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Applied Mechanics

Q1: A string of length 90 cm is fastened to two points A and B at the same level 60 cm apart. A ring weighing 120 g is slided on the string. A horizontal force P is applied to the ring such that it is in equilibrium vertically below B. The value of P is :

A 40 g

B 60 g

C 80 g

D 100 g

ANS:C - 80 g

Taking Moments at point B.
M1 * d1 = M2 * d2
m1 = 120g, d1 = 60cm d2 = 90cm.

Lengh of string. Hence subtitute 120g * 60cm = m2 * 90.
By solving m2 = 120 * 60/90 = > m = 80g. Correct answer.



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