Question 43
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
Reference: Past Exam Paper – June 2012 Paper 12 Q15 & November 2017
Paper 12 Q12
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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