A ladder with a length of 14.1 m and weight of 525.0 N rests against a frictionl

Question

A ladder with a length of 14.1 m and weight
of 525.0 N rests against a frictionless wall,
making an angle of 59.0 with the horizontal.
Find the horizontal force exerted on the
base of the ladder by Earth when a firefighter
weighing 854.0 N is 3.37 m from the bottom
of the ladder.

Answer in units of N

I cannot seem to figure out this one. I attempted to solve this problem as follows
Force= (525N*(14.1*cos59)+854N*(3.37m/2)) / (14.1*sin59)
The answer I received was 276.7878836N, but apparent;y that is incorrect. Any help would be appreciated.

Explanation / Answer

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: (525 N)(7.05m)cos59° + (854 N)(3.337m)cos59° - nwall(14.1 m)sin59° = 0 The last equation gives nwall = [(525 N)(7.05m)cos57° + (854 N)(3.37m)cos57°]/(14.1 m)sin57° = 303.0214 N

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