# Solid rigid equilibrium problem

1. Nov 9, 2009

### rAz:DD

1. The problem statement, all variables and given/known data
Hi. Given a homogeneous bar of gravity G,the length AB = 2L, resting without friction on
the two inclined planes with the angles alpha and beta to the horizontal, as shown in the following figure:
http://img44.imageshack.us/i/pic01ns.jpg/

Is is demanded:
1) The tetha angle for the equilibrium
2) The two reactions from the rests A and B (NA and NB)

2. Relevant equations
Sum Fix=0;
Sum Fiy=0;
Sum MiO=0;

3. The attempt at a solution
I've chosen X axis the line containing the bar, and the Y axis the line perpendicular to the line; O(0,0) represents the intersection of the axis in the middle of the bar.NA and NB are both on the Y axis, the moment (torque) of G is 0 , moment of NA and NB are both (2L/2)*force.

I can't identify the angles for the projections of G on the two axis. I don't need the full solution, just the projections of G, and , maybe, a hint if i did something wrong before this step.

2. Nov 9, 2009

### semc

Hello rAz, if i were to do this question, i would draw all the forces then express in the x and y direction. Take torque about either end and equate $$\Sigma$$$$\tau$$=F.d

3. Nov 9, 2009

### rAz:DD

Hi semc
How do i express G on the X Y axis (where x is the bar axis) ? I've got that tetha angle, but i can't find a right triangle there. I tried to consider the Y axis having the same direction as G, but it's the same problem with the forces of reaction from A and B.

4. Nov 10, 2009

### rAz:DD

5. Nov 10, 2009

### ehild

Extend the line of the rod till it intersects the horizontal. The angle it makes with the horizontal is equal to the angle between G and the normal of the rod (just like in case of a slope).

ehild

6. Nov 10, 2009

### rAz:DD

Thanks!
http://img23.imageshack.us/img23/4251/81187915.jpg [Broken]

I see a triangle with the first angle tetha, the second one 2*pi-alpha, therefor the needed angle is: 2*pi-tetha-(2*pi-alpha) = alpha-tetha.

Still, i belive this answer is wrong, because Gx and Gy should depend on the beta angle too. And the tetha angle depends on both alpha and beta. Im i doing something wrong?

Last edited by a moderator: May 4, 2017
7. Nov 10, 2009

### rl.bhat

The normal reactions marked at A and B are not correct. The normal reaction must be perpendicular to point of contact on the surface by the rod.
Suppose the angle made by rod at A less than θ, which force at the other end of the rod on the surface B pulls it to the equilibrium position?
Resolve Gx at B. One along the surface B and the other perpendicular to the surface B.
See whether it helps you to find θ.

8. Nov 10, 2009

### ehild

"I see a triangle with the first angle tetha, the second one 2*pi-alpha, "

well, it is pi-alpha...

"the needed angle is: 2*pi-tetha-(2*pi-alpha) = alpha-tetha."

the sum of angles in a triangle is pi...

But otherwise, the needed angle is alpha -theta.

"Still, i belive this answer is wrong, because Gx and Gy should depend on the beta angle too. "

Nevermind. The three angles are not independent.

But there is some other trouble, as rl.bhat noted. The directions of the reaction forces NA and NB do not look right from your drawing.

ehild