Q#71: [Forces > Moments] (Past Exam Paper – June 2009 Paper 1 Q13 & June 2014 Paper 11 Q13)
Spindle is attached at one end to centre of lever of length 1.20 m and at its other end to centre of disc of radius 0.20 m. String is wrapped round disc, passes over a pulley and is attached to 900 N weight.
What is the minimum force F, applied to each end of lever, that could lift the weight?
A 75 N B 150 N C 300 N D 950 N
Solution 71:
Answer: B.
Distance of Force F from the centre of the spindle = 1.20 / 2 = 0.60m.
The forces F cause a clockwise moment.
The 900N weight acts at a distance 0.20m (equal to the radius) from the spindle, causing an anticlockwise moment.
So for minimum force, the clockwise moment should be equal to the anticlockwise moment.
F (0.60) + F (0.60) = 900 (0.20).
1.20F = 180N
Minimum Force F = 180 / 1.20 = 150N
Post a Comment for "Spindle is attached at one end to centre of lever of length 1.20 m and at its"