A 8.43 -m ladder with a mass of 20.1 kg lies flat on the ground. A painter grabs

Question

A 8.43 -m ladder with a mass of 20.1 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 252 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.96 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

Given thatthe mass of ladder is m = 20.1kg

Length ofladder is L = 8.43m

The forceon the ground is F = 252N

Theangular accelaration is = 1.96 rad/s2

The nettorque on the ladder about the bottom is

= F*L - mg*L/2

= 252N*8.43m - 20.1kg*9.8m/s2*8.43m / 2

= 1294.089 or 1294 N.m

But the torque is

= I*

I = /

= 1294/1.96

= 660.2 kg.m2

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