008 (part 1 of 3) 10.0 points A ladder with a length of 12.0 m and weight of 515

Question

008 (part 1 of 3) 10.0 points A ladder with a length of 12.0 m and weight of 515.0 N rests against a frictionless wall, making an angle of 60.0 with the horizontal. Find the horizontal force exerted on the base of the ladder by Earth when a firefighter weighing 780.0 N is 3.15 m from the bottom of the ladder. Answer in units of N. 009 (part 2 of 3) 10.0 points Find the vertical force exerted on the base of the ladder by Earth. Answer in units of N. 010 (part 3 of 3) 10.0 points If the ladder is just on the verge of slipping when the firefighter is 7.20 m up, what is the coefficient of static friction between the ladder and the ground?

009 (part 2 of 3) 10.0 points Find the vertical force exerted on the base of the ladder by Earth. Answer in units of N.

010 (part 3 of 3) 10.0 points If the ladder is just on the verge of slipping when the firefighter is 7.20 m up, what is the coefficient of static friction between the ladder and the ground?

Explanation / Answer

Since the ladder is uniform, its center of gravity is right in the middle, 6 m from either end. The first condition for equilibrium gives:
nwall - Fx = 0
Fy - 515 N - 780 N = 0 or Fy = 1295 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:
(515 N)(6m)cos60° + (780 N)(3.15m)cos60° - nwall(12 m)sin60° = 0
The last equation gives
nwall = (1545+1228.5) / 12 sin 60= 266.88
Direction
towards the wall
Thus Fx = nwall =266.88 N, and Fy = 1295 N

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