A ladder of uniform construction stands on the floor and leans against a frictio

Question

A ladder of uniform construction stands on the floor and leans against a frictionless wall. The ladder has a length of 3.04 meters, a mass of 13.3 kg and it makes an angle of 24.8 degrees with the vertical wall. A student of mass 62.0 kg stands on a rung of the ladder which is 1.43 meters along the ladder measured from where the ladder touches the ground. The frictional force that acts at the bottom of the ladder to keep it from slipping is N

In the previous problem, the angle that the reaction force of the ground on the ladder makes is degrees with the horizontal.

Explanation / Answer

we have to find frictional force = N

taking moments about point where ladder touches the verticval wall we get as follows

N*3.04* cos24.8 = 13.3*9.8**0.5*3.04*cos24.8 + 62*9.8*1.61*cos24.8

N = frictional force = 386.958 N

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