Q#8 (Past Exam Paper – March 2017 Paper 12 Q13)
The vertical forces that the ground exerts on a stationary van are shown.
The van is 2.50 m long with the wheels at a distance of 0.600 m from the front of the van and 0.400 m from the rear of the van.
What is the horizontal distance of the van’s centre of gravity from the front of the van?
A 0.540 m B 0.960 m C 1.14 m D 1.36 m
Solution:
Answer: C.
Consider the moments about the front wheel (acting as the pivot). The 16 000 N force does not cause any moment.
Let the centre of gravity W be at a distance of y from the front wheel.
Since the van is stationary, the resultant force on the van is zero.
(Downward) Weight W = sum of upward forces
= 16 000 + 9 000 = 25 000 N
About front wheel,
Clockwise moment = W × y = 25 000 × y
Anti-clockwise moment = 9000 × 1.5
For equilibrium, sum of clockwise moment = sum of anticlockwise moment
25 000 × y = 9000 × 1.5
Distance y from front wheel = (9000 × 1.5) / 25 000 = 0.54 m
Distance from front of van = 0.6 + 0.54 = 1.14 m
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