OCR A Physics · Module 3: Forces and motion
Forces in action: practice questions with answers
6 exam-style questions on forces in action. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.
- Using the right equation2 marks
A uniform beam of weight 120 N and length 4.0 m rests on two supports. Support A is at the left end and support B is 1.0 m from the right end. Calculate the upward force at B.
FB = N
Show the answer and working
Answer FB = 80 N
Moments about A: clockwise 120 × 2.0 (weight at centre) = anticlockwise FB × 3.0, so FB = 80 N.
- Got 60 N? You assumed the load is shared equally. Take moments about A: FB × 3.0 = 120 × 2.0, so FB = 80 N.
- Got 60 N? You used the beam length (4.0 m) as the distance to B. B is 3.0 m from A, so FB = 120 × 2.0 ÷ 3.0 = 80 N.
- Resolving vectors2 marks
A 12 kg crate sits on a smooth slope inclined at 25° to the horizontal. Calculate the force, parallel to the slope, needed to hold it at rest.
F = N
Show the answer and working
Answer F = 49.8 N
Resolve weight along the slope: mg sin θ = 12 × 9.81 × sin 25° = 49.8 N.
- Got 107 N? You used the component perpendicular to the slope. The component down the slope is mg sin θ = 12 × 9.81 × sin 25° = 49.8 N.
- Got 118 N? You used the full weight. Only the component down the slope must be balanced: mg sin 25° = 49.8 N.
- Unit prefixes2 marks
A solid object of volume 2500 cm³ is fully submerged in water of density 1000 kg m⁻³. Calculate the upthrust on it.
U = N
Show the answer and working
Answer U = 24.5 N
V = 2500 × 10⁻⁶ = 2.5 × 10⁻³ m³. Upthrust = ρgV = 1000 × 9.81 × 2.5 × 10⁻³ = 24.5 N.
- Got 24500 N? You converted cm³ to m³ using 10⁻³. 1 cm³ = 10⁻⁶ m³, so V = 2.5 × 10⁻³ m³ and U = ρgV = 24.5 N.
- Got 0.0245 N? You converted cm³ to m³ using 10⁻⁹. The factor is 10⁻⁶, so V = 2.5 × 10⁻³ m³ and U = 24.5 N.
- Rearranging equations3 marks
A skydiver of mass 78 kg falls at a terminal speed of 50 m s⁻¹. Assume drag is proportional to speed squared. Calculate the drag force when the speed is 30 m s⁻¹.
D = N
Show the answer and working
Answer D = 276 N
At terminal speed drag = weight = 78 × 9.81 = 765 N. Drag ∝ v², so at 30 m s⁻¹: 765 × 0.36 = 275 N.
- Got 459 N? You treated drag as proportional to speed. Drag ∝ v², so D = 765 × (30 ÷ 50)² = 275 N.
- Got 165 N? You used v³. For this model drag ∝ v², so D = 765 × 0.6² = 275 N.
- Using the right equation1 mark
Which statement correctly describes a couple?
- A Two equal forces in the same direction acting at different points.
- B Two equal and opposite forces acting along the same line.
- C Two equal and opposite parallel forces with different lines of action.
Show the answer and working
Answer C: Two equal and opposite parallel forces with different lines of action.
A couple has no resultant force but a turning effect, because the lines of action are separated.
- Why not A? Forces in the same direction give a resultant force and no pure turning effect.
- Why not B? Equal and opposite forces along the same line cancel completely and produce no turning effect.
- Resolving vectors1 mark
A body is in equilibrium under three coplanar forces. Which statement must be true?
- A The three force vectors form a closed triangle when drawn head to tail.
- B The three forces must be equal in size.
- C Two of the forces must be perpendicular.
Show the answer and working
Answer A: The three force vectors form a closed triangle when drawn head to tail.
Zero resultant means the vectors add to nothing, so they close into a triangle.
- Why not B? Equilibrium needs a zero resultant, not equal forces; equal sizes only happen in special cases.
- Why not C? The forces can sit at any angles provided the vector sum is zero.