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a 8.32 -m ladder with a mass of 20.7 kg lies fiat on the ground. a painter grabs the top end of the ladder and pulls straight upward with a force of 269 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 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?

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The net torque acting on the ladder is 1122.08 Nm and the ladder's moment of inertia is 597.03 kgm².

In order to determine the net torque acting on the ladder, we need to calculate the torque exerted by the force applied by the painter and the torque due to the ladder's weight. The torque produced by the painter's force is given by the equation:

net torque = force × lever arm

net torque = 269 N × (8.32 m / 2)

net torque = 1122.08 Nm

The ladder's moment of inertia can be found using the equation:

angular acceleration = net torque / moment of inertia

1.88 rad/s² = 1122.08 Nm / moment of inertia

moment of inertia = 1122.08 Nm / 1.88 rad/s²

moment of inertia = 597.03 kgm²

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