Question

A 64 kg painter is standing three fourths of the distance up a ladder that is 3.0 m long. If the ladder makes an angle of 69o with the ground, what torque does the painter's weight exert on the ladder? Just need an explanation :) Answer:(5.1 x 102 N/m)

Answer 1

In general the equation for torque is:

\[\tau = rF \sin \theta\]
Where τ is the torque, r is the distance from the axis of rotation to the point where the force is applied, F is the applied force, and θ is the angle between the force and the lever arm (which is in the same direction as r).

The way the situation is set up, the axis of rotation is about the base of the ladder. The painter is ¾ * 3 = 2.25 m up the ladder, so r=2.25. You can figure out the force in a couple different ways. Remember, Fsinθ is all referring to the force. The reason we have the sinθ in there is to get the component of the force that is PERPENDICULAR to the lever arm. If we have an expression for the perpendicular component already, the sinθ disappears (because θ would be 90, making sinθ=1). In this case, we want the force perpendicular to the ladder.

The (perhaps) more intuitive way of finding the perpendicular force is by using mgcosθ, were θ=69, same as you would to find the perpendicular weight component of something on a ramp. Finding the value of:

\[\tau = rmg \cos \theta\]\[\tau = (2.25)(64)(9.8)( \cos 69)\]
Will give you 5.1 x 10^2 N m as your answer.

If you'd prefer to stay true to the original equation that uses sinθ you can, you just need to realize that θ=21, not 69, because the angle is in reference to the lever arm, not the ground.

\[\tau = rmg \sin \theta\]\[\tau = (2.25)(64)(9.8)( \sin 21)\]
Will also give you 5.1 x 10^2 N m as your answer.