A-LevelPhysicsForces, density and pressure9702/22 · Oct/Nov 2019 · Q48 Marks

A sphere in a liquid accelerates vertically downwards from rest. For the viscous force acting on the moving sphere, state:

(a)A sphere in a liquid accelerates vertically downwards from rest. For the viscous force acting on the moving sphere, state:
(a)(i)the direction[1]
(a)(ii)the variation, if any, in the magnitude.[1]
(b)A man of weight 750N stands a distance of 3.6m from end D of a horizontal uniform beam AD, as shown in Fig. 4.1. [Figure 4.1] The beam has a weight of 380N and a length of 9.0m. The beam is supported by a vertical force FB at pivot B and a vertical force FC at pivot C. Pivot B is a distance of 2.0m from end A and pivot C is a distance of 2.0 m from end D. The beam is in equilibrium.
(b)(i)State the principle of moments.[2]
(b)(ii)By using moments about pivot C, calculate FB. FB =[2]
(b)(iii)The man walks towards end D. The beam is about to tip when F♭ becomes zero. Determine the minimum distance x from end D that the man can stand without tipping the beam. X =[2]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[B1] (vertically) upwards/up
[B1] increases (with time/velocity/depth)
[B1] for a body in (rotational) equilibrium
[B1] sum/total of clockwise moments about a point = sum/total of anticlockwise moments about the (same) point
[C1] (FB × 5.0) or (380 × 2.5) or (750 × 1.6)
[A1] (FB × 5.0) = (380 × 2.5) + (750 × 1.6) AND FB = 430 N
[C1] taking moments about C: (380 × 2.5) = 750 × (2.0 −x) AND (2.0-x) = 1.3 AND x = 0.7 m OR
[(C1)] moments may be taken about other points, e.g. about D: (380 × 4.5) + (750 × x) = 1130 × 2.0
[(A1)] x = 0.7 m

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