# Calculate the tension in the string

by ssj
Tags: moments
 Mentor P: 41,323 You might want to study these examples: Torque Equilibrium Examples
 P: 36 Right to in problem 1 we basically made it perpendicular correct ? So for now lets call the lenght 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.
Mentor
P: 41,323
 Quote by ssj Right to in problem 1 we basically made it perpendicular correct ?
Yes.
 So for now lets call the lenght 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.
 P: 36 Yes I think i understand 200*L/2=L*T*sin30 Which becomes 200/sin30/2 400/2=T 200=T
 Mentor P: 41,323 Good! In the second problem it takes just half the tension to support the beam.
Mentor
P: 41,323
 Quote by ssj 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.
P: 36
 Quote by Doc Al 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.
Mentor
P: 41,323
 Quote by ssj 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.
P: 36
 Quote by Doc Al 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
Mentor
P: 41,323
 Quote by ssj Right this makes it 400*L/3=L*T*Sin30
No:
200*L + 200*L/2 = 200*(3L/2)
P: 36
 Quote by Doc Al 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?
Mentor
P: 41,323
 Quote by ssj 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.
P: 36
 Quote by Doc Al 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 ?
Mentor
P: 41,323
 Quote by ssj 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.
P: 36
 Quote by Doc Al 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 im stuck on is

I would think the answer is 300*L=L*T*Sin50 .
Mentor
P: 41,323
 Quote by ssj 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.
P: 36
 Quote by Doc Al 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 ?

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