Solve Magnitudal Tension Homework: Q & ACB Tension

[ride4life]

Homework Statement

A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P=750N, determine
a) the tension in cable ACB
and
b) the magnitude of load Q

Homework Equations

Using cos and sin rules...?

The Attempt at a Solution

I drew a free body diagram around C, with Q going down, a cable going to the right 25 degrees above the horizontal and a cable going to the left 55 degrees above the horizontal.
The cable on the left i assumed to be 750N.
So the sum of the forces along the X axis is (Tension of ACB)cos25-750cos55=0 and T came out to equal 474.7N.
The sum of the forces along the Y axis is (Tension of ACB)sin25+750sin55-Q=0 and I got Q to equal 241.6N
The answers came out to be in the kiloNewtons.

I also tried another way after finding out my answers were completely off. If it is 750N down, then the cable going from C to A would equal 915.6N (CA = 750/sin55). Using 915.6 as the force for my left cable on my free body diagram, the sum of the forces along the X axis is (Tension of ACB)cos25-915.6cos55=0 and the Tension came out to be 579.5N.
The sum of the forces along the Y axis is 579.5sin25+915.6sin55-Q=0 and Q=994.9N. These answers were also wrong.

So please, can anyone direct me in the correct direction?

 

[supratim1]

i think your first attempt is correct. i see nothing wrong. don't know if i am missing something.

 

[rl.bhat]

There are two string in the problem.
i) The string ACB, one end of which is fixed to the center of the pulley A. Each segment of the string has a tension T.
ii)The string DAC, one end of which is fixed to the center of the pulley C.
Net force in the Y direction is
T*cosθ1 + T*cosθ2) - Q = 0...(1)
String DAC at rest when
Q*cosθ1 - T = P. ...(2)
Here θ1 and θ2 are the angles made by the segments AC and CB with the vertical.
Substitute the value of Q from the eq.1 in eq1.2 and solve for T. From that find Q.

 

[ride4life]

There are two string in the problem.
i) The string ACB, one end of which is fixed to the center of the pulley A. Each segment of the string has a tension T.
ii)The string DAC, one end of which is fixed to the center of the pulley C.
Net force in the Y direction is
T*cosθ1 + T*cosθ2) - Q = 0...(1)
String DAC at rest when
Q*cosθ1 - T = P. ...(2)
Here θ1 and θ2 are the angles made by the segments AC and CB with the vertical.
Substitute the value of Q from the eq.1 in eq1.2 and solve for T. From that find Q.

I understand the first equation, but instead of using cos, I had sin55 and sin25. But I don't understand your second equation...

 

[ride4life]

Sweet! I have figured it out!

 

[joshizzlemofo]

can you help i have the same problem