Tension and net torque in a 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?

Homework Equations

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!

 

[berkeman]

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.

 

[berkeman]

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]

I've uploaded a linked image from google.
Thanks for the help!

 

[berkeman]

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?

 

[haruspex]

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?

Homework Equations

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.

 

[ryley]

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?

I'll upload a diagram from slader and have everything labeled

 

[ryley]

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.

 

[ryley]

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!

 

[haruspex]

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:

That said, your sum of vertical forces equation is open to challenge.

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]

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