Calculate the tension in the string

In summary, the conversation discusses calculating tension in a string in different situations and using basic trigonometry to solve problems involving torque. The conversation also provides guidance on solving equations and understanding the concept of moment and pivot points. The conversation concludes with a discussion on how the point of application of a force can affect the calculations.
  • #36
You treat the tension in problem 2 exactly as you did in problem 1.
 
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  • #38
Right to in problem 1 we basically made it perpendicular correct ? So for now let's call the length L meaning that 200*L=200*T*sin30, sin30 because that will give use the hypotenues which is T but the problem with the formula is that it gives the same answer as we got in question 1 meaning we must divide sin30 on each side meaning 200*L/sin30= 400 and once we find out the L we will get an new answer perhaps.
 
  • #39
ssj said:
Right to in problem 1 we basically made it perpendicular correct ?
Yes.
So for now let's call the length L meaning that 200*L=200*T*sin30,
If you're talking about problem 1, that should be:
200*L = L*T*sin30

sin30 because that will give use the hypotenues which is T but the problem with the formula is that it gives the same answer as we got in question 1 meaning we must divide sin30 on each side meaning 200*L/sin30= 400 and once we find out the L we will get an new answer perhaps.
Using the corrected equation, we get: T = 200/sin30 = 400 N. The L drops out and has no bearing on the answer for tension. If the beam were twice as long, you'd still get the same tension.

If you are talking about problem 2, the left side of the equation is different. The distance is not L, but L/2. The right side remains the same, since the cable has not moved and the angle is the same.

Rewrite the equation for problem 2 and solve for the tension. Again, the length of the beam (L) will drop out.
 
  • #40
Yes I think i understand 200*L/2=L*T*sin30
Which becomes 200/sin30/2
400/2=T
200=T
 
  • #41
Good! In the second problem it takes just half the tension to support the beam.
 
  • #43
ssj said:
My guess is something like 400*L/4=L*T*sin30
The right side is correct, but the left is not. Don't guess! You have two forces producing clockwise torques--add up the torque from each. You should know the torque from each one, since they are the same forces as appeared earlier.
 
  • #44
Doc Al said:
The right side is correct, but the left is not. Don't guess! You have two forces producing clockwise torques--add up the torque from each. You should know the torque from each one, since they are the same forces as appeared earlier.

So the torque from question 1 was 400N and question 2 was 200N meaning we have
600*L/2=L*T*sin30.
 
  • #45
ssj said:
So the torque from question 1 was 400N and question 2 was 200N meaning we have
600*L/2=L*T*sin30.
Those were the tensions, not the torques. The clockwise torque from problem 1 was 200*L; from problem 2 it was 200*L/2.
 
  • #46
Doc Al said:
Those were the tensions, not the torques. The clockwise torque from problem 1 was 200*L; from problem 2 it was 200*L/2.

Right this makes it 400*L/3=L*T*Sin30
 
  • #47
ssj said:
Right this makes it 400*L/3=L*T*Sin30
No:
200*L + 200*L/2 = 200*(3L/2)
 
  • #48
Doc Al said:
No:
200*L + 200*L/2 = 200*(3L/2)

How did you get "3L/2" on the right hand side and why isit L/2 on the left side ? Furthure more once I calcualted this euqation L=200 where do I go from here?
 
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  • #49
ssj said:
How did you get "3L/2" on the right hand side and why isit L/2 on the left side ? Furthure more once I calcualted this euqation L=200 where do I go from here?
This is not the equilibrium equation, this is just the sum of the clockwise torques (see post #43): 200*L + 200*(L/2) = 200*(3L/2) = 300*L

Now set this sum of clockwise torques equal to the counter-clockwise torque due to the tension to the get the equilibrium equation:
300*L = L*T*sin30

Now you can solve for the tension in problem 3.
 
  • #50
Doc Al said:
This is not the equilibrium equation, this is just the sum of the clockwise torques (see post #43): 200*L + 200*(L/2) = 200*(3L/2) = 300*L

Now set this sum of clockwise torques equal to the counter-clockwise torque due to the tension to the get the equilibrium equation:
300*L = L*T*sin30

Now you can solve for the tension in problem 3.

For the sum of the clockwise torques I understand that 200*L+200(L/2) but I don't understand 200*(3L/2) where did the "3L" come from ?
 
  • #51
ssj said:
For the sum of the clockwise torques I understand that 200*L+200(L/2) but I don't understand 200*(3L/2) where did the "3L" come from ?
L + L/2 = 3L/2
You can also write that as: (3/2)*L

You can also do this:
200*L + 200*(L/2) = 200*L + 100*L = 300*L

Please convince yourself that these are equivalent.
 
  • #52
Doc Al said:
L + L/2 = 3L/2
You can also write that as: (3/2)*L

You can also do this:
200*L + 200*(L/2) = 200*L + 100*L = 300*L

Please convince yourself that these are equivalent.

Ah yes I see if we replace L with 1 we get 1.5=1.5 meaning the formula becomes 300*L=L*T*Sin30 meaning T=600N.

The next question I am stuck on is http://img63.imageshack.us/img63/4524/yhse1.th.jpg [Broken]

I would think the answer is 300*L=L*T*Sin50 .
 
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  • #53
ssj said:
I would think the answer is 300*L=L*T*Sin50 .
The left-hand side (which represents clockwise torques) is correct--it's the same as the last problem. But the right-hand side is not correct. The angle is not the only thing that changes: the distance to the pivot point also changes--it's not L anymore.
 
  • #54
Doc Al said:
The left-hand side (which represents clockwise torques) is correct--it's the same as the last problem. But the right-hand side is not correct. The angle is not the only thing that changes: the distance to the pivot point also changes--it's not L anymore.

So this would make it 300*L=L/2*T*Sin30 ?
 
  • #55
ssj said:
So this would make it 300*L=L/2*T*Sin30 ?
You now have the correct distance on the right-hand side (L/2), but what happened to the angle?
 
  • #56
Doc Al said:
You now have the correct distance on the right-hand side (L/2), but what happened to the angle?

Sorry about that I meant Sin50.
 
  • #57
ssj said:
Sorry about that I meant Sin50.
Good! (I thought so.) Now you can solve for the tension--once again, the actual value of L doesn't matter.
 
  • #58
Doc Al said:
Good! (I thought so.)
Right so the equation now becomes 300*L=L/2*T*Sin50.
 
  • #59
ssj said:
Right so the equation now becomes 300*L=L/2*T*Sin50.
Yep.
 
  • #60
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  • #61
ssj said:
So if I replace L with 2 I get T=600/sin50 meaning T=783.
Correct. But realize you don't have to "replace" L with any number--the L just drops out:
300*L=L/2*T*Sin50 becomes 300=1/2*T*Sin50 if you cancel the L on both sides. (Which means dividing both sides by L.)

Now I am stuck with the following question ...
Another equilibrium problem. What are the conditions for equilibrium? Forces must add to zero and torques about any point must add to zero. Label the vertical support forces (at each end of the bridge) F1 & F2. Hint: A good point to use as a pivot is either end of the bridge.
 
  • #62
Doc Al said:
Another equilibrium problem. What are the conditions for equilibrium? Forces must add to zero and torques about any point must add to zero. Label the vertical support forces (at each end of the bridge) F1 & F2. Hint: A good point to use as a pivot is either end of the bridge.

Alright so if I lable the force between 4kN and 10kN F1 and from 10kN to 8kN F2 and place an pivot at the left hand side of the bridge would this be correct and what would the equation be for the clockwise rotation= counter clockwise rotation ?

As seen below
http://img211.imageshack.us/img211/5756/klcw7.th.jpg [Broken]
 
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  • #63
ssj said:
Alright so if I lable the force between 4kN and 10kN F1 and from 10kN to 8kN F2 and place an pivot at the left hand side of the bridge would this be correct and what would the equation be for the clockwise rotation= counter clockwise rotation ?
No. F1 is the upward force of the support on the left end of the bridge, F2 is the upward force on the right end of the bridge.

If you pick the left side of the bridge as your pivot, you will have the three axial loads producing the clockwise torques and the force F2 producing the counter-clockwise torques. Find those torques just like in all the previous problems.
 
  • #64
Would it be *F1 14kN*10.5M=8kN*5.25M*F2 ? am I correct ?
 
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  • #65
ssj said:
Would it be *F1 14kN*10.5M=8kN*5.25M*F2 ? am I correct ?
No. There are three forces (each at a different distance from the pivot) that create clockwise torques. And the force F2, which creates the counter-clockwise torque, is how far from the pivot?

I encourage you to study the examples discussed on the webpage I linked in post #37.
 
  • #66
Doc Al said:
No. There are three forces (each at a different distance from the pivot) that create clockwise torques. And the force F2, which creates the counter-clockwise torque, is how far from the pivot?

I encourage you to study the examples discussed on the webpage I linked in post #37.

Alright so is this correct ?
kl.jpg

Meaning that ths distance from the pivot is 5.25M+5.25M?
 
  • #67
No, the distance of F2 from the left pivot point is: 7m + 3.5m + 5.25m + 5.25m = 21m. (You didn't even have to do that addition: the problem states that the bridge supports are 21m apart.)
 
  • #68
Doc Al said:
No, the distance of F2 from the left pivot point is: 7m + 3.5m + 5.25m + 5.25m = 21m. (You didn't even have to do that addition: the problem states that the bridge supports are 21m apart.)

I think the total Clockwise force F1 would be equal to 22kN*21m=462 correct but what would the counter clockwise force be then?
 
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