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## Homework Statement

A uniform ladder of mass M and length L stands on rough horizontal ground and leans against a vertical wall with which it makes an angle θ. A man of mass m stands on the ladder a distance x from the bottom.

a) obtain expressions for the normal reaction and the friction force acting at the point of contact between the ladder and the ground.

## Homework Equations

The normal is obviously (m+M)g.

Other than that, it's static equilibrium, and the equation I used for balancing was the moments, where τ

_{NW}is the moment due to the normal F

_{NW}to the wall at the top of the ladder, and τ

_{Mg}and τ

_{mg}are the two moments of the ladder and man:

1. τ

_{NW}= τ

_{Mg}+ τ

_{mg}

## The Attempt at a Solution

Pretty simple. I have the ladder to be length L, so I set the origin of turning at the base of the ladder. I want to find F

_{NW}because it's equal to the frictional force F

_{F}at the base of the ladder.

The radius component for the moment τ

_{NW}is this Lcosθ. The moment is thus: F

_{NW}Lcosθ.

The radius component for τ

_{Mg}is (L/2)sinθ, and for τ

_{mg}is (x/L)sinθ.

So the two moments are: [(MgL)/2]sinθ and [(mgx)/L]sinθ

Equation 1. then becomes:

F

_{NW}Lcosθ = [(MgL)/2]sinθ + [(mgx)/L]sinθ

Which evidently becomes, solving for F

_{NW}:

F

_{NW}= (M/2 + mx/L

^{2})gtanθ

In the given answer, the mx/L

^{2}term is simply mx/L. But the position of the man along the ladder was given as x from the bottom, so this should be written x/L, right? Which means dividing out the L on the F

_{NW}side will always create a squared term.

Any insight appreciated.

Jack