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Applied Mechanics - Engineering

Q1:

A uniform rod 9 m long weighing 40 kg is pivoted at a point 2 m from one end where a weight of 120 kg is suspended. The required force acting at the end in a direction perpendicular to rod to keep it equilibrium, at an inclination 60° with horizontal, is

A 40 kg

B 60 kg

C 10 kg

D 100 kg

ANS:C - 10 kg

Equation moment about the pivot,

120 * 2cos(60) = 40 * 2.5 cos(60) + xcos(60)
or, 240 - 140 = 7x
x = 20kg.

Now, this is the weight of rod acting vertically downwards..to calculate the force perpendicular to the rod, resolve it in a direction perpendicular to the rod.

Hence,
Force = 20 cos (60) = 10kg.