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.

 

[Jhero]

Hello,

To find the equilibrium angle, you can use the principle of moments, which states that the sum of the moments about any point must be equal to zero for the body to be in equilibrium. In this case, we can choose point O as our reference point.

Let's start by analyzing the forces acting on the bar. We have the weight of the bar, W, acting downwards at its center of mass. We also have the reactions from the rests, NA and NB, acting upwards at points A and B respectively.

Now, we can take moments about point O. The weight of the bar does not create any moment about O since its line of action passes through O. Therefore, we only need to consider the moments created by NA and NB.

Let's start with the moment created by NA. We can calculate this as the force (NA) times the perpendicular distance from point O to the line of action of NA, which is L. Therefore, the moment created by NA is NL.

Similarly, the moment created by NB is NB * L.

Now, since the bar is in equilibrium, the sum of these two moments must be equal to zero. This gives us the following equation:

NL + NB * L = 0

Solving for NB, we get NB = -NL.

Now, we can also use the principle of forces to find the equilibrium angle. We can resolve all the forces acting on the bar into their components along the X and Y axes.

Along the X axis, we have the horizontal component of the weight, which is W * sin(theta), acting towards the left. We also have the horizontal component of the reaction at A, which is NA * cos(alpha), acting towards the right. Similarly, along the Y axis, we have the vertical component of the weight, which is W * cos(theta), acting downwards. We also have the vertical component of the reaction at B, which is NB * cos(beta), acting upwards.

Now, using the principle of forces, we can write the following equations:

Sum Fx = W * sin(theta) - NA * cos(alpha) = 0

Sum Fy = W * cos(theta) + NB * cos(beta) = 0

Solving for theta in the first equation, we get:

theta = sin^-1 (NA * cos(alpha) / W)

Similarly, solving for NB in the second equation and substit