Equilibrium of forces: past paper questions
Real questions on equilibrium of forces
The newest 20 of 50 written questions. Open one to see it on the real paper.
- 9702/42 · October/November 2025 · Q1b(iii)[2]
On Fig. 1.3, draw labelled arrows from the student to show the directions of the forces that act on the student to cause the resultant force in (b)(ii).
See it on the paper → - 9702/22 · May/June 2025 · Q2b(i)[3]
By taking moments about point Q, show that θ is 25°.
See it on the paper → - 9702/22 · May/June 2025 · Q2b(ii)[2]
On Fig. 2.2, draw a labelled scale vector triangle to represent the forces acting on the tree. The weight of the tree has been drawn to scale. weight Fig. 2.2
See it on the paper → - 9702/22 · May/June 2025 · Q2b(iii)[2]
The tree exerts a pressure of 150 kPa on the top of the post. Determine the surface area of the tree in contact with the post. area = m 2
See it on the paper → - 9702/22 · October/November 2024 · Q3b[2]
State the conditions necessary for an object to be in equilibrium.
See it on the paper → - 9702/22 · October/November 2024 · Q3c(i)[3]
By taking moments about A, determine the tension in the wire. tension = N (i)
See it on the paper → - 9702/23 · October/November 2024 · Q3c[3]
More water is gradually added to the container in (b), so that depth h in Fig. 3.1 gradually increases. The length x is continuously adjusted so that the system remains in equilibrium. On Fig. 3.2, sketch the variation of x with h. Use the space below for any working. 0.8 0.6 x / m 0.4 0.2 0 0.10 0.15 0.20 0.25 0.35 0....
See it on the paper → - 9702/42 · October/November 2024 · Q5a[1]
On Fig. 5.1, draw an arrow, from the centre of the sphere, to represent the direction of the resultant force acting on the sphere when it is in the position shown.
See it on the paper → - 9702/21 · May/June 2024 · Q1a[3]
Use SI base units to show that the drag coefficient has no units.
See it on the paper → - 9702/21 · May/June 2024 · Q1b[2]
Fig. 1.1 shows a sphere falling at terminal velocity in air. sphere terminal velocity Fig. 1.1 Assume that the upthrust on the sphere is negligible. On Fig. 1.1, draw and label arrows to show the directions of the two forces acting on the sphere.
See it on the paper → - 9702/21 · May/June 2024 · Q1c[2]
The mass of the sphere is 49 g. Calculate the drag force F D acting on the sphere. F D = N
See it on the paper → - 9702/21 · May/June 2024 · Q1d[3]
The sphere is falling in air at a terminal velocity of 25 in SI base units. The density of the air is 1.2 in SI base units. The diameter of the sphere is 0.060 in SI base units. Use your answer in (c) to calculate the drag coefficient C for the sphere. C =
See it on the paper → - 9702/22 · May/June 2024 · Q4c(i)[2]
The trapdoor is in equilibrium when F is 1.7 N. Calculate d. d = m
See it on the paper → - 9702/22 · May/June 2024 · Q4c(ii)[1]
Force F is decreased from 1.7 N. State the direction of the resultant moment about the pivot on the trapdoor. ,
See it on the paper → - 9702/23 · October/November 2023 · Q3a[2]
State the conditions for a system to be in equilibrium.
See it on the paper → - 9702/22 · May/June 2023 · Q2b(i)[3]
By taking moments about the pivot, calculate the tension T in the string. T = N
See it on the paper → - 9702/22 · May/June 2023 · Q2b(ii)[1]
The string is cut so that the beam is no longer in equilibrium. Calculate the magnitude of the resultant moment about the pivot acting on the beam immediately after the string is cut. resultant moment = N m
See it on the paper → - 9702/22 · February/March 2023 · Q3a[2]
State two conditions for an object to be in equilibrium.
See it on the paper → - 9702/22 · February/March 2023 · Q3b[1]
Show that the vertical component of the tension in the string is 13 N.
See it on the paper → - 9702/22 · February/March 2023 · Q3d[1]
Calculate the magnitude of the vertical component of the force exerted on the beam by the hinge. force = N
See it on the paper →
What examiners said
The majority of the candidates gave one of the conditions for equilibrium in terms of the resultant force being zero. The condition in terms of the resultant moment being zero was not always given. Some of the weaker candidates confused moment with momentum.
— Examiner report, October/November 2023, 9702/23 Q3(a) (Cambridge)
The principle of moments was usually applied correctly. Some candidates were unable to calculate the component of the tension perpendicular to the beam. Other candidates forgot that this component of the tension needed to be multiplied by its distance to the pivot in order to calculate its moment.
— Examiner report, May/June 2023, 9702/22 Q2(b)(i) (Cambridge)
The resultant moment about the pivot was usually calculated correctly. The most common error was to add the moments of the weights of the blocks rather than to subtract them.
— Examiner report, May/June 2023, 9702/22 Q2(b)(ii) (Cambridge)
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