# Find the forces exerted by the two cables

1. Nov 24, 2006

### petal5

I'd appreciate help with this problem as I don't know how to start it:

A beam with a mass of 1500 kg and a length of 2.0 m hangs from two cables,
such that it makes an angle of 65degrees with the vertical. Both of the cables are completely
vertical; one is attached at the lower end of the beam, while the other is attached
0.40 m from the higher end of the beam. (a) Find the forces exerted by the two
cables.

2. Nov 24, 2006

### Hootenanny

Staff Emeritus
Have you any thoughts on the problem?

3. Nov 24, 2006

### petal5

I'm thinking F1+F2=Mg but I'm not sure

4. Nov 24, 2006

### Hootenanny

Staff Emeritus
That is true, and it will be useful later in the problem. How about taking moments about the lower end of the beam?

5. Nov 24, 2006

### petal5

Sorry but what do you mean by 'talking moments'?

6. Nov 24, 2006

### Hootenanny

Staff Emeritus
Moments is equivalent to torques. Examine the torques on the beam using the lower end as the axis of rotation.

7. Nov 24, 2006

### petal5

Thanks for your help,I really appreciate it!
Do you mean to work out the following: T1=r1f1sinx , which would be T1=(0.4m)(F1)(sin25)

8. Nov 25, 2006

### Hootenanny

Staff Emeritus
You have the right idea, perhaps you have the benefit of the diagram, but from the text I disagree with your angle and distance. If you do have a diagram, would it be possible for you to post it on Imageshack or something similar?

9. Nov 25, 2006

### petal5

I've had another look at it and the angle and distance I posted above are incorrect.I've now come up with T1=r1f1sin65, T1=(0m)(F1)(0.9063).This is with using the lower end to take torques.However,I'm a bit confused as F1 turns out to be 0.
For T2 I've gotten (1.6m)(F2)(0.9063) therefore T2= -(1.45)F2
(I don't have a diagram by the way!)

10. Nov 25, 2006

### Hootenanny

Staff Emeritus
Of course T1 should be zero since you are taking torques about this point, this is your pivot point. Your on the right track but not there yet, you need to consider both the torque due to the tension and the weight of the beam. Thus;

$$T_{2}\cdot (1.6)\sin(65) - (1500\cdot g)\cdot (1)\sin(65) = 0$$

Do you understand why?

11. Nov 25, 2006

### petal5

In my last post I said that F1=0 and then you replied that T1=0.Are both F1 and T1 equal to 0?
Is the formula you mentioned above the formula for F2?

12. Nov 25, 2006

### Hootenanny

Staff Emeritus
That was poor practise from me, I apologise. You are taking torques about the base the beam, yes? Therefore, the tension is being applied through the axis of rotation, yes? Hence, the distance from the axis rotation to the applied force is zero. Does that make more sense?
Yes, in words the above equation is "the magnitude of the torque exerted by the tension in the second string minus the magnitude torque exerted by the weight of the beam must be zero.

Do you follow?

13. Nov 25, 2006

### petal5

14. Nov 25, 2006

### Hootenanny

Staff Emeritus
You do not need to calculate F2, you need to calculate T2, the equation for which I posted in post #10. Can you solve that for T2?

15. Nov 25, 2006

### petal5

But part (a) of the problem is: Find the forces exerted by the two cables.Then part (b) is:What are the torques exerted by each of the two cables and by the centre of mass of the beam about a pivot point at then lower endc of the beam.So I seem to need F2??

Last edited: Nov 25, 2006
16. Nov 25, 2006

### Hootenanny

Staff Emeritus
The force exerted by the cable is the tension in the cable i.e. T2

17. Nov 25, 2006

### petal5

Thanks again for all your help.Filling in for the equation in post #10 I get T2= 13321.274

18. Nov 25, 2006

### Hootenanny

Staff Emeritus
Now, you can use your equation;
To calculate T1. (I'm not checking your arithmetic though )
My pleasure.