Question

A 9.98 -m ladder with a mass of 22.9 kg lies flat on the ground. a painter grabs the top end of the ladder and pulls straight upward with a force of 242 n. at the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.57 rad/s 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?

Answer 1

The net torque acting on the ladder when the painter pulls it upward is calculated using the distance from the pivot to the force application point and is found to be 1207.58 N·m.

Step-by-step explanation:

Calculating Net Torque on the Ladder

To calculate the net torque acting on the ladder, we will use the formula τ = rFsin(θ), where τ is the torque, r is the distance from the pivot point to the point where the force is applied, F is the force applied, and θ is the angle between the force vector and the lever arm. Since the force is applied upward at one end of the ladder and the pivot is at the other end, the angle θ is 90 degrees, making sin(θ) equals 1.

The center of gravity of the ladder is at its midpoint, so the distance r = 9.98 m / 2 = 4.99 m. The applied force is F = 242 N. Thus, the net torque is τ = 4.99 m * 242 N * sin(90°) = 4.99 m * 242 N * 1 = 1207.58 N·m.