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Tension of a rope in a pulley system

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A 1.0 kg mass and a 4.0 kg mass are attached to a lightweight cord that passes over a frictionless pulley. The hanging masses are free to move. What is the acceleration of the larger mass? What is the tension on the cord?

    2. Relevant equations

    F(net force)=F(tension)-weight

    F=ma


    3. The attempt at a solution
    I found the acceleration using F=ma. The system has 3 kg of mass going down (4kg-1kg), which is equal to 29.4N of force (found by multiplying 3kg by 9.8m/s^2). The whole system has a mass of 5 kg, so I substituted in to F=ma.

    29.4N=5.0 kg + a

    The acceleration is 5.88 m/s^2

    Now comes the part I am confused about. The tension of the cord should be
    F(net force)=tension-weight, or rearranged to be:

    F(tension)=F(net force)+weight or (Tension=ma+mg)

    I can substitute in what i know:
    Tension =m(5.88m/s^2)+m(9.8m/s^2)

    I am at a loss of which mass to use. Do i use the mass that i found in working on acceleration (3 kg), or the mass of the whole system (5 kg)?
     

    Attached Files:

    Last edited: Nov 12, 2009
  2. jcsd
  3. Nov 12, 2009 #2
    can u draw the force diagrams of each mass? that will help u a lot :)
     
  4. Nov 12, 2009 #3
    I attached a quick diagram, i don't know if that's detailed enough, but that is basically what's going on. (i'm bad at drawing with a mouse, so it probably doesn't look that spectacular)- the 4kg mass is pulling the whole system down at 5.88m/s^2, which is great since that is the first half of the question.

    My problem is which mass do i use to find the tension? the mass of 3 kg (because of the pulley 4kg-1kg=3kg, the 3kg was then used to find the amount of downward force, which turned out to be 29.4N) or the mass of the whole system, 5kg?
     
    Last edited: Nov 12, 2009
  5. Nov 14, 2009 #4
    ok, u are done basically, u can use any of the masses actually, so let's say we concentrate on the heavier one

    the heavy mass' equation of force will be:

    -Mg+T=-Ma

    -(4 kg)(9.8 m/s^2)+T=-(4kg)(5.88 m/S^2)

    that's it, u've found it :)

    can u find it now using the smaller mass? :)
     
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