Q#43 (Past Exam Paper – June 2012 Paper 12 Q15 & November 2017 Paper 12 Q12)
The diagram shows a solid cube with weight W and sides of length L. It is supported at rest by a frictionless spindle that passes through the centres of two opposite vertical faces. One of these faces is shaded.
The spindle is now removed and replaced at a distance L / 4 to the right of its original position.
From June 2012 Paper 12 Q15:
When viewing the shaded face, what is the torque of the couple that will now be needed to stop the cube from toppling?
A WL / 2 anticlockwise
B WL / 2 clockwise
C WL / 4 anticlockwise
D WL / 4 clockwise
From November 2017 Paper 12 Q12:
When viewing the shaded face, what is the torque of the couple that will now be needed to keep the cube at rest?
A WL / 4 anticlockwise
B WL / 4 clockwise
C WL / 2 anticlockwise
D WL / 2 clockwise
Solution:
June 2012 Paper 12 Q15 – Ans: D.
November 2017 Paper 12 Q12 – Answer: B.
The weight acts downwards at the centre of the cube.
The spindle acts as the pivot.
When the spindle is displaced, the weight causes an anticlockwise moment about the pivot.
The perpendicular distance of the weight from the pivot (spindle) is L/4.
Moment = force × perpendicular distance
Moment due to weight = W × L/4 = WL / 4 (anticlockwise)
The cube would topple anticlockwise when released, so the torque (moment) needed to stop it from turning must be clockwise with a magnitude equal to WL / 4 (since, for equilibrium, the clockwise moment is equal to the anticlockwise moment).
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