Calculate the tension in the string

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  • #1
ssj
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1. Calculate the tension in the string T in each of the following situations.
In all calculations ignore the mass of the beam.


http://img146.imageshack.us/img146/7291/43422126ru2.th.jpg [Broken]


2. Moment=F X Perpendicular Distance and Basic Trigonometry



3.I have tried 200N X 0.5m=100 100/Cos30

Help is much apperchated.
 
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  • #2
Doc Al
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I have tried 200N X 0.5m=100 100/Cos30
I understand the left side of this equation: the clockwise torque due to the 200 N force. The right side should be the counter-clockwise torque due to the tension of the cable.

Correct this equation, rewriting the right hand side in terms of T (tension). Then solve for T.
 
  • #3
ssj
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I understand the left side of this equation: the clockwise torque due to the 200 N force. The right side should be the counter-clockwise torque due to the tension of the cable.

Correct this equation, rewriting the right hand side in terms of T (tension). Then solve for T.

I think i have found the solution by calculating the "opp" tan30 X 0.5 = 0.2886 from there onward we do 200 X 0.5= 100/0.2886 =346.5/cos 30 = 400 N

However now I am stuck on an similar equation of http://img374.imageshack.us/img374/3332/22917237wh9.th.jpg [Broken]
 
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  • #4
Doc Al
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I think i have found the solution by calculating the "opp" tan30 X 0.5 = 0.2886 from there onward we do 200 X 0.5= 100/0.2886 =346.5/cos 30 = 400 N
Don't really understand what you did there with tangents and cosines. All you need is:
0.5*200 = 0.5*T*sin(30)
which gives you T = 400.

However now I am stuck on an similar equation of...
Solve it in the same way. The only difference is the moment arm associated with the 200 N force.

I strongly suggest you adopt a systematic approach, like the one I used.
 
  • #5
ssj
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But why would you use sine30 wouldnt that work out the "opp"?. Due to the fact the T is on the hypoteanus.
 
  • #6
Doc Al
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But why would you use sine30 wouldnt that work out the "opp"?. Due to the fact the T is on the hypoteanus.
You want the component of T perpendicular to the beam, which in this case means [itex]T\sin {30}[/itex].
 
  • #7
ssj
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But why would T need to become perpendicular ? Also for question 2 how am I ment to work it out with out the given length ?
 
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  • #8
Doc Al
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But why would T need to become perpendicular ?
Because you are finding the torque (moment) it generates about the pivot point. Do you understand the definition of moment?
 
  • #9
ssj
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Because you are finding the torque (moment) it generates about the pivot point. Do you understand the definition of moment?

The moment is the turning force right ?
 
  • #10
Doc Al
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The moment is the turning force right ?
You could call it that, yes.
 
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  • #11
ssj
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Im stuck on the second question I posted because they havent stated the length what do I do ?
 
  • #12
Doc Al
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Im stuck on the second question I posted because they havent stated the length what do I do ?
You are given all the lengths you need. (The force is at the midpoint of the beam.)
 
  • #13
ssj
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So the length is the mid-point between 0 and 0.5 ? meaning its 0.25
 
  • #14
Doc Al
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So the length is the mid-point between 0 and 0.5 ? meaning its 0.25
That's correct. But realize that you don't need the length of the beam: You can just use L or L/2. (The actual length doesn't matter.)
 
  • #15
ssj
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But doesnt the formular state Motion=Force * perpendicular distance ? This would mean the formula is 200*L = L*T*sin30 correct ?
 
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  • #16
Doc Al
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But doesnt the formular state Motion=Force * perpendicular distance ? This would mean the formula is 200*L = L*T*sin30 correct ?
Sure. Notice that the length cancels from both sides of that equation.
 
  • #17
ssj
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So your left with 200/sin30= T meaning 400= T ?
 
  • #18
Doc Al
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So your left with 200/sin30= T meaning 400= T ?
Right, as I stated in post #4. (This is for your first problem.)
 
  • #19
ssj
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This is for the second problem though, wouldnt the Motion be different due to the force being applyed to the mid-point? I seem to of gotten T=400 in my previous post.
 
  • #20
Doc Al
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This is for the second problem though, wouldnt the Motion be different due to the force being applyed to the mid-point? I seem to of gotten T=400 in my previous post.
These calculations apply to the first problem.

For the second problem, the method is exactly the same: Set clockwise and counter-clockwise torques equal. Of course the torque exerted by the 200 N force is different in the second problem, since it's applied at the midpoint instead of at the end.
 
  • #21
ssj
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Right, so I am correct when 200*L=L*T*Sin30 the "L"'s cancel out leaving 200=T*Sin30
We then Divide sin30 on each side leaving 200/sin30=T meaning 400=T is this correct ?
 
  • #22
Doc Al
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Right, so I am correct when 200*L=L*T*Sin30 the "L"'s cancel out leaving 200=T*Sin30
We then Divide sin30 on each side leaving 200/sin30=T meaning 400=T is this correct ?
You are correct for the first problem, but not if you are talking about the second problem. (In the second problem, the moment arm for the 200 N force is no longer L.)
 
  • #23
ssj
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What is it then in the second problem?
 
  • #24
Doc Al
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What is it then in the second problem?
You tell me. That force is applied at the midpoint of the beam, so what do you think?
 
  • #25
ssj
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I think that it becomes 200*0.25=0.25*t*sin30 is this equation correct ? 0.25 because 0.5/2= 0.25 also now the center of gravity has changed.
 

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