Turning effect of forces

1. Overview

Forces do not always move objects in a straight line; they can also cause objects to rotate or "turn" around a fixed point. Understanding the turning effect of forces (moments) is essential for explaining how everything from simple tools like scissors to complex structures like cranes remain stable and functional.

Key Definitions

• Moment: A measure of the turning effect of a force about a specific point.
• Pivot (Fulcrum): The fixed point about which an object rotates.
• Perpendicular Distance: The shortest distance from the pivot to the line of action of the force (measured at 90°).
• Equilibrium: A state where an object has no resultant force and no resultant moment acting upon it.
• Principle of Moments: For an object in equilibrium, the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about that same point.

Core Content

The Moment of a Force

A moment occurs whenever a force is applied to an object that is fixed at a pivot.

• Everyday Examples:
  • Pushing a door handle (farthest from the hinges to make it easier to open).
  • Using a spanner/wrench to loosen a bolt.
  • Sitting on a seesaw.

Calculating the Moment

The size of a moment depends on two things: the size of the force applied and the distance from the pivot. Equation: $Moment = Force \times perpendicular\ distance\ from\ pivot$

Simple Principle of Moments

When a beam is balanced (in equilibrium) with one force on each side:

• The Clockwise Moment must equal the Anticlockwise Moment.
• $F_1 \times d_1 = F_2 \times d_2$

Worked Example: A child weighing 300 N sits 2.0 m from the pivot of a seesaw. Where must a 400 N child sit on the other side to balance it?

1. Anticlockwise Moment = $300\ N \times 2.0\ m = 600\ Nm$
2. Clockwise Moment = $400\ N \times d$
3. Set them equal: $600 = 400 \times d$
4. $d = 600 / 400 = 1.5\ m$

Conditions for Equilibrium

For an object to be in total equilibrium:

1. No Resultant Force: The sum of upward forces equals the sum of downward forces.
2. No Resultant Moment: The sum of clockwise moments equals the sum of anticlockwise moments.

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Extended Content (Extended Only)

Complex Principle of Moments

In more complex scenarios, there may be multiple forces acting on either side of the pivot. To solve these, you must sum all moments on each side. $\text{Sum of Clockwise Moments} = \text{Sum of Anticlockwise Moments}$ $(F_1 \times d_1) + (F_2 \times d_2) = (F_3 \times d_3) ...$

Worked Example (Multiple Forces): A uniform meter rule is pivoted at the 50 cm mark. A 2 N weight is placed at the 10 cm mark and a 3 N weight is placed at the 20 cm mark. Where must a 5 N weight be placed to balance the rule?

1. Identify distances from pivot (50 cm): $d_1 = 40\ cm$ (0.4 m), $d_2 = 30\ cm$ (0.3 m).
2. Total ACW Moment: $(2 \times 0.4) + (3 \times 0.3) = 0.8 + 0.9 = 1.7\ Nm$.
3. CW Moment must be 1.7 Nm: $5\ N \times d = 1.7$.
4. $d = 1.7 / 5 = 0.34\ m$ (or 34 cm from the pivot).

Experiment: Demonstrating No Resultant Moment

1. Balance a uniform meter rule on a pivot (triangular prism) at its center of gravity.
2. Hang different known masses from threads on both sides of the pivot.
3. Adjust the positions of the masses until the rule is perfectly horizontal (in equilibrium).
4. Measure the distance ($d$) of each mass from the pivot.
5. Calculate the force ($W = mg$) for each mass.
6. Calculate the sum of clockwise moments ($F \times d$) and the sum of anticlockwise moments.
7. Result: The sums will be equal (within experimental error), proving there is no resultant moment in equilibrium.

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Key Equations

• $M = F \times d$
  • $M$: Moment (Newton-metres, Nm)
  • $F$: Force (Newtons, N)
  • $d$: Perpendicular distance from pivot (Metres, m)
• For Equilibrium: $\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}$

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Common Mistakes to Avoid

• ❌ Wrong: Assuming forces must be equal on both sides to balance.
• ✓ Right: A smaller force can balance a larger force if it is applied at a greater distance from the pivot.
• ❌ Wrong: Measuring the distance from the end of the beam or the start of the ruler.
• ✓ Right: Always measure the distance from the Pivot Point to the line of action of the force.
• ❌ Wrong: Forgetting the weight of the beam itself.
• ✓ Right: If the beam is "uniform," its weight acts at its center (the geometric middle). If the pivot is not at the center, the beam's weight creates its own moment!
• ❌ Wrong: Using mass (kg) in the moment equation.
• ✓ Right: Always convert mass to weight (Force) using $W = m \times g$ (where $g \approx 9.8$ or $10\ m/s^2$).

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Exam Tips

1. Check Units: Exam questions often give distances in centimeters (cm). Convert them to meters (m) to get the standard unit of Nm, or ensure you keep units consistent on both sides of the equation.
2. Identify the Pivot: Before starting any calculation, clearly mark the pivot point. All distances must be measured from this specific point.
3. Draw a Table: For complex problems, list "Force," "Distance from Pivot," and "Direction (CW/ACW)" for every force acting on the object to ensure you don't miss any.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A uniform metre rule is pivoted at its centre. A weight of 2.0 N is hung at the 20 cm mark.

(a) Define the moment of a force. [2]

(b) Calculate the magnitude of the force that must be applied at the 90 cm mark to balance the metre rule horizontally. [3]

Worked Solution:

(a)

1. The moment of a force is a measure of its turning effect about a pivot. The moment of a force is a measure of its turning effect. [Definition of moment]

2. It is equal to the product of the force and the perpendicular distance from the pivot to the line of action of the force. It is equal to the force multiplied by the perpendicular distance from the pivot. [Completion of definition]

How to earn full marks:

• State that moment is a measure of the turning effect.
• Include that it is a force multiplied by a perpendicular distance from the pivot.

(b)

1. Calculate the distance of the 2.0 N weight from the pivot. Distance = 50 cm - 20 cm = 30 cm = 0.30 m [Correctly calculates the distance]

2. Calculate the moment due to the 2.0 N weight. Moment = Force × Distance = 2.0 N × 0.30 m = 0.60 Nm [Applies the moment equation correctly]

3. Calculate the distance of the applied force from the pivot. Distance = 90 cm - 50 cm = 40 cm = 0.40 m [Correctly calculates the distance]

4. Equate the moments and solve for the force. Force × 0.40 m = 0.60 Nm Force = $\frac{0.60 Nm}{0.40 m}$ = 1.5 N [Equates the clockwise and anticlockwise moments to find the force]

How to earn full marks:

• Correctly calculate the distance of each force from the pivot.
• State the correct moment equation and substitute values.
• Calculate the force with correct units.
• $\boxed{Force = 1.5 N}$

Common Pitfall: Remember that the distance used in the moment equation must be the perpendicular distance from the pivot to the line of action of the force. Also, be sure to convert all distances to meters before performing calculations to ensure your final answer has the correct units.

Exam-Style Question 2 — Short Answer [6 marks]

Question:

(a) State the principle of moments. [2]

(b) Calculate the weight of the first child. Assume $g = 9.8 N/kg$. [2]

(c) Calculate the mass of the second child required to balance the see-saw horizontally. [2]

Worked Solution:

(a)

1. For an object in equilibrium, the sum of the clockwise moments about a point. The sum of the clockwise moments... [Beginning of the principle of moments]

2. ...is equal to the sum of the anticlockwise moments about the same point. ...is equal to the sum of the anticlockwise moments. [Completion of the principle of moments]

How to earn full marks:

• State that it concerns an object in equilibrium.
• Mention that the sum of clockwise moments equals the sum of anticlockwise moments about the same point.

(b)

1. State the weight equation. Weight = mass × gravitational field strength [Recall the weight equation]

2. Substitute and calculate the weight. Weight = 30 kg × 9.8 N/kg = 294 N [Applies the equation correctly]

How to earn full marks:

• State the correct equation.
• Calculate the weight with correct units.
• $\boxed{Weight = 294 N}$

(c)

1. Calculate the moment due to the first child. Moment = Force × Distance = 294 N × 1.5 m = 441 Nm [Correctly calculates the moment]

2. Calculate the force required to balance the see-saw. Force = $\frac{Moment}{Distance} = \frac{441 Nm}{2.0 m}$ = 220.5 N [Applies the moment equation correctly]

3. Calculate the mass of the second child. Mass = $\frac{Force}{g} = \frac{220.5 N}{9.8 N/kg}$ = 22.5 kg [Correctly calculates the mass]

How to earn full marks:

• Calculate the moment due to the first child.
• Calculate the force required on the other side.
• Calculate the mass with correct units.
• $\boxed{Mass = 22.5 kg}$

Common Pitfall: Students often forget that the force in the moment equation is the weight of the object, not the mass. Make sure to calculate the weight using $W = mg$ before calculating the moment. Also, double-check that your final answer is in the correct units (kg for mass, N for weight).

Exam-Style Question 3 — Extended Response [9 marks]

Question:

A mechanic is using a wrench to tighten a bolt on a car engine. The wrench is 0.30 m long, and the mechanic applies a force of 150 N at the end of the wrench, perpendicular to the wrench.

(a) Calculate the moment of the force applied by the mechanic. [2]

(b) Explain why it is easier to loosen a tight bolt using a longer wrench compared to a shorter wrench, assuming the same force is applied. [3]

(c) The mechanic then uses a crowbar to lift a heavy engine component. The crowbar is 1.2 m long and is pivoted 0.2 m from the load. The engine component has a weight of 800 N. Calculate the force the mechanic needs to apply to the end of the crowbar to lift the engine component. [4]

Worked Solution:

(a)

1. State the moment equation. Moment = Force × Perpendicular Distance [Recall the moment equation]

2. Substitute and calculate the moment. Moment = 150 N × 0.30 m = 45 Nm [Applies the equation correctly]

How to earn full marks:

• State the correct equation.
• Calculate the moment with correct units.
• $\boxed{Moment = 45 Nm}$

(b)

1. Explain that a longer wrench increases the perpendicular distance from the pivot. A longer wrench increases the perpendicular distance from the pivot to the point where the force is applied. [States the effect of a longer wrench]

2. Relate the increased distance to an increased moment. Since moment = force × perpendicular distance, a larger distance results in a larger moment for the same applied force. [Explains the relationship between moment and distance]

3. Relate the increased moment to the ease of loosening the bolt. A larger moment provides a greater turning effect, making it easier to loosen the tight bolt. [Explains the effect of the increased moment]

How to earn full marks:

• State that a longer wrench increases the perpendicular distance.
• Explain how this increases the moment for the same force.
• Explain how the increased moment makes it easier to loosen the bolt.

(c)

1. Calculate the distance of the load from the pivot. Distance of load = 0.2 m [Correctly identifies the distance]

2. Calculate the distance of the applied force from the pivot. Distance of force = 1.2 m - 0.2 m = 1.0 m [Correctly calculates the distance]

3. Calculate the moment due to the weight of the engine component. Moment due to weight = 800 N × 0.2 m = 160 Nm [Correctly calculates the moment]

4. Equate the moments and solve for the force. Force × 1.0 m = 160 Nm Force = $\frac{160 Nm}{1.0 m}$ = 160 N [Equates the clockwise and anticlockwise moments to find the force]

How to earn full marks:

• Calculate the correct distance of each force from the pivot.
• Calculate the moment due to the weight.
• Equate moments and find the applied force.
• Calculate the force with correct units.
• $\boxed{Force = 160 N}$

Common Pitfall: In part (c), many students mistakenly use the entire length of the crowbar as the distance for the applied force. Remember to only use the distance from the pivot to the point where the force is applied. Always draw a diagram to help visualize the distances involved.

Exam-Style Question 4 — Extended Response [8 marks]

Question:

A student is investigating the turning effect of a force using a metre rule. The metre rule is suspended from a pivot at the 50 cm mark. A weight, W, is hung at the 10 cm mark. The student then hangs a second weight, F, on the other side of the pivot to balance the metre rule horizontally.

(a) Describe how the student can ensure that the metre rule is balanced horizontally. [2]

(b) The student records the following data:

• Weight W = 2.0 N
• Distance of W from the pivot = 40.0 cm
• Distance of F from the pivot = 30.0 cm

Calculate the weight F needed to balance the metre rule. [3]

(c) The student repeats the experiment with different values of W and measures the corresponding values of F. The student plots a graph of F (y-axis) against W (x-axis) and obtains a straight line.

Suggest what the gradient of the graph represents. Explain your reasoning. [3]

Worked Solution:

(a)

1. Observe the metre rule's position. The student should observe the position of the metre rule. [States the initial observation]

2. Adjust the position of weight F until the metre rule is horizontal. The student should adjust the position of weight F, or its magnitude, until the metre rule is horizontal. The student can use a set square against the bench to ensure the rule is level. [Describes the adjustment process]

How to earn full marks:

• State that the student should observe the metre rule.
• Describe the adjustment process until the metre rule is horizontal.

(b)

1. Calculate the moment due to weight W. Moment due to W = 2.0 N × 0.40 m = 0.80 Nm [Correctly calculates the moment]

2. Equate the moments and solve for F. Moment due to F = 0.80 Nm F = $\frac{0.80 Nm}{0.30 m}$ = 2.67 N (or 2.666... N) [Equates moments and finds the force]

How to earn full marks:

• Calculate the moment due to weight W.
• Equate moments and find the force.
• Calculate the force with correct units.
• $\boxed{F = 2.67 N}$

(c)

1. State the relationship between F and W based on the principle of moments. According to the principle of moments, $W \times d_W = F \times d_F$, where $d_W$ is the distance of W from the pivot and $d_F$ is the distance of F from the pivot. [States the equation relating F and W]

2. Rearrange the equation to express F in terms of W. Rearranging the equation, we get $F = (\frac{d_W}{d_F}) \times W$ [Rearranges the equation]

3. Explain the gradient represents the ratio of the distances. Therefore, the gradient of the graph of F against W represents the ratio $\frac{d_W}{d_F}$, which is the ratio of the distance of weight W from the pivot to the distance of weight F from the pivot. [Explains what the gradient represents]

How to earn full marks:

• State the relationship between F and W.
• Rearrange the equation to express F in terms of W.
• Explain that the gradient represents the ratio of the distances.

Common Pitfall: In part (c), students often struggle to connect the physics principles to the graph. Remember that the gradient of a straight-line graph represents the constant of proportionality between the y-axis and x-axis variables. In this case, the gradient directly relates to the ratio of the distances from the pivot.