A uniform ladder 5.0 m long rests against a frictionless,vertical wall with its

Question

A uniform ladder 5.0 m long rests against a frictionless,vertical wall with its lower end 3.0 m from the wall. The ladderweighs 135 N. The coefficient of staticfriction between the foot of the ladder and the ground is 0.40. Aman weighing 720 N climbs slowly up theladder.
(a) What is the maximum frictional force thatthe ground can exert on the ladder at its lower end?


(b) What is the actual frictional force when the man has climbed1.0 m along the ladder?


(c) How far along the ladder can the man climb before the ladderstarts to slip?
A uniform ladder 5.0 m long rests against a frictionless,vertical wall with its lower end 3.0 m from the wall. The ladderweighs 135 N. The coefficient of staticfriction between the foot of the ladder and the ground is 0.40. Aman weighing 720 N climbs slowly up theladder.
(a) What is the maximum frictional force thatthe ground can exert on the ladder at its lower end?


(b) What is the actual frictional force when the man has climbed1.0 m along the ladder?


(c) How far along the ladder can the man climb before the ladderstarts to slip?

Explanation / Answer

the forces acting on the ladder are Fx = w2 + (-fk) = w2 + (-k*n) -------------(1) and Fy = n + (-w1) = 0 or n = w1 -----------(2) from (1) and (2) Fx = w2 + (-k* w1) w2 = w * sin and w1 = w * cos or Fx = w * sin + (-k* w *cos) (a)the maximum frictional force that the ground can exert onthe ladder at its lower end fk = k* n = k* w* cos k = 0.40 w = 135 N cos = (3.0/5.0) (b)the actual frictional force when the man has climbed 1.0 malong the ladder fk = k* n n = (w1 + w') = (w * cos + w') or fk = k* (w * cos +w') w' = 720 N (c)when the ladder starts to slip w = w1 + w' or W = W1 where W = w * S and W1 = (w1 + w') * S' or w * S = (w1 + w') * S' or S' = [(w * S)/(w1 + w')] = [(w * S)/(w * cos +w')] S = 5.0 m and S' gives the distance where the ladder starts toslip

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