Wrecking ball torque question

[Rijad Hadzic]

Homework Statement

At one point during its swing, a wrecking ball exerts a tension force of 9800 N on its cable, which makes an angle of 30 degrees with the horizontal. The crane's 9 m long boom is at an angle of 55 degrees with the horizontal. What is the torque exerted by the wrecking ball on the crane about an axis perpendicular to the page and passing through the point P shown in picture

here is a pic of the problem:
http://imgur.com/a/c4B0X

Homework Equations

Torque = r * applied force

The Attempt at a Solution

Well.. I have some questions first with this one.

Am I finding the torque at the point the cable meets the crane, or at point P? I guess I really don't know much about cranes at all I am not sure how they work so this adds to the confusion..

Also how am I suppose to find the length of the cable??

 

[TomHart]

Am I finding the torque at the point the cable meets the crane, or at point P?

The problem statement says point P.

Also how am I suppose to find the length of the cable??

I don't think it's necessary if you are just finding the torque from the wrecking ball around point P.

 

[Rijad Hadzic]

The problem statement says point P.

I don't think it's necessary if you are just finding the torque from the wrecking ball around point P.

So do I just multiply r (9 meters) * the force 9800 N?

the components doesn't matter right?

 

[TomHart]

You have to take into account the angle of the force relative to the boom.

 

[Rijad Hadzic]

You have to take into account the angle of the force relative to the boom.

I see that the angle of the force is 5 degrees off from the ideal 90 degrees. But how do I know which component to use now? The x component makes sense to me but I'm not sure..

 

[TomHart]

I'm not sure how you are defining your x direction. But you should use the component of the 9800 N force that is perpendicular to the boom.

Edit: Look at force F2 in figure P12.45. Let's assume that force F2 is at an angle 30 degrees below horizontal. The component of the force that is perpendicular to that rod would be (F2)sin30.

 

[Rijad Hadzic]

I'm not sure how you are defining your x direction. But you should use the component of the 9800 N force that is perpendicular to the boom.

Is 9800sin(30) * 9 m the same answer you got?

 

[TomHart]

30 degrees is the angle that the force makes with the horizontal, not the angle that the force makes with the boom.

Edit: What angle does the force make with the boom?

 

[Rijad Hadzic]

30 degrees is the angle that the force makes with the horizontal, not the angle that the force makes with the boom.

Edit: What angle does the force make with the boom?

85 degrees.

so I would use the component parallel with the boom, so my answer would be

9800sin(85) * 9 m , correct?

 

[TomHart]

85 degrees.

so I would use the component parallel with the boom, so my answer would be

9800sin(85) * 9 m , correct?

85 degrees is right. That is the angle between the boom and the force. And I believe 9800(sin85)(9) is the correct answer. But that is the perpendicular component, not the parallel component.