# Find largest angle

The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical walls. If the coefficient of static friction between each bar and the wall is 0.4, determine the largest angle θ for which the assembly will remain at rest.

Attempt at solution,

so i use the fbd of the whole body, and take summation of forces on x axis, and i got Na = Nc (normal at point a is equal to point c). then, i take summation of forces on y axis, so thats

Fa - 6N -8N +Fb = 0 (where fa and fb are friction forces on a and b), and since fa = fb (their normals are equal), i use f = uN and yielded Na = 17.5 N.

I then take summation of moments at point B( bar of A to B) and heres eqn

(6)(300)(cos(x)) - 17.5 (600)(cos(x)) + 7(600)(sin(x)) = 0 and

the value of angle is 64 degress. the book says 10 degress. Where did I go wrong?

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Generally we discourage questions that just ask for someone to check your work. Once you have identified the specific concept that you're not sure about, that's the point at which it's appropriate to ask a question here. – David Z Feb 26 '12 at 15:47

It's not necessary that both friction forces are $\mu_sN$. $\mu_sN$ is the maximum value of static friction. Since the situation is unsymmetric, for the limiting case for $\theta$, one of the frictions can be less that $\mu_sN$. I'd say that this is the left-hand one (I could be wrong). Take both as $f_1$,$f_2$ and try the problem.
NO, that's the point. $f_{static}\leq\mu_sN$. ONE of them has to reach max value. Let's take a non-limiting case, where $f\neq\mu_sN$ for both. Then, it is pretty obvious that $f_1\neq f_2$ as system is asymmetric. If we slowly change the angle, at that point ONE of the frictions will reach $\mu_sN$ (the limiting case). The other may still be quite less. – Manishearth Feb 26 '12 at 13:37
The normals are equal. Friction is not. MAXIMUM friction is $\mu_sN$. It can be less as well. – Manishearth Feb 26 '12 at 13:43