uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0 m from the base of the wall. What must be the magnitude of the force of static friction supplied by the floorto keep the ladder from slipping

Fx1 (6 m) sin 60 = 300 (3 m) cos 60  balancing torques about floor

Fx1 = 900 * 1/2 / 5.20 = 86.6 N  this is the horizontal force that must be supplied by the wall to balance torques about the floor

This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.

Fx1 = Fx2 = 86.6 N

prolly a beacuse the momentuem the car has increases while going down hill

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