A 9.91 -m ladder with a mass of 22.3 kg lies flat on the ground. A painter grabs

Question

A 9.91 -m ladder with a mass of 22.3 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 257 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.88 rad/s2 about an axis passing through the bottom end of the ladder. The ladder's center of gravity lies halfway between the top and bottom ends. (a) What is the net torque acting on the ladder? (b) What is the ladder's moment of inertia?

Explanation / Answer

this is method values may change The net torques acting are due to the painter and the weight of the ladder. So torque = 269N*8.54m - 21.7kg*9.8m/s^2*8.54m/2 = 1389N-m = 1390N-m Using torque = I*alpha we get I = torque/alpha = 1389N/1.95rad?s^2 = 712 kg-m^2

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