Aperson with mass m stands on a ladder of mass m that leans against a frictionless vertical wall and makes an angle of 53° measured from the horizontal. the ladder does not slip on the horizontal surface due to a static frictional force between the ladder and the ground. some of the forces exerted on the ladder can produce torques on the ladder. using the contact point between the ladder and the ground as a reference, which force(s) exerted on the ladder produce clockwise torques?

clockwise torque is due to weight of the person + weight of the ladder

Explanation:

As we know that the forces that act on the ladder is given as

1) friction force

2) Normal force at the ends of the ladder

3) Weight of the person standing on the ladder

4) Weight of the ladder

So here we know that our reference point is at the bottom position of the ladder

So we will have

Torque due to weight of the person and weight of the ladder is counter balanced by the torque due to normal force and torque due to friction force

So here we have

clockwise torque is due to weight of the person + weight of the ladder

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.