Solve for Tension in Equilibrium: Frictionless Rod with 50lb Load at x=4.5

[baird.lindsay]

Homework Statement

This is a question from statics course.
Collar A is connected to a 50lb load on a frictionless horizontal rod. Determine magnitude of P to maintain equilibrium when x=4.5.

Im confused on the concept of Tension

I originally got the answer 11.25 lb like the images below but my solution manual is different.
The solution manual says:

tan alpha= 20/4.5 = 77.3 degrees

sum of F sub X=0
-P + T cos 77.3
P= 50lb (cos 77.3)
= 10.98lb (answer manual)

Is this because the tension in the rope is equivalent to the mass of the hanging weight? Is there a way to solve for 10.98lb using method like the one below ...i.e. adding the vector components and solving for the unknowns? confused. Thanks!

but this one below is what makes sense to me.
http://imageshack.us/photo/my-images/15/yci9.png/

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http://imageshack.us/photo/my-images/534/y53e.png/

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Homework Equations

sum of x =0
tan=o/a

The Attempt at a Solution

 

[SteamKing]

Your preferred method is correct until you reach the statement:

sin 77.32 / 50 = sin 12.68 / P ---> incorrect statement

If you draw a proper free body diagram of the collar, you will see that the tension in line AB = 50 lb. You know from trigonometry that the horizontal force P must equal 50 * cos a, where the angle 'a' is between line AB and the horizontal, which is 77.32 degrees, when x = 4.5 in.

 

[baird.lindsay]

Your preferred method is correct until you reach the statement:

sin 77.32 / 50 = sin 12.68 / P ---> incorrect statement

If you draw a proper free body diagram of the collar, you will see that the tension in line AB = 50 lb. You know from trigonometry that the horizontal force P must equal 50 * cos a, where the angle 'a' is between line AB and the horizontal, which is 77.32 degrees, when x = 4.5 in.

I am confused because when I draw a free body diagram. I draw it through the center of the collar and move the weight vector to the negative y axis. I have the tension force in the first quadrant , I have the P force on the -x axis , and I have the weight force on the negative y axis.

That doesn't look right because the forces don't add up.. I get T sub x cos 77.32-P=0 and T sub y sin 77.32 - weight =0

 

[SteamKing]

If B is a frictionless sheave, then the tension in the line must be the same on both sides of the sheave. Since the weight C of 50 lbs is static, then the tension in line AB must also be 50 lbs. Once this fact is established, then the force P can be worked out using trig, which is determined by the distances given in the setup.