A 9.46 -m ladder with a mass of 24.0 kg lies flat on the ground. A painter grabs

Question

A 9.46 -m ladder with a mass of 24.0 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 246 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.89 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

length of ladder= 9.46 -m

mass= 24.0 kg

force= 246 N.

angular acceleration = 1.89 rad/s2

F = 246 N applied perpendicular to ladder, ladder weight = mg

a) T = (FxL) - (mg x L/2)

.. T = (246N x 9.46m) - (24kg x 9.8 x 9.46m / 2)

Torque,T = 1214.664 Nm

b) Torque = moment of inertia x angular acceleration

.. T(Nm) = I(kg.m2) x a(rad/s2) ..

.. I = T / a= 1214.66 / 1.89 .. .

I = 642.68 kg.m2

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