What is the tension in the cable?

[huybinhs]

Homework Statement

Problem's pic:
http://i92.photobucket.com/albums/l18/bonbons06/steelBeam01.png

Steel construction beams, with an industry designation of W12 × 22, have a weight of 22 lb per foot. A new business in town has hired you to place its sign on a 4.0 m long steel beam of this type. The design calls for a beam to extend outward horizontally from the front brick wall (see figure). It is to be held in place by a 5.0 m long steel cable. The cable is attached to one end of the beam and to the wall above the point where the beam is in contact with the wall. During an initial stage of construction, the beam is not to be bolted to the wall, but to be held in place solely by friction.

What is the tension in the cable? (lb)


2. The attempt at a solution

The beam weighs 22 lb./ ft * 13..115 ft = 288.5 lbs
The friction force between wall and the beam supports 144.25 lbs

The cable, the beam, and the wall make a 3-4-5 right triangle.
The tension of the cable pulls at angle with a sin = 3/5 and cos = 4/5.
The bottom right end of the cable supports ½ the weight of the beam.
The vertical component of the Tension must support 144.25 lbs

Tension * sin θ = vertical component of the Tension
Tension * sin θ = = Tension vertical
Tension * 3/5 = 144.25
Tension = 240.4166 lbs => INCORRECT. What's wrong here?

Please advise! Thanks!

 

[rl.bhat]

The frictional force between wall and the beam supports 144.25 lbs
How did you get this value?
In the equilibrium condition
Vertical component of the tension + frictional force = Weight of the beam.

 

[huybinhs]

The frictional force between wall and the beam supports 144.25 lbs
How did you get this value?
In the equilibrium condition
Vertical component of the tension + frictional force = Weight of the beam.

The cable supports ½ the weight of the beam and the wall supports the other half.

 

[huybinhs]

The frictional force between wall and the beam supports 144.25 lbs
How did you get this value?
In the equilibrium condition
Vertical component of the tension + frictional force = Weight of the beam.

Weight of the beam = (22 * 4) / 0.304 = 289.47 lbs.

 

[huybinhs]

Still getting the same answer and it's incorrect :(

 

[rl.bhat]

Frictional force = μ*Fn = μ*Tcosθ.
What is the μ value?

 

[PhanthomJay]

I think sometimes the problems try to get cute and mix up SI and USA units intentionally, while other times, it may be unintentional. Try assuming the beam length is 4 feet and the cable length is 5 feet, and see what you get.