Question

A ladder is resting with one end on a vertical wall and the other end on a horizontal floor. Is it more likely to slip when a man stands near the bottom or near the top?

Answer 1

The ladder tends to slip more when the man stands near the wall. 

Let us first draw a diagram to understand the problem. AB is the ladder with end A resting on the ground and end B against the wall. OA = d, OB = h, and W is the weight of the man on the ladder at distance x from end A horizontally. R and R' are normal forces at the ends A and B also F and F' are friction forces at A and B.

Assuming µ same at both contact surfaces, 

F =µR, F' = µR' 

Equating forces vertically, 

R+F' =W 

R+µR' =W 

Equating forces horizontally 

F = R' 

Taking Moments of all forces about B, we have 

W(d-x)+Fh-Rd = 0 →W(d-x)+µRh-Rd = 0 

→W(d-x) = R(d-µh) 

→R=W(d-x)/(d-µh) 

Overturning Torque (anticlockwise) about A

 = Wx 

Restoring torque (Clockwise) 

= R'h+F'd 

So net overturning torque T 

=Wx-(R'h+F'd) 

=Wx-(Fh+µR'd) 

=Wx-(Fh+µFd) 

=Wx-F(h+µd) 

=Wx-µR(h+µd)

 =Wx-µW(d-x)(h+µd)/(d-µh) 

={Wxd-Wxµh-µW(dh+µd²-xh-µxd)}/(d-µh) 

={Wxd-Wxµh-µWdh+µ²d²W+µxhW+µ²Wxd)}/(d-µh) 

={Wxd(1+µ²)+Wµd(µd-h)}/(d-µh) 

In this expression all except x are constant and we can see that as x increases T also increases. 

So, the ladder tends to slip more when the man stands near the wall.