Question

Given: drum weight = 300 N, coefficient of static friction between the drum and the floor µs = 0.5, a = 0.6 m, b = 0.8 m, θ = 57°. Find the smallest value of P in newtons that will cause impending motion (tipping or slipping) of the drum.

Put a minus sign in front of your answer if your condition is tipping.

Answer 1

The friction force = mu*(N + P*3/5) = 0.5*300 + P*0.9/5 = 150+ P*0.9/5 N

Horizontal component of pushing force = P*4/5

equating both for slipping we get P = 241.93 N

For tipping balancing torque about farther corner from man, touching the ground

P*4/5*b = 300*a/2 + P*3/5*a

solving this we get P = 321.42N

Thus smallest value will be that of slipping P = 241.93 N

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