CIE AS & A Level Physics 4.1 Turning effects of forces Study Notes- 2025-2027 Syllabus
CIE AS & A Level Physics 4.1 Turning effects of forces Study Notes – New Syllabus
CIE AS & A Level Physics 4.1 Turning effects of forces Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics Study Notes syllabus with Candidates should be able to:
- understand that the weight of an object may be taken as acting at a single point known as its centre of gravity
- define and apply the moment of a force
- understand that a couple is a pair of forces that acts to produce rotation only
- define and apply the torque of a couple
Centre of Gravity
The centre of gravity (C.G.) of an object is the single point at which the entire weight of the object may be considered to act, regardless of the object’s orientation.
Explanation:
- In a uniform gravitational field, the weight of an object can be assumed to act at a single point — its centre of gravity.
- If supported at this point, the object will remain in equilibrium (balanced horizontally).
- In a non-uniform gravitational field (e.g., near very large masses), the weight may not act through a single point — but for most practical A Level cases, the field is considered uniform.
Key Points:
- The centre of gravity coincides with the geometric centre for uniform, symmetrical bodies (like a cube or sphere).
- For irregular shapes, it lies closer to the denser or heavier regions.
- The stability of an object depends on the position of its centre of gravity relative to its base of support.
Determining the Centre of Gravity Experimentally:
- Hang the object from one point and allow it to settle vertically.
- Draw a vertical line downwards from the suspension point (along the plumb line).
- Repeat from a different suspension point.
- The intersection of the two lines gives the position of the centre of gravity.
Example
Describe how you could determine the centre of gravity of an irregularly shaped flat metal plate.
▶️ Answer / Explanation
Hang the plate freely from one of its holes and let it rest. Suspend a plumb line next to it and mark the vertical line of the plumb line on the plate. Repeat the process by hanging the plate from a different point and draw another vertical line. The point of intersection of the two lines gives the centre of gravity of the plate.
Moment of a Force
The moment of a force (or torque) about a point is the product of the force and the perpendicular distance from the point (or axis) to the line of action of the force.
\( \mathrm{Moment = Force \times Perpendicular\ Distance} \)
\( \mathrm{M = Fd} \)
- \( \mathrm{M} \): moment of force (N·m)
- \( \mathrm{F} \): applied force (N)
- \( \mathrm{d} \): perpendicular distance from the pivot or axis to the line of action of the force (m)
Explanation:
- The moment of a force measures its ability to cause rotation about a point or axis.
- It depends on both the magnitude of the force and how far it acts from the pivot.
- A larger distance or greater force produces a greater turning effect (moment).
Direction / Sense of Moment:
- Moments causing clockwise rotation are considered positive (by convention).
- Moments causing anticlockwise rotation are considered negative (or vice versa, depending on the system used).
Principle of Moments:
For a body in rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point.
\( \mathrm{\sum M_{clockwise} = \sum M_{anticlockwise}} \)
Units:
Moment is measured in newton-metres (N·m).
Applications:
- Opening a door (the handle provides a large perpendicular distance).
- Using a wrench or spanner to turn a bolt.
- Balancing beams, see-saws, and levers.
Example
A force of \( \mathrm{8.0\,N} \) acts at right angles to a spanner \( \mathrm{0.25\,m} \) long. Find the moment of the force about the pivot.
▶️ Answer / Explanation
Using \( \mathrm{M = Fd} \):
\( \mathrm{M = 8.0 \times 0.25 = 2.0\,N·m.} \)
The moment of the force about the pivot is \( \mathrm{2.0\,N·m} \) in the direction of rotation caused by the force.
Example
A uniform beam 2.0 m long is pivoted at its centre. A force of \( \mathrm{30\,N} \) acts downward at one end. At what distance from the other end must a \( \mathrm{20\,N} \) force act upward to keep the beam horizontal?
▶️ Answer / Explanation
For equilibrium, \( \mathrm{\sum M_{clockwise} = \sum M_{anticlockwise}} \).
\( \mathrm{(30)(1.0) = (20)(x)} \)
\( \mathrm{x = \dfrac{30}{20} = 1.5\,m.} \)
The \( \mathrm{20\,N} \) force must act \( \mathrm{1.5\,m} \) from the other end (i.e., \( \mathrm{0.5\,m} \) beyond the centre) to balance the beam.
Couple
A couple is a pair of equal and opposite forces whose lines of action do not coincide. A couple acts to produce rotation only, without causing any resultant linear motion of the body.
Explanation:
- The two forces are equal in magnitude, opposite in direction, and parallel but act along different lines.
- Because the forces are opposite, their linear effects cancel — there is no resultant (translational) force.
- However, they produce a turning effect (rotation) about a point or axis.
Characteristics of a Couple:
- Consists of two forces that are equal, opposite, and parallel.
- The forces act along different lines of action, separated by a perpendicular distance.
- The resultant force = 0, but the resultant moment ≠ 0.
- A couple can rotate a body without moving its centre of mass.
Examples of Couples in Daily Life:
- Turning a steering wheel.
- Opening or closing a bottle cap using both hands.
- Using a wrench to tighten or loosen a nut.
- Pedalling a bicycle (each foot applies a couple to the crank).
Example
Explain why turning a steering wheel involves a couple and why the wheel does not translate sideways when turned.
▶️ Answer / Explanation
When you apply equal and opposite forces with both hands on opposite sides of the steering wheel, the two forces form a couple. They produce a rotational effect (moment) about the wheel’s centre. Because the forces are equal and opposite, their linear effects cancel, resulting in pure rotation without translation.
Torque of a Couple
The torque of a couple is the product of one of the forces and the perpendicular distance between the lines of action of the two forces.
\( \mathrm{Torque\ of\ couple = F \times d} \)
- \( \mathrm{F} \): magnitude of one of the forces (N)
- \( \mathrm{d} \): perpendicular distance between the lines of action of the forces (m)
- Unit: \( \mathrm{N·m} \)
Explanation:
- The torque of a couple measures its rotational effectiveness — i.e., how strongly it tends to rotate the body.
- The torque is independent of the position of the axis of rotation — it depends only on the size of the forces and their separation.
- For equilibrium, the sum of all torques (moments due to couples) acting on a body must be zero.
Properties of the Torque of a Couple:
- Produces angular acceleration but no linear acceleration.
- Acts in a plane perpendicular to the axis of rotation.
- Has a direction given by the right-hand rule (for vectors): Curl the fingers in the direction of rotation — the thumb points in the direction of the torque vector.
Note:
The torque of a couple is sometimes called the moment of a couple.
Example
Two equal and opposite forces of \( \mathrm{5.0\,N} \) act on opposite sides of a steering wheel of diameter \( \mathrm{0.40\,m.} \) Find the torque of the couple acting on the wheel.
▶️ Answer / Explanation
The perpendicular distance between the forces is equal to the diameter of the wheel:
\( \mathrm{d = 0.40\,m.} \)
Torque of the couple:
\( \mathrm{\tau = F \times d = 5.0 \times 0.40 = 2.0\,N·m.} \)
The couple produces a torque of \( \mathrm{2.0\,N·m} \), causing the wheel to rotate.
Example
Two identical spanners are used to loosen different bolts. On one, the mechanic applies a \( \mathrm{15\,N} \) force at a distance of \( \mathrm{0.20\,m.} \); on the other, a \( \mathrm{10\,N} \) force at \( \mathrm{0.30\,m.} \). Which produces a greater turning effect?
▶️ Answer / Explanation
Calculate torque for each case:
First: \( \mathrm{\tau_1 = 15 \times 0.20 = 3.0\,N·m.} \)
Second: \( \mathrm{\tau_2 = 10 \times 0.30 = 3.0\,N·m.} \)
Both couples produce the same torque of \( \mathrm{3.0\,N·m} \). Hence, both produce the same rotational effect despite different force–distance combinations.