# Homework - Calculating normal force with additional accelerations (besides gravity)

I'm working on a problem, and I'm getting too muddled in trying to figure what is the normal force.

If the whole thing was at rest, the normal force would simply be the angled reaction for $mg$, but since an acceleration is acting on the system, I'm thinking it will "relieve" some of the normal force (i.e. act against the horizontal $mg$ force to lessen the amount that the normal force has to accommodate for [if that makes sense?]). But I think I'm getting confused because I have an acceleration to work with, not a force.

I finally cracked and peeked in the back of the book. Using the value it gives for $O_y$, it seems that the normal force should be 438.357N. This is assuming that $\Sigma F_y = 0 = -mg + N cos\theta + O_y$. But if I take the sum of the moments about O (which will eliminate $O_y$, $O_x$, and I'm thinking any force related to $a$?), I get that $2 mg cos \theta= N$, so then $N =$ 588.6N.

So I guess I'm officially confused... how does that normal force work with that acceleration at $O$?

• the problem didn't say there is gravity.. – philip_0008 Dec 4 '16 at 1:49
• Hmm, is this one of those options where static equilibrium removes the $mg$ factor? I hadn't thought of that, but I still don't know how to cope with the $a$ to the right. – Asinine Dec 4 '16 at 1:50
• perhaps treat it as if $a$ is $g$ (or $-g$) – philip_0008 Dec 4 '16 at 1:52
• This is not so difficult, think of 'a' acting like 'g' but in the -x- axis.So there's a 'weight, of ma in the horizontal direction. – user98038 Dec 4 '16 at 8:09

Use an accelerating frame of reference which is moving along with the apparatus. Introduce pseudo-force $ma$ acting on the CM of the rod, horizontally to the left, where $m$ is the mass of the rod. The force of gravity $mg$ is still acting vertically in the accelerating frame.