AS Physics: Forces, Density and Pressure — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Forces, Density and Pressure revision notes

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Questions

1. Define a moment, and state both conditions for a body to be in equilibrium. [4]

2. Explain what a couple is and why it produces rotation without translation. [3]

3. A uniform beam of weight 240 N and length 3.0 m rests on two supports, one at each end. A load of 500 N is placed 1.0 m from the left support.

(a) Take moments about the left support to find the force on the right support. [4] (b) Find the force on the left support. [2] (c) Explain why taking moments about the left support is a sensible choice. [2]

4. A block of density 2700 kg m⁻³ and volume 3.5 × 10⁻⁴ m³ is fully submerged in water of density 1000 kg m⁻³. (g = 9.81 m s⁻²)

(a) Calculate its weight. [2] (b) Calculate the upthrust. [2] (c) Calculate the apparent weight in water. [1] (d) State Archimedes’ principle and explain the physical origin of upthrust. [3]

5. Explain why a steel ship floats although steel is denser than water. [2]

6. Calculate the hydrostatic pressure increase (gauge pressure) at a depth of 25 m in seawater of density 1030 kg m⁻³, and explain why the shape of the container is irrelevant. [4]

7. Describe how a closed vector triangle can be used to find an unknown force acting on an object in equilibrium under three coplanar forces. [3]

8. A uniform beam of weight 40 N is pivoted at its centre (so the beam’s own weight produces no moment about the pivot), with a 25 N weight hung 0.6 m from the pivot on one side. How far from the pivot must a 15 N weight be hung on the other side to balance it? [3]

9. Two forces of 8 N each, acting 0.5 m apart, form a couple. Calculate the torque of the couple. [2]

10. Explain why the pressure difference between two points in a fluid depends only on the vertical depth difference between them, not on how far apart they are horizontally. [2]

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Answers

1. A moment is the force multiplied by the perpendicular distance from the pivot to the line of action of the force [1] [1]. Equilibrium requires resultant force = 0 [1] and resultant moment about any point = 0 [1].

2. A couple is two equal, opposite, parallel forces not acting along the same line [1]. The forces cancel, so there is no resultant force and hence no change to the body’s translational motion [1] — a body already moving continues at that same velocity — but they produce moments in the same rotational sense, giving a nonzero resultant moment and so an angular acceleration [1].

3. (a) Taking moments about the left support: Clockwise: (240 × 1.5) + (500 × 1.0) = 360 + 500 = 860 N m [1] [1] Anticlockwise: R × 3.0 [1] R = 860 ÷ 3.0 = 287 N [1]. (b) Total upward = total downward: L + 287 = 240 + 500 [1] L = 453 N [1]. (c) The unknown force at the left support has zero moment about that point [1], so it drops out of the equation and only one unknown remains [1].

4. (a) m = ρV = 2700 × 3.5 × 10⁻⁴ = 0.945 kg [1] W = mg = 0.945 × 9.81 = 9.27 N [1]. (b) Upthrust = ρ_water × g × V = 1000 × 9.81 × 3.5 × 10⁻⁴ [1] = 3.43 N [1]. (c) 9.27 − 3.43 = 5.84 N [1]. (d) The upthrust equals the weight of fluid displaced [1]. Pressure increases with depth, so the upward pressure on the bottom face exceeds the downward pressure on the top face [1], and the difference gives a net upward force [1].

5. The ship’s average density, including the enclosed air, is less than that of water [1], so it displaces its own weight of water before becoming fully submerged [1].

6. Δp = ρgh = 1030 × 9.81 × 25 [1] [1] = 2.53 × 10⁵ Pa [1]. This is the gauge pressure — the increase above the pressure at the surface, not the absolute pressure (which would also include atmospheric pressure). The pressure increase depends only on depth, density and g [1] — not on the shape or total volume of the container.

7. Draw each of the three forces as an arrow, tip-to-tail, in turn and to scale [1]; if the object is truly in equilibrium, the three arrows form a closed triangle, returning to the starting point with no gap [1]. Any unknown force’s magnitude or direction can then be found by measuring or calculating the missing side of that triangle [1].

8. By the principle of moments: 25 × 0.6 = 15 × d [1], so 15 = 15d [1], giving d = 1.0 m [1].

9. Torque = one force × perpendicular distance between the forces = 8 × 0.5 = 4.0 N m [1] [1].

10. Pressure at a point in a fluid arises from the weight of the fluid column above it [1]; that weight depends only on the vertical height of fluid above the point, not on horizontal position, so two points at the same depth experience equal pressure regardless of how far apart they are sideways [1].

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Where marks are usually lost

• Omitting “perpendicular” from the definition of a moment.
• Giving only one condition for equilibrium.
• Forgetting the beam’s own weight acts at its centre.
• Saying upthrust depends on the object’s weight rather than the fluid displaced.
• Drawing the vector triangle open rather than closed, or failing to keep the arrows tip-to-tail in a consistent direction.
• Confusing torque of a couple (one force × distance between the forces) with a single moment (force × distance to a pivot).