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## Homework Statement

The question is:

A uniform ladder of mass of 50kg is palced with its foot on a horizontal floor and its top against a vertical wall. Determine the minimum value of themass which must be placed against the foot of the ladder to retain it at an angle of 45degrees to the floor if the coefficient of friction between all relevant surfaces is 0.2.

## Homework Equations

Fs=μN

g=9.81

ladder length = L, and eighter side of mass L/2( in equations)

Joint A=ladder contact with wall

Joint B=ladder contact with floor(supported by mass)

## The Attempt at a Solution

I thought the best way was to split the ladder from the mass keeping it upright when summing the forces...

Joint A

∑fy=0=P+Fs-50.g.(L/2)cos45

∑fx=0=N-50.g(sin45)=o......N= 347N

Fs=μN=0.2x347=69.4

then i decided to take moments

M.A=0=50.g.L/2.cos45

Joint B

∑fy=o=N-50.g.sin45=o

N=347N

∑fx=0= -Fs-P-50.g.cos45=0

P= -Fs-347=-416.4N

I dont know where to go with it next as i cnnot get the answer that was provided of m=58.4kg

Any help would be much appriciated. Thanks.