What is the Tension in the Tower Crane Cable?

[SoylentBlue]

Homework Statement

The beam ABC is1000LBs and its center of mass is given as 6.5ft to right of B. B is a pin joint. A & D are frictionless rollers. Find tension in cable.

Homework Equations

Call the tension in the cable T.
Trig shows us that the left-hand triangle is 45 degrees at ABD and the right-hand triangle is 15.5 degrees at BCD.
Torque due to wt of load and mass of beam are clockwise and negative; torques due to T are ccw and positive.

The Attempt at a Solution

So the torque at B should be: -(6.5ft)(1000Lb) + -(12ft)(3000LB) + (18ft)(sin15.5)T + (5ft)(sin45)T=0
This yields a T of 5114; the book shows the answer as 6770.
What am I missing?

 

[mfb]

The part of the torque you calculated for (A) actually goes in the opposite direction, and there is a second part.

I can confirm the book's answer.

 

[SoylentBlue]

Thank you for helping! I am trying to work through this statics course on my own, so the Internet is, um, my professor.

So the cable from CD pulls up and to the left, yielding ccw rotation. Shouldn't the tension from D to A pull down and push to the right, also yielding a ccw rotation?

Oh...is the second part that you reference the torque from AE? This yields an additional torque of 5T.

 

[mfb]

Shouldn't the tension from D to A pull down and push to the right, also yielding a ccw rotation?

It pulls A up and to the right.

Oh...is the second part that you reference the torque from AE? This yields an additional torque of 5T.

Right.

 

[SoylentBlue]

Oh, you can look at the pulley at the top of the crane the same way you'd look at a cable draped across the top of a suspension bridge's tower; so the tension on the right opposes the tension on the left. Now I see. I just needed to brush up on Pulleys 101.

If you add in the torque from AE, then the final value comes out almost exactly to the 6770 that the book shows.

Again, thank you for helping!