# Find the forces exerted by the two cables (1 Viewer)

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#### 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.

#### Hootenanny

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Have you any thoughts on the problem?

#### petal5

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

#### Hootenanny

Staff Emeritus
Gold Member
I'm thinking F1+F2=Mg but I'm not sure
That is true, and it will be useful later in the problem. How about taking moments about the lower end of the beam?

#### petal5

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

#### Hootenanny

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Gold Member
Sorry but what do you mean by 'talking moments'?
Moments is equivalent to torques. Examine the torques on the beam using the lower end as the axis of rotation.

#### 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)

#### Hootenanny

Staff Emeritus
Gold Member
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)
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?

#### 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!)

#### Hootenanny

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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?

#### 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?

#### Hootenanny

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

#### petal5

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?

Ye,the above makes sense to me.However,I am still unsure as to how to work out F2

#### Hootenanny

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Gold Member
Ye,the above makes sense to me.However,I am still unsure as to how to work out F2
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?

#### 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:

#### Hootenanny

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

#### petal5

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

#### Hootenanny

Staff Emeritus
Gold Member
Now, you can use your equation;
I'm thinking F1+F2=Mg
To calculate T1. (I'm not checking your arithmetic though )
petal5 said:
Thanks again for all your help
My pleasure.

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