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?

  • $\begingroup$ 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. $\endgroup$ – 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.

Also, I can't understand which summation of moments you took. I can't see where the 7 came in the third term, though I have to admit that I haven't tried solving the equation. If you're conserving torque for a single rod, take note that the rods exert a normal force on each other at point B, of unknown direction.

Aside from that the problem is in FPS and you're solving it in SI units without conversion.

  • $\begingroup$ try taking sumation of forces hrizontally (whole body). ull see tht both normal forces are equal to eac other. this in turn makes both friction equal. (frictions should reach max values and obey f = uN since the whole body is in tht state) $\endgroup$ – WantIt Feb 26 '12 at 13:34
  • $\begingroup$ 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. $\endgroup$ – Manishearth Feb 26 '12 at 13:37
  • $\begingroup$ but, why is that if you look at the figure, then take sumation of forces horizontally, it turns out that only Na and Nc are the horizontal forces. so we have Na - Nc = 0, thus Na = Nc, correct me if im wrong (the two normals are equal) $\endgroup$ – WantIt Feb 26 '12 at 13:41
  • $\begingroup$ The normals are equal. Friction is not. MAXIMUM friction is $\mu_sN$. It can be less as well. $\endgroup$ – Manishearth Feb 26 '12 at 13:43
  • $\begingroup$ lets bring this into discussion chat.stackexchange.com/rooms/2613/statics tnx $\endgroup$ – WantIt Feb 26 '12 at 14:14

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