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  1. S

    Vsquared= usquared+ 2as

    Yes, but what does that stand for is there no specific term ?
  2. S

    Vsquared= usquared+ 2as

    Homework Statement In the forumla vsquared= usquared+ 2as what does vsquared represent ?
  3. S

    Calculate the tension in the string

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

    Calculate the tension in the string

    Alright so is this correct ? Meaning that ths distance from the pivot is 5.25M+5.25M?
  5. S

    Calculate the tension in the string

    Would it be *F1 14kN*10.5M=8kN*5.25M*F2 ? am I correct ?
  6. S

    Calculate the tension in the string

    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...
  7. S

    Calculate the tension in the string

    So if I replace L with 2 I get T=600/sin50 meaning T=783. Now im stuck with the following question http://img63.imageshack.us/img63/5889/klrk3.th.jpg [Broken]. I cant seem to understand how to work the answer out.
  8. S

    Calculate the tension in the string

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

    Calculate the tension in the string

    Sorry about that I meant Sin50.
  10. S

    Calculate the tension in the string

    So this would make it 300*L=L/2*T*Sin30 ?
  11. S

    Calculate the tension in the string

    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 http://img63.imageshack.us/img63/4524/yhse1.th.jpg [Broken] I would think the answer is 300*L=L*T*Sin50 .
  12. S

    Calculate the tension in the string

    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 ?
  13. S

    Calculate the tension in the string

    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?
  14. S

    Calculate the tension in the string

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

    Calculate the tension in the string

    So the torque from question 1 was 400N and question 2 was 200N meaning we have 600*L/2=L*T*sin30.
  16. S

    Calculate the tension in the string

    What about this question http://img217.imageshack.us/img217/7070/poqn3.th.jpg [Broken] My guess is something like 400*L/4=L*T*sin30
  17. S

    Calculate the tension in the string

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

    Calculate the tension in the string

    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...
  19. S

    Calculate the tension in the string

    I cant 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 ?
  20. S

    Calculate the tension in the string

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

    Calculate the tension in the string

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

    Calculate the tension in the string

    Alright so the perpendicular distance is changing its now at the mid point the half of 0.5 is 0.25 and due to the fact that the force has moved half way I assume the distance is now half as well thus making it 0.25.
  23. S

    Calculate the tension in the string

    Wouldnt this mean that the pivot is slightly more changing the angle as the force is closer to the motion.
  24. S

    Calculate the tension in the string

    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.
  25. S

    Calculate the tension in the string

    What is it then in the second problem?
  26. S

    Calculate the tension in the string

    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 ?
  27. S

    Calculate the tension in the string

    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.
  28. S

    Calculate the tension in the string

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

    Calculate the tension in the string

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

    Calculate the tension in the string

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