Q#7 (Past Exam Paper – November 2017 Paper 11 Q14)
The diagram shows a motorised vehicle for carrying one person.
The vehicle has two wheels on one axle. The passenger stands on a platform between the wheels.
The weight of the machine is 600 N. Its centre of mass is 200 mm in front of the axle. The wheel radius is 400 mm.
When stationary, a passenger of weight 600 N stands with his centre of mass 200 mm behind the axle to balance the machine.
The motor is now switched on to provide a horizontal force of 90 N at the ground to move the vehicle forwards.
How far and in which direction must the passenger move his centre of mass to maintain balance?
A 60 mm backwards
B 60 mm forwards
C 140 mm backwards
D 140 mm forwards
Solution:
Answer: B.
To maintain balance,
Anti-clockwise moment = Clockwise moment
The weight of the machine exerts a clockwise moment (= 600×200).
The weight of the person exerts an anti-clockwise moment (= 600×200).
To move forward, the machine exerts a force of 90N in the direction of motion. This provides an anti-clockwise moment (= 90×400).
When in motion,
(600×200) + (90×400) = (600×200)
When the motor is on, the anti-clockwise moment is greater than the clockwise moment.
To maintain balance,
Anti-clockwise moment = Clockwise moment
So, to maintain balance, the anti-clockwise moment should be reduced so that it equals the clockwise moment.
The moment exerted by the weight of the person can be changed by moving his centre of mass. To reduce the anti-clockwise moment, the person should bring his centre of mass towards the pivot (axle). That is, he needs to move forwards. This reduced the distance of his weight from the axle.
Let the distance moved forwards to maintain balance be x.
[600 × (200 – x)] + (90×400) = (600×200)
200 – x = [(600×200) – (90×400)] / 600 = 140
x = 200 – 240 = 60 mm
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