Rotation Dynamics: Finding Tension and Force Components in a Hanging Beam System

[shane1]

A 1220N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960N crate hangs from the far end of the beam. Using the data shown in the drawing, find (A) the magnitude of the tension in the wire and (B) the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam.
http://img479.imageshack.us/img479/1285/physics5ie.png
There is the picture that I have drawn. I have no idea where to start on this question. Any help would be greatly apreciated.
Shane

 

[Doc Al]

Start by identifying the forces acting on the beam. The beam is in equilibrium: What does that tell you about the forces?

 

[shane1]

If forces are in equilibrium then, the sum of all forces must equal 0. Also would I need to deal with the business of summing torques seeing as they are in equilibrium. If I did where would I get lever arm values?

 

[Doc Al]

If forces are in equilibrium then, the sum of all forces must equal 0. Also would I need to deal with the business of summing torques seeing as they are in equilibrium.

Right!

If I did where would I get lever arm values?

Maybe you don't need them.

 

[shane1]

Also do you guys think that when it says "the wire" is it reffering to the bottom part? The top part? or the whole thing?

 

[lightgrav]

By now they figure you know the tension in the BOTTOM wire ... "trivial".
It's the tension in the TOP wire that is the challenge (this chapter).

The way you get the lever-arm values is to call the beam length "L".

My suggestion: Sum Torques around the place with the most unknown F's
(so they're multiplied by zero lever-arm) ... like the hinge at the wall.

 

[Paul_Bunyan]

Hi guys,

Sorry for resurrecting a super old post, but ironically enough I have some more questions about this exact same problem...

So, if I give the beam a length value "L," I can find the torque caused by the crate force with 1906*L*Cos(30), is that right?

But then I'm stuck on how to find the other torque value and how to get the other lever-arm perpendicular to the force.

Maybe it's just late and I'm not thinking clearly, but any advice would be greatly appreciated!

-Paul