Question

A 9.76-m ladder with a mass of 22.6 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 254 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.99 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?

Answer 1

Answer:

Part a)

$\tau_{net} = 1397.1 Nm$

Part b)

$I = 702.06 kg m^2$

Explanation:

Part a)

Torque on the rod is due to two forces

(i) Due to weight of the rod at mid point

(ii) Due to external force applied at the end

Now we have

$\tau_{net} = FL - mg\frac{L}{2}$

$\tau_{net} = 254(9.76) - (22.6)(9.81)(\frac{9.76}{2})$

$\tau_{net} = 1397.1 Nm$

Part b)

As per Newton's law we know that

$\tau = I\alpha$

now we have

$1397.1 Nm = I(1.99 rad/s^2)$

$I = 702.06 kg m^2$