Calculate the tension in the string

1. Sep 28, 2007

ssj

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.

Last edited by a moderator: May 3, 2017
2. Sep 28, 2007

Staff: Mentor

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. Sep 28, 2007

ssj

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]

Last edited by a moderator: May 3, 2017
4. Sep 28, 2007

Staff: Mentor

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.

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. Sep 28, 2007

ssj

But why would you use sine30 wouldnt that work out the "opp"?. Due to the fact the T is on the hypoteanus.

6. Sep 28, 2007

Staff: Mentor

You want the component of T perpendicular to the beam, which in this case means $T\sin {30}$.

7. Sep 28, 2007

ssj

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 ?

Last edited: Sep 28, 2007
8. Sep 28, 2007

Staff: Mentor

Because you are finding the torque (moment) it generates about the pivot point. Do you understand the definition of moment?

9. Sep 28, 2007

ssj

The moment is the turning force right ?

10. Sep 28, 2007

Staff: Mentor

You could call it that, yes.

Last edited: Sep 28, 2007
11. Sep 28, 2007

ssj

Im stuck on the second question I posted because they havent stated the length what do I do ?

12. Sep 28, 2007

Staff: Mentor

You are given all the lengths you need. (The force is at the midpoint of the beam.)

13. Sep 28, 2007

ssj

So the length is the mid-point between 0 and 0.5 ? meaning its 0.25

14. Sep 28, 2007

Staff: Mentor

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. Sep 28, 2007

ssj

But doesnt the formular state Motion=Force * perpendicular distance ? This would mean the formula is 200*L = L*T*sin30 correct ?

Last edited: Sep 28, 2007
16. Sep 28, 2007

Staff: Mentor

Sure. Notice that the length cancels from both sides of that equation.

17. Sep 28, 2007

ssj

So your left with 200/sin30= T meaning 400= T ?

18. Sep 28, 2007

Staff: Mentor

Right, as I stated in post #4. (This is for your first problem.)

19. Sep 28, 2007

ssj

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. Sep 28, 2007

Staff: Mentor

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.