine was introduced by English mathematician and physicist George Atwood in the eighteenth century.
The machine allowed scientists to study uniformly accelerated motion using accelerations much smaller than gravitational acceleration.
Instead of trying to measure an object falling freely at approximately 9.8 m/s², researchers could choose two similar masses and create a much smaller, easier-to-measure acceleration.
Common Mistakes
Mistake 1: Assuming tension equals weight
The masses are accelerating, so tension is generally not equal to either mass's weight.
Mistake 2: Giving the masses different acceleration magnitudes
For an ideal taut, non-stretching string:
|a₁| = |a₂|
Mistake 3: Using only the difference in masses as the total mass
The driving force depends on the mass difference, but both masses must be accelerated.
Therefore:
Driving force = (m₂ − m₁)g
while:
Total mass = m₁ + m₂
Mistake 4: Adding tension when analyzing the whole system
For the combined two-mass system, tension is internal and cancels.
Mistake 5: Forgetting direction
The heavier mass accelerates downward while the lighter mass accelerates upward.
Mistake 6: Using different tension values in an ideal system
With an ideal massless string and frictionless, massless pulley, the tension is the same throughout the string.
A Strategy for Solving Atwood-Machine Problems
A reliable method is:
- Identify the two masses.
- Determine which mass is heavier.
- Predict the direction of acceleration.
- Draw a separate free-body diagram for each mass.
- Label tension upward and weight downward.
- Apply Newton's Second Law to each object.
- Write one equation for each mass.
- Add the equations to eliminate tension.
- Solve for acceleration.
- Substitute the acceleration into either object's equation.
- Solve for tension.
- Check whether the result makes physical sense.
For m₂ > m₁, you should normally find:
m₁g < T < m₂g
and:
0 < a < g
Key Terms
Atwood machine: A system of two masses connected by a string passing over a pulley.
Tension: A pulling force transmitted through a string, rope, or cable.
Weight: The gravitational force acting on an object.
Pulley: A wheel that allows a rope or cable to change direction.
Connected objects: Objects whose motions are linked by a string, rope, cable, or other connection.
Driving force: The unbalanced external force causing a system to accelerate.
Ideal pulley: A pulley assumed to have negligible mass and friction.
Ideal string: A string assumed to have negligible mass and no stretching.
Key Equations
Weight:
Fg = mg
Newton's Second Law:
Fnet = ma
For the lighter mass moving upward:
T − m₁g = m₁a
For the heavier mass moving downward:
m₂g − T = m₂a
Ideal Atwood-machine acceleration:
a = ((m₂ − m₁)g)/(m₁ + m₂)
Tension using the lighter mass:
T = m₁(g + a)
Tension using the heavier mass:
T = m₂(g − a)
For equal masses:
a = 0
Key Takeaways
- An Atwood machine consists of two masses connected by a string passing over a pulley.
- Each hanging mass experiences weight downward and tension upward.
- Separate free-body diagrams should normally be drawn for each mass.
- In an ideal system, the tension is the same throughout the string.
- Connected masses have the same magnitude of acceleration but move in opposite directions.
- The heavier mass moves downward while the lighter mass moves upward.
- Newton's Second Law can be applied separately to each mass.
- When the two masses are treated as one system, tension is an internal force and cancels.
- The difference between the weights provides the driving force.
- The total mass of both objects must be accelerated.
- For an ideal Atwood machine, a = ((m₂ − m₁)g)/(m₁ + m₂).
- The acceleration is smaller than gravitational acceleration.
- Once acceleration is known, either mass can be used to calculate tension.
- If the masses are equal, the system has zero acceleration.
- Atwood-machine principles apply to pulley systems, counterweights, elevators, cranes, and other lifting systems.
5. Elevators and Apparent Weight
Learning outcomes
- I can define apparent weight.
- I can explain why apparent weight changes during acceleration.
- I can draw free-body diagrams for elevator systems.
- I can calculate apparent weight in accelerating systems.
- I can relate apparent weight to everyday experiences.
What Is Apparent Weight?
Your actual weight is the gravitational force acting on you.
It is calculated using:
Fg = mg
where:
- Fg = gravitational force or weight (N)
- m = mass (kg)
- g = gravitational field strength, approximately 9.8 N/kg near Earth's surface
Your actual weight does not suddenly change when an elevator starts moving.
However, how heavy you appear to be can change.
This is called apparent weight.
Apparent weight is the support force acting on an object, usually the normal force.
For a person standing on a scale:
Apparent weight = Normal force
or:
Wapparent = N
Actual Weight vs Apparent Weight
Actual weight and apparent weight are not always the same.
Actual weight:
Fg = mg
It depends mainly on mass and gravitational field strength.
Apparent weight:
Wapparent = N
It depends on how strongly the supporting surface pushes on the object.
If you stand on a scale in an accelerating elevator, the scale measures the normal force, not the gravitational force directly.
That is why the scale reading can change even though your mass remains constant.
Forces Acting on a Person in an Elevator
Consider a person standing on the floor of an elevator.
Two main forces act on the person:
↑ Normal force, N
● Person
↓ Weight, mg
The gravitational force always acts downward.
The elevator floor pushes upward on the person's feet. This upward contact force is the normal force.
The relationship between these forces determines the person's acceleration and apparent weight.
Elevator at Rest
Suppose the elevator is stationary.
Acceleration is zero:
a = 0
Therefore:
Fnet = 0
The forces must balance:
N = mg
So:
Apparent weight = Actual weight
For example, consider a 60 kg person.
Actual weight:
Fg = mg
Fg = 60 × 9.8
Fg = 588 N
Therefore:
N = 588 N
The person's apparent weight is:
588 N
Elevator Moving at Constant Velocity
Now suppose the elevator is moving upward at a constant velocity.
Because velocity is constant:
a = 0
Therefore:
Fnet = 0
and:
N = mg
The same is true if the elevator moves downward at constant velocity.
This is an important point:
Motion itself does not change apparent weight. Acceleration does.
An elevator can be moving quickly upward or downward, but if its velocity is constant:
N = mg
Elevator Accelerating Upward
Suppose the elevator accelerates upward.
The person's acceleration is also upward.
Therefore, the upward force must be greater than the downward force:
N > mg
Taking upward as positive:
Fnet = N − mg
Using Newton's Second Law:
N − mg = ma
Therefore:
N = mg + ma
Factor out m:
N = m(g + a)
Since:
Apparent weight = N
we have:
Wapparent = m(g + a)
The person feels heavier.
Why Do You Feel Heavier?
When an elevator accelerates upward, the floor must push upward on you strongly enough to accelerate you upward.
That means the normal force becomes greater than your weight.
Your body senses this increased contact force.
Therefore, you feel heavier.
Your actual gravitational weight has not increased.
Instead:
N has increased.
This is the source of the increased apparent weight.
Worked Example 1: Accelerating Upward
A 70 kg person stands on a scale in an elevator accelerating upward at 2.0 m/s².
Calculate the apparent weight.
Use:
N = m(g + a)
N = 70(9.8 + 2.0)
N = 70(11.8)
N = 826 N
Therefore:
Apparent weight = 826 N
The person's actual weight is:
mg = 70 × 9.8
mg = 686 N
So the scale reads more than the person's actual weight.
Elevator Accelerating Downward
Now suppose the elevator accelerates downward.
Weight is greater than the normal force:
mg > N
If downward acceleration has magnitude a:
mg − N = ma
Rearranging:
N = mg − ma
Therefore:
N = m(g − a)
So:
Wapparent = m(g − a)
The person feels lighter.
Worked Example 2: Accelerating Downward
A 70 kg person is in an elevator accelerating downward at 2.0 m/s².
Calculate the apparent weight.
Use:
N = m(g − a)
N = 70(9.8 − 2.0)
N = 70(7.8)
N = 546 N
Therefore:
Apparent weight = 546 N
Actual weight remains:
686 N
The person therefore feels lighter.
Comparing the Three Main Situations
For a person of mass m:
| Elevator Condition | Apparent Weight |
|---|---|
| No acceleration | N = mg |
| Accelerating upward | N = m(g + a) |
| Accelerating downward | N = m(g − a) |
Therefore:
Upward acceleration → apparent weight increases
Zero acceleration → apparent weight equals actual weight
Downward acceleration → apparent weight decreases
Direction of Motion Is Not Enough
A common mistake is to look only at whether the elevator is moving upward or downward.
What matters is the direction of acceleration, not simply the direction of velocity.
For example, an elevator can be:
moving upward but accelerating downward
This means the elevator is slowing down while moving upward.
In this situation:
N < mg
and the passenger feels lighter.
Similarly, an elevator can be:
moving downward but accelerating upward
This means it is slowing down while moving downward.
Then:
N > mg
and the passenger feels heavier.
Velocity and Acceleration in Elevators
There are several possible combinations.
Moving upward and speeding up:
Acceleration upward → feel heavier
Moving upward and slowing down:
Acceleration downward → feel lighter
Moving downward and speeding up:
Acceleration downward → feel lighter
Moving downward and slowing down:
Acceleration upward → feel heavier
Moving at constant velocity:
Acceleration = 0 → normal apparent weight
The key question should always be:
Which direction is the acceleration?
A Complete Elevator Journey
Consider a typical elevator traveling from a lower floor to a higher floor.
Stage 1: Starting upward
The elevator accelerates upward.
N > mg
Passenger feels heavier.
Stage 2: Moving upward at constant velocity
Acceleration = 0.
N = mg
Passenger feels normal weight.
Stage 3: Slowing near the upper floor
The elevator is still moving upward but accelerates downward.
N < mg
Passenger feels lighter.
Stage 4: Stopped
Acceleration = 0.
N = mg
Passenger feels normal weight again.
Free-Body Diagrams for Elevator Problems
Free-body diagrams are extremely useful when solving apparent-weight problems.
For a passenger:
Accelerating upward:
↑↑ N
●
↓ mg
N > mg
No acceleration:
↑ N
●
↓ mg
N = mg
Accelerating downward:
↑ N
●
↓↓ mg
N < mg
The relative arrow lengths can be used to show which force is larger.
Using Newton's Second Law
The safest method is not simply to memorize separate formulas.
Instead, begin with:
Fnet = ma
Choose a positive direction.
If upward is positive:
ΣFy = N − mg
Therefore:
N − mg = ma
Then use the sign of acceleration.
If acceleration is upward:
a is positive
If acceleration is downward:
a is negative
This single equation can solve all elevator apparent-weight situations.
Worked Example 3: Using Signed Acceleration
A 50 kg passenger is in an elevator accelerating downward at 3.0 m/s².
Take upward as positive.
Therefore:
a = −3.0 m/s²
Use:
N − mg = ma
N − (50)(9.8) = 50(−3.0)
N − 490 = −150
N = 340 N
Apparent weight = 340 N
This method avoids needing separate equations for upward and downward acceleration.
Calculating Acceleration from Apparent Weight
Sometimes the scale reading is given and the acceleration must be determined.
Suppose a 60 kg person stands on a scale reading 720 N.
Calculate the elevator's acceleration.
Actual weight:
mg = 60 × 9.8
mg = 588 N
Use:
N − mg = ma
720 − 588 = 60a
132 = 60a
a = 2.2 m/s²
Because N > mg, the net force is upward.
Therefore:
Acceleration = 2.2 m/s² upward
Worked Example 4: A Lower Scale Reading
A 75 kg passenger stands on a scale reading 600 N.
Find the elevator's acceleration.
Weight:
mg = 75 × 9.8
mg = 735 N
Use:
N − mg = ma
600 − 735 = 75a
−135 = 75a
a = −1.8 m/s²
The negative sign means downward.
Therefore:
Acceleration = 1.8 m/s² downward
The passenger may actually be moving upward or downward—we cannot determine that from this information alone.
We only know the acceleration is downward.
What Does a Scale Actually Measure?
A bathroom scale does not directly measure the gravitational force acting on you.
Instead, it responds to the contact force between you and the scale.
In an ordinary stationary situation:
N = mg
so the scale can convert the normal force into a mass reading.
For example, a scale may detect approximately 686 N and display:
70 kg
But if the scale is accelerating, the normal force changes.
The displayed value can therefore differ from the person's actual mass unless the scale accounts for the acceleration.
Apparent Weight in Free Fall
Now consider an extreme situation.
Suppose the elevator and passenger are both falling freely with:
a = g
For downward acceleration:
N = m(g − a)
Therefore:
N = m(g − g)
N = 0
The apparent weight is zero.
The passenger is still affected by gravity.
Actual weight is still:
Fg = mg
But there is no supporting normal force.
Therefore:
Apparent weight = 0
This condition is called weightlessness or apparent weightlessness.
Weightlessness Does Not Mean No Gravity
This is a very important distinction.
During free fall:
Gravity is still acting.
In fact, gravity is the force causing the person to accelerate downward.
The person feels weightless because there is no support force pushing against them.
Therefore:
Weightlessness does not mean Fg = 0.
It means:
N = 0.
Apparent Weight Greater Than Normal
If an elevator accelerates upward strongly, the normal force can become significantly greater than the person's weight.
For example, suppose:
m = 80 kg
a = 4.0 m/s² upward
Then:
N = m(g + a)
N = 80(9.8 + 4.0)
N = 80(13.8)
N = 1104 N
Actual weight:
mg = 80 × 9.8
mg = 784 N
The person therefore experiences an apparent weight about 41% greater than normal.
Apparent Weight as a Multiple of Normal Weight
Sometimes apparent weight is described using g-force.
If:
N = mg
the person experiences approximately:
1 g
If:
N = 2mg
the apparent weight is twice normal:
2 g
If:
N = 0
the person experiences apparent weightlessness:
0 g apparent weight
This idea becomes particularly important in:
- aircraft
- spacecraft
- roller coasters
- racing vehicles
- amusement rides
Apparent Weight on Roller Coasters
The same physics occurs on amusement rides.
When the seat pushes strongly against you, your apparent weight increases.
You may feel pressed into the seat.
When the support force decreases, your apparent weight decreases.
You may feel lighter or feel as though you are lifting out of your seat.
Although circular-motion calculations can make roller coaster situations more complicated, the sensation of "heavier" or "lighter" is still closely related to changes in the normal force.
Apparent Weight in Aircraft
Aircraft can also produce noticeable changes in apparent weight.
During certain maneuvers, passengers may experience:
- greater-than-normal apparent weight
- reduced apparent weight
- brief periods of apparent weightlessness
Aircraft used for weightlessness training follow special curved flight paths called parabolic flights.
During part of the maneuver, the aircraft and passengers are essentially falling together.
This produces a very small normal force and therefore apparent weightlessness.
Astronauts and Apparent Weightlessness
Astronauts orbiting Earth appear weightless.
This is not because Earth's gravity has disappeared.
Gravity at the altitude of a low-Earth-orbit spacecraft is still substantial.
The spacecraft and astronauts are continuously falling toward Earth together while moving sideways fast enough to remain in orbit.
Because they fall together, there is very little normal support force between the astronaut and the spacecraft.
They therefore experience apparent weightlessness, commonly described as microgravity.
Apparent Weight in a Car
You can even experience similar effects in a car.
When a car travels over a hill or through a dip, the normal force from the seat can change.
You may feel:
- lighter when passing over the top of a hill
- heavier when passing through a dip
Again, your actual gravitational weight has not suddenly changed.
The support force from the seat has changed.
Apparent Weight and Mass
Mass does not change when apparent weight changes.
Suppose a person's mass is:
70 kg
Whether the person feels heavier, lighter, or weightless:
Mass = 70 kg
Their actual gravitational weight near Earth's surface remains approximately:
686 N
What changes is the normal force.
This distinction between mass, actual weight, and apparent weight is essential.
Comparing Mass, Weight, and Apparent Weight
| Quantity | Meaning | Typical Equation | Unit |
|---|---|---|---|
| Mass | Amount of matter/inertia | m | kg |
| Weight | Gravitational force | Fg = mg | N |
| Apparent weight | Support force | Wapparent = N | N |
Mass is not a force.
Weight and apparent weight are forces.
Elevator Cables
Elevator problems can also involve the tension in the cable supporting the elevator.
For an elevator car of mass M:
↑ T
●
↓ Mg
If the elevator accelerates upward:
T − Mg = Ma
Therefore:
T = M(g + a)
If the elevator accelerates downward:
T < Mg
If it moves at constant velocity:
T = Mg
The same Newton's Second Law reasoning applies to both passengers and the elevator itself.
Worked Example 5: Elevator Cable Tension
An elevator has a total mass of 1200 kg and accelerates upward at 1.5 m/s².
Calculate the cable tension.
Use:
T − Mg = Ma
Therefore:
T = M(g + a)
T = 1200(9.8 + 1.5)
T = 1200(11.3)
T = 13,560 N
The cable tension is:
13.6 kN
approximately.
Did You Know?
The brief "heavy" and "light" sensations you experience when an elevator starts or stops are direct evidence of Newton's Second Law.
You can investigate this using a scale in an elevator.
As the elevator begins accelerating upward, the reading increases.
During constant-speed motion, it returns to normal.
As the elevator slows near the upper floor, the reading decreases.
The entire sequence can be explained by changes in the normal force.
Common Mistakes
Mistake 1: Saying actual weight changes in an elevator
Near Earth's surface, mg remains essentially constant. It is apparent weight that changes.
Mistake 2: Thinking upward motion means greater apparent weight
The important quantity is acceleration, not velocity.
Mistake 3: Thinking downward motion means lower apparent weight
An elevator moving downward but slowing has upward acceleration, so apparent weight is greater than normal.
Mistake 4: Saying weightlessness means no gravity
Apparent weightlessness occurs when the normal force becomes zero or nearly zero.
Mistake 5: Confusing mass with scale reading
Mass remains constant even when the support force changes.
Mistake 6: Forgetting the direction of acceleration
Always identify the acceleration before choosing the force equation.
A Strategy for Solving Apparent-Weight Problems
- Identify the person or object being analyzed.
- Draw a free-body diagram.
- Draw weight downward:
Fg = mg
- Draw the normal force upward:
N
- Determine the direction of acceleration.
- Choose a positive direction.
- Apply:
ΣF = ma
- Solve for N.
- Remember:
Apparent weight = N
- Check whether the answer makes sense.
If acceleration is upward:
N > mg
If acceleration is zero:
N = mg
If acceleration is downward:
N < mg
If the object is in free fall:
N = 0
Key Terms
Actual weight: The gravitational force acting on an object.
Apparent weight: The support force experienced by an object, usually the normal force.
Normal force: A contact force exerted perpendicular to a supporting surface.
Acceleration: The rate of change of velocity.
Free fall: Motion in which gravity is the only significant force acting.
Weightlessness: A condition in which apparent weight is zero or nearly zero.
g-force: A way of comparing apparent weight or acceleration effects with normal gravitational conditions.
Key Equations
Actual weight:
Fg = mg
Apparent weight:
Wapparent = N
Using upward as positive:
N − mg = ma
Accelerating upward:
N = m(g + a)
Accelerating downward with acceleration magnitude a:
N = m(g − a)
No acceleration:
N = mg
Free fall:
N = 0
For the elevator car itself:
T − Mg = Ma
Key Takeaways
- Apparent weight is the support force acting on an object, usually the normal force.
- A scale measures the normal force rather than gravitational weight directly.
- Actual weight is calculated using Fg = mg.
- Actual weight remains essentially constant during ordinary elevator motion.
- Apparent weight changes when the elevator accelerates.
- Upward acceleration produces N > mg, so you feel heavier.
- Downward acceleration produces N < mg, so you feel lighter.
- With zero acceleration, N = mg.
- The direction of velocity alone does not determine apparent weight.
- An elevator moving upward can produce either increased or decreased apparent weight depending on its acceleration.
- During free fall, N = 0, producing apparent weightlessness.
- Weightlessness does not mean that gravity has disappeared.
- Similar changes in apparent weight occur in elevators, cars, aircraft, spacecraft, and amusement rides.
- Free-body diagrams and Newton's Second Law provide a reliable way to solve apparent-weight problems.