An 80 kg man is 1/4 of the way up a 10m uniform ladder that is resting against a frictionless wall. If the ladder has a made of 20 kg, and it will make an angle of 60 degrees to the ground, find the force of friction on the ground on the foot of the ladder?

Question

anonymous · Answer

$$\sum_{}^{} \tau  = 0$$Choose a pivot point, say at the foot of the ladder. Then Clockwise torques are: $$80kg*g*(2.5m)\cos(60) + 20kg*g*(5m)\cos(60) = F _{wall}*(10m) \sin (60)$$. From Sum (forces) = 0, we have $$F _{wall} = f _{k}$$and $$N = (20kg + 80kg)g$$Play with the algebra and solve for the frictional force

anonymous · Answer

The right hand side of the torque equation = the counter-clocwwise torques

anonymous · Answer

why do you use cos60 instead of sin?

anonymous · Answer

The angle between the ground and the ladder is 60 deg. We want the moment arm (torque arm) to be the perpendicular distance from the rotation point (base of the ladder) to the line of action of the force... that is the distance along the ladder * cos 60