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Torque and tension question

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data

    J5T6pCf.png
    2. Relevant equations


    3. The attempt at a solution
    seems like Torque = r f sin theta should be used here but we don't know r.

    Am I suppose to find the x and y components of the tension of the chain?

    off topic : if the chain was moved so it would just be above the end of the rod, just like this : nFy8K4l.png , would it put less tension on the chain?
     
  2. jcsd
  3. Apr 20, 2015 #2

    BvU

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    For variables you don't know, adopt a symbol and go to work. Good chance the thing will drop out towards the end.
     
  4. Apr 20, 2015 #3

    Orodruin

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    You do not always need to know all of the quantities. Why don't you start by simply assuming that the length of the rod is R and do things analytically from there. A good practice is to never try to insert numbers until you have arrived at your final expression.
     
  5. Apr 20, 2015 #4
    No the chain is in the proper angle to help you calculate the torque of the tension of chain easily ##T_1=31.2r## where r is the length of rod. This torque will be countered by the torque of the weight. Can you find what is the angle ##\phi_1## that the weight makes with the line of rod and then find the torque of weight? You then can see that r is simplified when equating the two torques.
     
  6. Apr 20, 2015 #5

    BvU

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    Problem solving systematics urges you to list the relevant equations and the given/known data.
    You do know them already and you do use them already, but still it's a good way to get a grasp on an exercise.

    In this case you want ##\Sigma\;\tau = 0## and indeed need ##\tau = \vec r\times\vec F = r F \sin\theta## but this ##\theta## is not the ##\theta## in the drawing that comes with your exercise.

    Oh, am I a slow typist. And I need to do some paid work too, so I leave you to the other good helpers...
     
  7. Apr 20, 2015 #6
    i'm not sure where you mean ϕ1. by weight, do you mean the right end of the rod?
     
  8. Apr 20, 2015 #7
    No i mean the force of weight due to gravity... Can you draw a diagram with all the forces acting on the rod?
     
  9. Apr 20, 2015 #8
    should be 78 degrees
     
  10. Apr 20, 2015 #9
    Ok correct, now what is the torque of weight?
     
  11. Apr 20, 2015 #10
    r f sin 78

    f = mg

    ?
     
  12. Apr 20, 2015 #11
    Not exactly. The force of weight is considered to be applied at the c.o.m of the rod. Where is the c.o.m of rod?
     
  13. Apr 20, 2015 #12
    at the middle of the rod
     
  14. Apr 20, 2015 #13
    ok so you have the weight acting on the middle of the rod, at an agle of 78, so what is its torque?
     
  15. Apr 20, 2015 #14
    r/2 x mg x sin 78

    ?
     
  16. Apr 20, 2015 #15
    yes correct.
     
  17. Apr 20, 2015 #16
    do we need the torque of the chain? or is that given by the tension in the chain?
     
  18. Apr 20, 2015 #17
    yes you need the torque of the chain. I already said in post #4 what it is. Equate the two torques and solve for m. What do you get?
     
  19. Apr 20, 2015 #18
    torque of chain = r 31.2

    r 31.2N = r mg sin 78

    m = 3.544kg

    ?
     
  20. Apr 20, 2015 #19
    Didnt we say the torque of weight is (r/2)mgsin(78)
     
  21. Apr 20, 2015 #20
    whoops, yes
     
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