# Three forces in equilibrium. Uniform rod

by afrocod
Tags: equilibrium, forces, uniform
 P: 31 1. The problem statement, all variables and given/known data A uniform rod AB of weight W rests in equilibrium with the end A in contact with a smooth vertical wall and the end B in contact with a smooth plane inclined at 45 to the wall. Find the reactions at A and B in terms of W. 2. Relevant equations The sum of the three forces equal zero. 3. The attempt at a solution Well having a look at the answer is W and √2W, it's obviously a square and the rod must be perpendicular to the inclined plane. So how could I know that it was perpendicular to the plane. What indicated that or how can I prove that assumption to myself?
HW Helper
P: 3,464
 Quote by afrocod Well having a look at the answer is W and √2W, it's obviously a square and the rod must be perpendicular to the inclined plane. So how could I know that it was perpendicular to the plane. What indicated that or how can I prove that assumption to myself?
What do you mean 'its a square'? And why would you say the rod must be perpendicular to the inclined plane? Please expand on your thinking.
 P: 31 Well I just meant half the square. The right triangle with unit length and hypotenuse square root of two. So one reaction is equal in magnitude and direction to W and the other is inclined at 45 and thus the hypotenuse with (square root of 2)W. I drew a lot of diagrams to help me and the the only way this makes sense to me is if rod AB is perpendicular to the inclined plane. I know nobody has answered because I haven't shown any working out but all I've done is draw diagrams. I can't make any headway into the problem. All I can do is think about why those answers would make sense. Thanks for replying by the way. Also I'm just doing this for my own enjoyment. Nobody is doing my homework for me, if you help me out.
HW Helper
P: 3,464
Three forces in equilibrium. Uniform rod

 Quote by afrocod So one reaction is equal in magnitude and direction to W and the other is inclined at 45 and thus the hypotenuse with (square root of 2)W.
Yes, equal in magnitude, but not equal in direction. I think you meant to say perpendicular in direction.

 Quote by afrocod I drew a lot of diagrams to help me and the the only way this makes sense to me is if rod AB is perpendicular to the inclined plane.
I guess this is where you're going wrong. The wall and the plane are frictionless, so this straightaway tells us something about the direction of any contact force with them.
HW Helper
P: 3,464
 Quote by afrocod Thanks for replying by the way. Also I'm just doing this for my own enjoyment. Nobody is doing my homework for me, if you help me out.
Ah, no worries. and I get enjoyment from doing physics questions also. which is why I first came to this site. I also like helping people out on questions. But usually I see if I can do the question myself first, since then it is easier to explain.
P: 31
 Quote by BruceW I guess this is where you're going wrong. The wall and the plane are frictionless, so this straightaway tells us something about the direction of any contact force with them.
Yes, this is what I was thinking. That there is some piece of intuition given those particular facts that I don't know. I'll take a stab but I really haven't a clue.

Does it tell me that when the rod is in equilibrium on a frictionless surface that the angles between the rod and the vertical is equal to the rod and the inclined plane?
 P: 31 Oh wait, wait hang on. Does it mean that the rod is lying flat on the incline because there was no friction to keep it up? If it does, I cannot believe that was the answer. I've been looking at this thing for hours. I feel like an absolute moron.
 P: 31 That doesn't seem to work either because W would be √2 times the other length's and not one equal to W and the other √2W.
 HW Helper P: 3,464 Yes, it's neither of those things. It might be a 'piece of intuition given those facts' which you are not aware of. It is to do with the definition of contact forces and friction. In fact, it really doesn't have anything to do with the rod, it is just to do with the contact force which a surface will provide. I'm going to try to give hints without actually saying it outright. You may have used this principle before, but maybe forgotten about it. OK, so say we have a surface, and its normal points in a certain direction. If the surface is frictionless, then what direction must the contact force be in?
 P: 31 Perpendicular to the surface.
 HW Helper P: 3,464 yeep. that's it. No friction means the contact force is perpendicular to the surface.
 P: 31 Isn't it always perpendicular to the surface though?
 HW Helper P: 3,464 usually the contact force is defined to include frictional forces, so it is not always perpendicular to the surface. (since frictional forces are parallel to the surface).
 P: 31 Ah, I see the solution now. It all seems so simple in retrospect. Thanks for all the help.
 HW Helper P: 3,464 mm, no worries

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