Finding the tension of a beam?

Question

A 1250 N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A W = 1980 N crate hangs from the far end of the beam.

a) Calculate the magnitude of the tension in the wire.


(b) Calculate the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam.
Fx =
Fy =

Here is a drawing
http://www.webassign.net/CJ/p9-66alt.gif

there are so much information in this problem that I don't know where to start. Any help would be greatly appreciated.

Thank You.

Answer 1

All right, don't panic! Just take a deep breath and start by drawing a free body diagram. FBDs are your friend, and they should be your first step in any statics problem. You draw a simplified version of the system, and then you draw all of the forces acting on it. In this case, the simplified system is the beam. The forces are as follows:

Tx is the x-component of tension in the wire.
Ty is the y-component of tension in the wire.
Fx is the x-component of the reaction force at the wall.
Fy is the y-component of the reaction force at the wall.
1250 N is the gravitational force acting downwards on the beam at its center of gravity, due to its own weight.
1980 N is the gravitational force acting downards on the beam at its far end, due to the weight of the crate.

At first it looks like you have four unknowns. However, you actually only have three: rewrite Tx as T*cos(theta) and Ty as T*sin(theta), where theta is 50 degrees.

You can now solve, because you have three unknowns (T, Fx, and Fy) and three equations: summation of forces in the x-direction equal to zero, summation of forces in the y-direction equal to zero, and summation of moments equal to zero. Start by assigning a coordinate system. I recommend +x to the right, +y upwards, and +moment in a counter-clockwise direction.

Now do the moment balance (summation of moments (or torques) is equal to zero) about the point at which the beam contacts the wall. We choose this point because Fx and Fy will have a moment arm of zero, meaning that the only unknown in the equation will be T. The length of the beam is not given, because it ends up not affecting the answer. Call it L if you like; it will just cancel out in the end.

Once you know T, you can make a note of the values of Tx and Ty. Then you finish by doing the force balances (summation of forces is equal to zero) in the x- and y-directions. Now that you know Tx and Ty, the only unknown in the x-direction is Fx and the only unknown in the y-direction is Fy.