A 16.0-m uniform ladder weighing 520 N rests against a frictionless wall. The la

Question

A 16.0-m uniform ladder weighing 520 N rests against a frictionless wall. The ladder makes a 63.0° angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 820-N firefighter has climbed 3.70 m along the ladder from the bottom. Horizontal Force magnitude N direction Vertical Force magnitude N direction (b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?

Explanation / Answer

Since the ladder is uniform, its center of gravity is right in the middle,8 m from either end. The first condition for equilibrium gives:
nwall - Fx = 0
Fy - 520 N - 820 N = 0 or Fy = 1340 N
Direction
up


Since the wall is frictionless, there is no vertical force from the wall. The second condition for equilibrium, choosing the base of the ladder as the axis of rotation, gives:
(520 N)(3.7m)cos63° + (820 N)(3.7m)cos63° - nwall(16 m)sin63° = 0
The last equation gives
= 157 N
Direction
towards the wall
Thus Fx = nwall = 157N, and Fy = 1340 N

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