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Tension

Discussion in 'MCAT Discussions' started by Mr. Z, Aug 11, 2002.

  1. Mr. Z

    Mr. Z Senior Member 7+ Year Member

    451
    0
    Apr 19, 2002
    Philadelphia
    For some reason I am having a hell of a time trying to figure out how to handle tension problems. Does anybody have any good "surefire" ways to handle them?

    Here is a problem...

    You have two masses, mass 1 = M, and mass 2 =2M, hanging across a massless, frictionless pulley. What is the tension in the rope?
     
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  3. rCubed

    rCubed taiko master 10+ Year Member

    416
    0
    Jul 14, 2002
    NY
    did u try drawing a free body diagram, like they did back in physics class?

    are both of the masses hanging down or is one on a table and the other hanging down?
     
  4. Mr. Z

    Mr. Z Senior Member 7+ Year Member

    451
    0
    Apr 19, 2002
    Philadelphia
    they are both hanging down, nothing is touching the system
     
  5. rCubed

    rCubed taiko master 10+ Year Member

    416
    0
    Jul 14, 2002
    NY
    hmmm...i'm not sure, but i think it goes something like this


    the tension in the string should be enough so that the 2M mass can accelerate downwards and the smaller mass can go up.

    the smaller mass will experience a force mg downwards and a force of T upwards. since it moves up, then T-mg= ma

    the larger mass experiences T upwards but accelerates downwards due to gravity's force, 2mg, so ma= 2mg-T

    solving these, i think u get something like 1.3mg=T

    hope that helps
     
  6. rCubed

    rCubed taiko master 10+ Year Member

    416
    0
    Jul 14, 2002
    NY
    just remember,ma is the net force making a body move, so its the difference between the two forces acting on it, if they're acting in opposite directions
     
  7. Mr. Z

    Mr. Z Senior Member 7+ Year Member

    451
    0
    Apr 19, 2002
    Philadelphia
    that's the correct answer, thanks, i appreciate it.
     
  8. limit

    limit Molesting my inner-child 10+ Year Member

    570
    1
    Jun 21, 2000
    New York City
    Always, when analyzing a free body diagram,
    ma = (dominant force) - (resistive force)

    For example, a 2 kg mass and 1 kg mass compete for who gets the right to slide down, and who gets hoisted into the air.

    For the 2kg mass: ma = mg - T
    (mg = pulling down, T = resisting the pull)

    For the 1kg mass: ma = T - mg
    (T = pulling up, mg = resisting)

    The equations come out to be as follows:
    (m2)a = (m2)g - T
    (m1)a = T - (m1)g

    Solving for a in one equation and plugging it into the other yields:
    T = 4g/3 = 13N

    Hope I didn't screw up anywhere
     

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