1. ## Statics problem

The uniform ladder in Fig. 2 weighs 0.6kN. If
x = 4m, how large must the coefficient of friction against the ground be for the ladder not to slip down? If the
coefficient of friction is m = 0.5, show that the ladder will only fall when the man climbs up to x = 2.8m or a height
of 3.5m above the ground!

Now I think I'm correct in saying, at A, there's a vertical reaction pointing upwards and a horizontal reaction pointing to the right? At B there's a horizontal reaction pointing left? I worked out the reactions and got FAy=1.6Kn, FAx=1.04Kn, FBx=1.04Kn. Just got stuck at the last part where its asking for the coefficient of friction. Could anyone explain how to answer the last part of the question? Cheers.

2. Hello Haris
Originally Posted by Haris
The uniform ladder in Fig. 2 weighs 0.6kN. If
x = 4m, how large must the coefficient of friction against the ground be for the ladder not to slip down? If the
coefficient of friction is m = 0.5, show that the ladder will only fall when the man climbs up to x = 2.8m or a height
of 3.5m above the ground!

Now I think I'm correct in saying, at A, there's a vertical reaction pointing upwards and a horizontal reaction pointing to the right? At B there's a horizontal reaction pointing left? I worked out the reactions and got FAy=1.6Kn, FAx=1.04Kn, FBx=1.04Kn. Just got stuck at the last part where its asking for the coefficient of friction. Could anyone explain how to answer the last part of the question? Cheers.
I take it you're referring to the last part of the first question, which asks for the coefficient of friction when $x = 4$; i.e. when the man is at the top of the ladder.

I agree with your results, which are obtained by resolving horizontally, vertically and taking moments (about A, for instance). Then you simply say that if the ladder is on the point of slipping at A, then

$F_{Ax} = \mu F_{Ay}$

$\Rightarrow \mu = \frac{1.04}{1.6}=0.65$

3. I understand the first part of the last question. Could you please explain how to calculate the last part now? Cheers.

4. Hello Haris
Originally Posted by Haris
I understand the first part of the last question. Could you please explain how to calculate the last part now? Cheers.
In the final part, $\mu = 0.5$ and the ladder is on the point of slipping. So, now we get:

Resolve vertically: $F_{Ay}=1.6$ (as in part 1)

Resolve horizontally: $F_{Ax} = F_{Bx}$ (also as in part 1)

But now: $F_{Ax} = \mu F_{Ay} = 0.8 \Rightarrow F_{Bx} = 0.8$

Take moments about A: $0.6\times2+1\times x = F_{Bx}\times 5=0.8\times 5 = 4$

$\Rightarrow x = 4-1.2=2.8$

So, by similar triangles, the man's height above the ground is $\frac{2.8\times5}{4}=3.5$ m