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A solid cylinder of mass 6 kg is lightly held with its axis vertical in jaws of a vice and is on the point of slipping downwards.the magnitude of the force exerted by each of the jaw is 70 N Calculate the coefficient of friction between vice and the cylinder

Answers (1)

Given that,

M=mass of the cylinder
F= applied force = 70 N
g = acceleration by gravity = 9.8 m/s²
 
the weight of the packet is:
w = m * g =58.8 N
 
This force must be "canceled out" by friction between the jaws of the vice and the cylinder. Each jaw will provide for half of the weight, so 
(58.8 N) / 2 = (70 N) × µ 
µ = 0.42

Posted by

Satyajeet Kumar

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