Equilibrium — OCR A-Level Physics
Test yourself on Equilibrium with OCR A-Level practice questions.
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Equilibrium explained
The moment of a force is the turning effect of that force about a pivot.
Read the full explanation
It is calculated as the product of the force and the perpendicular distance from the pivot to the line of action of the force: moment = Fx, where F is the force in newtons and x is the perpendicular distance in metres. The unit is the newton metre (N m). A larger force or a larger perpendicular distance gives a larger moment. If the force is not perpendicular to the lever, use the perpendicular component of the force or the perpendicular distance to the line of action. Moments are vector quantities in the sense that clockwise and anticlockwise moments oppose each other.
(b) couple; torque of a couple; torque = Fd
A couple is a pair of equal and opposite parallel forces whose lines of action do not coincide. A couple produces rotation without any resultant translational force. 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: torque = Fd, where F is the magnitude of one force and d is the perpendicular separation. The unit is the newton metre (N m). Because the two forces are equal and opposite, they cancel translationally but their moments add to give a pure turning effect. A steering wheel turned by two hands is a common example.
(c) the principle of moments
The principle of moments states that for an object in rotational equilibrium, the sum of clockwise moments about any pivot equals the sum of anticlockwise moments. A moment is the product of a force and the perpendicular distance from the pivot to the force's line of action: moment = F × d, measured in newton metres (N m). For example, a 10 N force acting 0.50 m from a pivot produces a moment of 5.0 N m. When several forces act, calculate each moment separately, then equate the totals. This principle applies to any pivot point, provided the object is in equilibrium. It is used to solve problems involving seesaws, beams and balanced rulers, and is fundamental to understanding static equilibrium.
(d) centre of mass; centre of gravity; experimental determination of centre of gravity
The centre of mass of an object is the point where its entire mass may be considered to be concentrated. The centre of gravity is the point where the entire weight of the object acts. For a uniform gravitational field, these points coincide. For a symmetrical object of uniform density, the centre of mass lies at the geometric centre. For an irregular lamina, the centre of gravity can be found experimentally by suspending it freely from a pin and drawing a vertical line (using a plumb line) from the suspension point; repeating from a different point gives intersecting lines, and their intersection locates the centre of gravity. This method relies on the object hanging in equilibrium with its centre of gravity directly below the suspension point.
(e) equilibrium of an object under the action of forces and torques
An object is in equilibrium when the resultant force acting on it is zero and the resultant torque (moment) about any point is zero. This means the object has no linear acceleration and no angular acceleration. For coplanar forces, the conditions are: the vector sum of all forces is zero (ΣF = 0), and the sum of clockwise moments about any point equals the sum of anticlockwise moments (Στ = 0). For example, a beam balanced on two supports has upward forces from the supports equal to the total downward weight, and the moments about any support balance. These conditions apply to both static and dynamic equilibrium (constant velocity).
(f) condition for equilibrium of three coplanar forces; triangle of forces.
For three coplanar forces to be in equilibrium, their lines of action must be concurrent (meet at a single point) and the vector sum of the forces must be zero. This means the three forces can be represented in magnitude and direction by the sides of a closed triangle, taken in order. This is known as the triangle of forces. For example, if three forces act on an object and it remains in equilibrium, drawing the forces head-to-tail will form a closed triangle. The condition is a special case of the general equilibrium conditions: resultant force zero and resultant torque zero. It is used to solve problems involving three forces acting on a point or a rigid body.
Your focus
- Define moment of a force and state its unit.
- Calculate a moment using moment = Fx with perpendicular distance.
- Apply the principle of moments to simple equilibrium problems.
Show all 18 objectives
- Define a couple and explain why it produces no resultant translational force.
- Calculate the torque of a couple using torque = Fd.
- Distinguish between the moment of a single force and the torque of a couple.
- State the principle of moments and define the moment of a force.
- Calculate the moment of a force using the perpendicular distance from the pivot.
- Apply the principle of moments to solve equilibrium problems involving coplanar forces.
- Define centre of mass and centre of gravity and explain when they coincide.
- Describe an experiment to determine the centre of gravity of an irregular lamina.
- Explain why the centre of gravity lies directly below the suspension point when the object is in equilibrium.
- State the two conditions for equilibrium of a rigid body under coplanar forces.
- Apply the conditions to solve problems involving forces and torques.
- Distinguish between static and dynamic equilibrium.
- State the condition for equilibrium of three coplanar forces.
- Describe the triangle of forces and use it to solve equilibrium problems.
- Apply trigonometry to find unknown forces or angles in a triangle of forces.
Equilibrium exam tips
Marking Points
- Moment of a force is the turning effect about a pivot.
- moment = Fx, where F is force and x is perpendicular distance from pivot to line of action.
- The unit of moment is the newton metre (N m).
- Increasing force or perpendicular distance increases the moment.
- Clockwise and anticlockwise moments are treated as opposing when applying equilibrium.
- A couple consists of two equal and opposite parallel forces whose lines of action do not coincide.
- A couple produces rotation with no resultant translational force.
- torque = Fd, where F is one force and d is the perpendicular distance between the lines of action.
- The unit of torque is the newton metre (N m).
- The moments of the two forces add because they act in the same rotational sense.
- State that the principle of moments applies to an object in rotational equilibrium.
- Define moment as force multiplied by the perpendicular distance from the pivot to the line of action of the force.
- Use the correct unit for moment: newton metre (N m).
- Apply the principle by equating the sum of clockwise moments to the sum of anticlockwise moments about a chosen pivot.
- Calculate individual moments correctly using the perpendicular distance, not the distance along the force direction.
- Define centre of mass as the point where the total mass of an object may be considered to be concentrated.
- Define centre of gravity as the point where the total weight of an object may be considered to act.
- State that for a uniform gravitational field, the centre of mass and centre of gravity coincide.
- Describe the experimental method: suspend the lamina freely from a pin, draw a vertical line using a plumb line, repeat from another point, and locate the intersection.
- Explain that the intersection of the lines gives the centre of gravity because the object hangs in equilibrium with its centre of gravity directly below the suspension point.
- State that for equilibrium, the resultant force on the object must be zero.
- State that for equilibrium, the resultant torque (moment) about any point must be zero.
- Apply the condition ΣF = 0 in component form (e.g., sum of horizontal forces = 0, sum of vertical forces = 0).
- Apply the condition Στ = 0 by equating clockwise and anticlockwise moments about a chosen pivot.
- Recognise that equilibrium can be static (object at rest) or dynamic (object moving with constant velocity).
- State that for three coplanar forces to be in equilibrium, their lines of action must be concurrent.
- State that the vector sum of the three forces must be zero.
- Describe the triangle of forces: the three forces can be represented by the sides of a closed triangle taken in order.
- Use the triangle of forces to find an unknown force or angle by applying trigonometry (e.g., sine rule, cosine rule).
- Recognise that the triangle of forces is a consequence of the general equilibrium conditions.
Examiner Tips
- 💡Identify the pivot and draw the perpendicular distance before substituting values.
- 💡Convert all lengths to metres before using moment = Fx.
- 💡Check the direction of each moment and state clockwise or anticlockwise.
- 💡Show the equation, substitution and unit in your working.
- 💡State that the two forces are equal in magnitude and opposite in direction.
- 💡Identify the perpendicular distance between the lines of action, not between the points of application.
- 💡Use torque = Fd and give the unit N m.
- 💡Check whether the question asks for a moment or a torque and use the correct term.
- 💡Always identify the pivot point clearly before calculating moments.
- 💡Check that all distances are perpendicular to the force; if not, resolve the force or use the perpendicular component.
- 💡Verify that the sum of clockwise moments equals the sum of anticlockwise moments as a final check.
- 💡Remember that the plumb line always hangs vertically, so the line drawn is vertical.
- 💡Ensure the lamina can swing freely; any friction at the pivot may cause error.
- 💡Use a sharp pencil and a fine plumb line to improve accuracy of the intersection point.
- 💡Draw a clear free-body diagram showing all forces and their directions.
- 💡Choose a pivot point that eliminates unknown forces from the moment equation to simplify calculations.
- 💡Check that both force and moment conditions are satisfied; one alone is not sufficient for equilibrium.
- 💡Sketch the triangle of forces clearly, labelling the forces and angles.
- 💡Use the sine rule or cosine rule to solve for unknown magnitudes or angles.
- 💡Check that the directions of the forces in the triangle correspond to the actual directions in the problem.
Common Mistakes
- Using the distance along the lever rather than the perpendicular distance to the line of action.
- Forgetting to convert distances from centimetres to metres before calculating.
- Adding moments in the same direction instead of recognising clockwise and anticlockwise opposition.
- Using the wrong unit, such as N, instead of N m.
- Confusing a couple with two forces that act along the same line: a couple requires separated, parallel lines of action.
- Using the distance between the points of application rather than the perpendicular distance between the lines of action.
- Thinking the forces cancel completely: they cancel translationally but produce a net turning effect.
- Forgetting that torque has the same unit as moment, the newton metre.
- Using the distance along the line of action instead of the perpendicular distance from the pivot; correction: always use the perpendicular distance from the pivot to the line of action.
- Forgetting to convert distances to metres before calculating moments; correction: convert all lengths to metres to keep units consistent (e.g., 50 cm = 0.50 m).
- Assuming the principle of moments only applies when the pivot is at the centre of the object; correction: the principle holds for any chosen pivot point when the object is in equilibrium.
- Confusing centre of mass with centre of gravity; correction: centre of mass relates to mass distribution, centre of gravity relates to weight distribution; they coincide in a uniform gravitational field.
- Thinking that the centre of gravity must lie within the material of the object; correction: it can lie outside, e.g., for a ring or horseshoe.
- Drawing only one line in the experimental determination; correction: at least two lines from different suspension points are needed to find the intersection.
- Assuming that equilibrium only requires balanced forces and ignoring torques; correction: both resultant force and resultant torque must be zero.
- Forgetting to consider all forces, including weight and reaction forces; correction: draw a free-body diagram to identify all forces and their points of application.
- Choosing a pivot that simplifies the problem but then incorrectly omitting the moment of a force that acts through that pivot; correction: forces acting through the pivot have zero moment about that pivot, so they can be omitted from the moment equation.
- Assuming that any three forces in equilibrium must form a triangle without checking that their lines of action are concurrent; correction: the forces must be concurrent for the triangle of forces to apply.
- Drawing the triangle with forces not taken in order (head-to-tail); correction: the forces must be arranged head-to-tail to form a closed triangle.
- Forgetting that the triangle of forces only applies to three forces; correction: for more than three forces, a polygon of forces is used.