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Pulley and cords tension physics

  1. Sep 18, 2013 #1
    The pulley in (Figure 1) is suspended by a cord C. Suppose that m1 = 1.0kg and m2 = 3.9kg .
    Determine the tension in the cord that supports the pulley C after the masses are released and before one hits the ground. Ignore the mass of the pulley and cords.

    I'm having trouble getting started please help.

    i know the tension connecting the masses is the same i just dont know how to solve for it.
     
  2. jcsd
  3. Sep 18, 2013 #2
    To solve these types of problems, I would always recommend starting by drawing force diagrams. This will help you answer such questions as: What upward force is acting on m1? What downward force is acting on m1? What is the net force on m1? What about m2?
     
  4. Sep 18, 2013 #3
    so weight would be the downward force for both m1 and m2 and tension would be the upward force. correct?
     
  5. Sep 18, 2013 #4
    Correct. And, as you mentioned earlier, tension (T) is the same for both.

    Now set up your equations.
     
  6. Sep 18, 2013 #5
    would the tension for m1 and m2 be 5.8?
     
  7. Sep 18, 2013 #6
    Where are you getting 5.8? That's not the correct answer. How did you set up your equations?
     
  8. Sep 18, 2013 #7
    thats the problem i have is getting the equations.

    would the first be Fnet=T-mg=ma?
     
  9. Sep 18, 2013 #8
    Yes. Do that for both masses.
     
  10. Sep 18, 2013 #9
    would the second equation be mg-T=ma or would it be the same as the first?
     
  11. Sep 18, 2013 #10
    Technically yes, it would be mg - T = ma. Let "a1" be the acceleration of the first mass and "a2" be the acceleration of the second mass.

    We have the equations:
    (1) T - m1*g = m1*a1
    (2) T - m2*g = m2*a2

    Both m1 and m2 undergo the same magnitude of acceleration, but in opposite directions, therefore a2 = -a1.

    Thus, equation 2 becomes T - m2*g = -m2*a1
    Or, as you put it: m2*g - T = m2*a1
     
  12. Sep 18, 2013 #11
    so would T-9.8=a1? correct
     
  13. Sep 18, 2013 #12
    would T=15.6N
     
  14. Sep 18, 2013 #13
    Yep, that's the answer.
     
  15. Sep 18, 2013 #14
    my mastering physics tells me its wrong? is there more to the problem?
     
  16. Sep 18, 2013 #15
    Ah, I see. I just re-read the original question: it's "the tension in the cord that supports the pulley". I was confused slightly by the wording of the question, and didn't have access to the problem's diagram.

    The net force on the pulley is 0. Draw a force diagram for the pulley; it is acted upon by an upward force (which you are trying to find) and the tensions below it (which you have already solved for).
     
  17. Sep 18, 2013 #16
    so tension would be the downward and would upward force be the norma force?
     
  18. Sep 18, 2013 #17
    The upward force should be the tension in the cord that holds the pulley up. The downward force is the tensions in the ropes (technically, 2*15.6N since there are two ropes hanging from the pulley). And the net force is 0.
     
  19. Sep 18, 2013 #18
    oh okay i get it. the tension in the cord holding the pulleys has to be equal to the tensions of both the cords below to bring the net force = to 0.
     
  20. Sep 18, 2013 #19
    Yes. Exactly.
     
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