# Tension and net torque in a cable

• ryley
In summary, the equation that yields the distance D is T sin(θ) where θ is the angle at which the tension is applied to the cable.
ryley

## Homework Statement

In figure 12-34, a uniform beam of weight 500N and length 3m is suspended horizontally, on the left it is hinged to the wall and on the right end of the beam is a cable attached to support it. the cable is attached distance D up the wall from the hinge. The least amount of tension that will snap the cable of 1200N.

a) What is the value of D that corresponds to that tension?

Σt net=0

## The Attempt at a Solution

I've figured out to use the torque net equation to find the angle of θ and then solve for the distance D. What I'm not understanding is why I can't just use the Fnet y=0 to solve for θ?

Fnety=Tsinθ- mg
θ=arcsin(mg/T)
Doing it this way yields a different result than if I solve for θ using this method

t net z= mgr⊥-Tsinθr⊥

I understand just seeing how the numbers are different that it will change the answer but I don't understand why I can use the force net y to solve for θ and then use that to solve for the length of D.

Thanks for the help!

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I think I understand what you are asking, but I'm not sure. Can you post a copy of your FBD for the beam, showing the tension on the cable at the angle theta? Thanks.

ryley
You can use the Upload button in the lower right of the Edit window to upload a PDF or JPEG copy of your diagram.

ryley
Thanks for the help!

ryley said:
I've figured out to use the torque net equation to find the angle of θ and then solve for the distance D. What I'm not understanding is why I can't just use the Fnet y=0 to solve for θ?

Fnety=Tsinθ- mg
θ=arcsin(mg/T)
Doing it this way yields a different result than if I solve for θ using this method

t net z= mgr⊥-Tsinθr⊥
Can you label the diagram with some of your quantities? And I'm not seeing the length of the beam showing up in any torque equation yet, or am I missing it?

ryley
ryley said:

## Homework Statement

In figure 12-34, a uniform beam of weight 500N and length 3m is suspended horizontally, on the left it is hinged to the wall and on the right end of the beam is a cable attached to support it. the cable is attached distance D up the wall from the hinge. The least amount of tension that will snap the cable of 1200N.

a) What is the value of D that corresponds to that tension?

Σt net=0

## The Attempt at a Solution

I've figured out to use the torque net equation to find the angle of θ and then solve for the distance D. What I'm not understanding is why I can't just use the Fnet y=0 to solve for θ?

Fnety=Tsinθ- mg
θ=arcsin(mg/T)
Doing it this way yields a different result than if I solve for θ using this method

t net z= mgr⊥-Tsinθr⊥

I understand just seeing how the numbers are different that it will change the answer but I don't understand why I can use the force net y to solve for θ and then use that to solve for the length of D.

Thanks for the help!
I don't understand how you are getting different results. Both equations yield mg = T sin(θ).

That said, your sum of vertical forces equation is open to challenge. You seem to be assuming there is no vertical force from the hinge. This is true, but it is not clear how you have established that.

Edit: misread the question. See later post.

Last edited:
ryley
haruspex said:
I don't understand how you are getting different results. Both equations yield mg = T sin(θ).

That said, your sum of vertical forces equation is open to challenge. You seem to be assuming there is no vertical force from the hinge. This is true, but it is not clear how you have established that.
I thought since the hinge was chosen as the pivot point that the moment arm for it would be zero and then that would cancel it out?

haruspex said:
I don't understand how you are getting different results. Both equations yield mg = T sin(θ).

That said, your sum of vertical forces equation is open to challenge. You seem to be assuming there is no vertical force from the hinge. This is true, but it is not clear how you have established that.
I also noticed that the equation used in the solution has θ=arcsin(mg/2T). Tension being multiplied by 2, which I can see from the algebra but the fact I get two different answers isn't making sense. I'm sure its not my calculator, since I end up with two different equation which doesn't make sense.

Oh actually I see now my mistake, since it is in equilibrium the hinge vertical force would be equal to the tension vertically and so that's how id end up with arcsin(mg/2T) NOT just arcsin(mg/T). Is what I'm thinking making any sense? Sorry I couldn't get a better diagram!

ryley said:
Oh actually I see now my mistake, since it is in equilibrium the hinge vertical force would be equal to the tension vertically and so that's how id end up with arcsin(mg/2T) NOT just arcsin(mg/T). Is what I'm thinking making any sense? Sorry I couldn't get a better diagram!
Sorry, I misread the question. I thought it was a weight attached to the end of the beam.
This part at least was valid:
haruspex said:
That said, your sum of vertical forces equation is open to challenge.
ryley said:
I thought since the hinge was chosen as the pivot point that the moment arm for it would be zero and then that would cancel it out?
I wasn't objecting to your moments equation (though I should have because you had L instead of L/2). It was in regard to the sum of vertical forces. Unless you have shown otherwise, you should allow for a vertical force from the hinge. This is unaffected by your choice of axis for moments.
Had it been a weight at the end of the beam there would have been no vertical force at the hinge, but with the weight in the middle there will be.

ryley
haruspex said:
Sorry, I misread the question. I thought it was a weight attached to the end of the beam.
This part at least was valid:I wasn't objecting to your moments equation (though I should have because you had L instead of L/2). It was in regard to the sum of vertical forces. Unless you have shown otherwise, you should allow for a vertical force from the hinge. This is unaffected by your choice of axis for moments.
Had it been a weight at the end of the beam there would have been no vertical force at the hinge, but with the weight in the middle there will be.

Yes I see what youre saying, thanks for the help! I'll try and get a picture uploaded of my solutions tomorrow when I'm home to make sure were on the same page. Again thanks for the help, its much harder to figuer this out when its just all on your own :p

## 1. What is tension in a cable?

Tension in a cable refers to the pulling force that is applied to the cable. It is the force that keeps the cable taut and allows it to transmit loads.

## 2. How is tension calculated in a cable?

Tension in a cable is calculated using the formula T = F/A, where T is the tension, F is the force applied, and A is the cross-sectional area of the cable. It is typically measured in units of newtons (N) or pounds (lbs).

## 3. What factors affect tension in a cable?

The tension in a cable is affected by several factors, including the weight of the object the cable is supporting, the angle at which the cable is suspended, and the length and material of the cable itself.

## 4. How does net torque affect tension in a cable?

Net torque refers to the overall twisting force acting on an object. In the case of a cable, net torque can cause an increase or decrease in tension depending on the direction of the torque and the location of the cable's attachment points. A positive net torque will increase tension, while a negative net torque will decrease tension.

## 5. Can tension in a cable be greater than the weight of the object it is supporting?

Yes, tension in a cable can be greater than the weight of the object it is supporting. This is because tension is not only affected by the weight of the object, but also by other factors such as the angle and length of the cable. In certain situations, the tension in a cable may need to be greater than the weight of the object to ensure stability and safety.

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