Calculate the tension in the string

[ssj]

So the length L must be 0.5 and this means we must 0.5/2 because of L/2 correct?

 

[Doc Al]

So the length L must be 0.5 and this means we must 0.5/2 because of L/2 correct?

Right.

 

[ssj]

Alright this means that the equation becomes 200*L/2=L/2*T*sin30 ?

 

[Doc Al]

Alright this means that the equation becomes 200*L/2=L/2*T*sin30 ?

Nope. Why do you have L/2 on both sides of the equation? Realize that the equation as you've written it gives the same answer as you got for problem 1. Is that reasonable?

 

[ssj]

I can't seem to work it out correctly the force is L/2 away from the pivot and in order to work out the tension I must do something must I make T perpendicular ?

 

[Doc Al]

You treat the tension in problem 2 exactly as you did in problem 1.

 

[Doc Al]

You might want to study these examples: Torque Equilibrium Examples

 

[ssj]

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.

 

[Doc Al]

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.

 

[ssj]

Yes I think i understand 200*L/2=L*T*sin30
Which becomes 200/sin30/2
400/2=T
200=T

 

[Doc Al]

Good! In the second problem it takes just half the tension to support the beam.

 

[ssj]

What about this question
http://img217.imageshack.us/img217/7070/poqn3.th.jpg

My guess is something like 400*L/4=L*T*sin30

 

[Doc Al]

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.

 

[ssj]

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.

 

[Doc Al]

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.

 

[ssj]

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

 

[Doc Al]

Right this makes it 400*L/3=L*T*Sin30

No:
200*L + 200*L/2 = 200*(3L/2)

 

[ssj]

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?

 

[Doc Al]

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.

 

[ssj]

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 ?

 

[Doc Al]

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.

 

[ssj]

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

I would think the answer is 300*L=L*T*Sin50 .

 

[Doc Al]

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.

 

[ssj]

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 ?

 

[Doc Al]

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?

 

[ssj]

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.

 

[Doc Al]

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.

 

[ssj]

Good! (I thought so.)

Right so the equation now becomes 300*L=L/2*T*Sin50.

 

[Doc Al]

Right so the equation now becomes 300*L=L/2*T*Sin50.

Yep.

 

[ssj]

Yep.

So if I replace L with 2 I get T=600/sin50 meaning T=783.

Now I am stuck with the following question http://img63.imageshack.us/img63/5889/klrk3.th.jpg .

I can't seem to understand how to work the answer out.