Solid rigid equilibrium problem

[rAz:DD]

Homework Statement

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)

Homework Equations

Sum Fix=0;
Sum Fiy=0;
Sum MiO=0;

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.

Looking forward for any reply.

 

[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

 

[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.

 

[rAz:DD]

Can someone please help me?

 

[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

 

[rAz:DD]

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

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 believe 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. I am i doing something wrong?

 

[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.

 

[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 believe 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 N_A and N_B do not look right from your drawing.

ehild