# Boom problem help!

1. Dec 15, 2003

### formulajoe

a pole 8 m long with a mass of 10 kg, pivots at point A. there is a ball weighing 10 kg hanging from the end of the boom. the boom makes a 30 deg angle with the upper portion of the mast, which has a cable going to boom 4 m above point A. im supposed to calculate the tension in the cable and the total force exerted by the pivot at point A.
i have absolutely no idea what i am supposed to do here.

2. Dec 15, 2003

### StephenPrivitera

I don't understand your description of the problem. If you can make it more clear I can help.

3. Dec 16, 2003

### formulajoe

a boom is connected to a mast. a cable is connected to the boom and the mast. the cable is located 4 m above the pivot point where the boom is connected to the mast. im supposed to calculate the tension in the cable. it doesnt really say how the boom pivots.

4. Dec 16, 2003

### gnome

Is A the point where the boom is connected to the mast?

And is A also the pivot point?

5. Dec 16, 2003

### formulajoe

yes, and the boom makes a 30 deg angle with the cable.

6. Dec 16, 2003

### gnome

then you have to set up 3 equations & solve simultaneously:

one equation for the torques being applied to the beam. You can use the pivot point as the axis for the torques, or you can pick any other point on the beam & figure out the torques around that point. What's important is, since the beam is in equilibrium, the sum of the torques about any axis of rotation must = 0

another equation for the vertical components of the forces

another equation for the horizontal components of the forces

again, since the beam is in equilibrium, the sum of the vertical components must be 0 and the sum of the horizontal components must be 0

7. Dec 16, 2003

### formulajoe

how would i find the acceleration for the force?
edit: and you said torques? i only see 1 torque.

8. Dec 16, 2003

### gnome

What acceleration? This thing is in equilibrium -- therefore no acceleration.

Only 1 torque? If there were only 1 torque, the thing would be rotating (and accelerating), wouldn't it?

9. Dec 16, 2003

### formulajoe

force in x direction = mass of boom plus mass of ball * g * cos 120 right?
that would be the tension on the cable.
i see 1 torque at point a, where is the other torque, where the cable connects to the boom?

10. Dec 16, 2003

### formulajoe

tension in cable = 1/2 * 10 + 10/sin 30

11. Dec 16, 2003

### formulajoe

i worked it out and im pretty sure ive got an answer, but i got two and im not real sure on which one is correct. sum of torques =
Fx*sin theta*L - W1(L/2) - W2*L. solve for Fx and i get a tension in the cable of 30 N.
Fy = (w1+w2)*g which ends up around 200 N.
but with this , i get a totaly force at point a of tan 30(30/200) which isnt right. what am i missing?

12. Dec 16, 2003

### gnome

If I understand you correctly, the cable connects to the MAST at a point 4 m above A.

and there are TWO 30-degree angles, one where the cable meets the mast, and one where the cable meets the boom.

So we have to figure out where the cable connects to the boom?

Or am I wrong about the angles and are you given where the cable connects to the boom?

13. Dec 16, 2003

### gnome

Notice: if there is a 30-degree angle where the boom meets the mast AND a 30-degree angle where the cable meets the boom, then the cable doesn't attach at the end of the boom, right?

Then the distance from A to the attachment point is 8cos30 so you will have to use that as the hypotenuse in order to figure out the moment arm of the torque applied by the tension.

OK?

14. Dec 16, 2003

### formulajoe

30 deg angle between boom and cable. 60 deg angle between boom and mast. it looks kinda like this
|---
| /
| /
|/
not quite to right, but the / represents the boom, the | represent the mast and the - represents the cable.

15. Dec 16, 2003

### gnome

Then they made it real easy for you. The direction of the tension is exactly perpendicular to the wall, right? So, now, if the tension is T, how big is the torque applied by the tension, as a function of T?

16. Dec 16, 2003

### formulajoe

the torque would be T * 8 *sin 60

17. Dec 16, 2003

### gnome

NO!!

You said that the cable is attached to the wall 4 m above A, and the cable is perpendicular to the wall. So the moment arm of T about A is just 4 m.

That's it for me. I have a final tomorrow. Good luck.

18. Dec 17, 2003

### formulajoe

where does the other torque come from and what are the forces in each of the torques?

19. Dec 17, 2003

### gnome

The cable provides the counterclockwise torque.

The weight of the boom provides the clockwise torque.

Now, can you solve for T?

(After you do that, you still have some more work to do. You have to figure out the force acting at point A.)