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Hang vertically downward are connected by a pulley

  1. Jun 13, 2003 #1
    Find the work done by gravity when two masses (42kg and 75kg) which hang vertically downward are connected by a pulley. The masses are initially at rest. The mass move under the influence of gravity for 2.0s. Assume no friction in the pulley.

    My solution:
    Considering the larger mass m1, the work done by gravity on m1 is
    W1=w1*d, where W1 is the work done on m1 by gravity, w1 is the force of gravity acting on m1, and d is the distance traversed by m1 in the time period. d is the same for both m1 and m2 (the smaller mass) since both move under the influence of gravity and therefore have the same acceleration. The net force on m1 is the force of gravity on m1 minus the tension of the rope (if downward is to be the positive direction). The tension of the rope is equivalent in magnitude to the weight of m2. Thus, F(net)1=g*(m1-m2)=m1*a. This distance traversed by m1 is then d=g/(2m1)*(m1-m2)(t^2). The work done by gravity on m1 is then
    W1=g*m1*g/(2m1)*(m1-m2)(t^2)=(g^2)/2*(m1-m2)(t^2)=6300J
    Without calculating, I can see that the work done by gravity on the other mass should be equivalent in magnitude to this answer.

    Book's answer: 1.8x10^3J

    Where did I go wrong?
    Thanks in advance for the help.
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Jun 13, 2003 #2

    FZ+

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    This is obviously wrong, as m2 is accelerating upwards, showing a net resultant force on m2 of m2*a.

    Try instead to work out an expression for T for each mass, and equate the Ts to get the acceleration.
     
  4. Jun 13, 2003 #3

    Tom Mattson

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    Re: Work

    Hi Stephen, and welcome to PF.

    OK

    This is a mistake. Remember that m2 is accelerating, so Newton's 2nd law for m2 reads is:

    ΣFy=T-m2g=m2a2.

    T does not equal m2g!

    Try implementing that correction into your solution and see what you come up with.
     
  5. Jun 14, 2003 #4
    Thanks for the help. To show you I learned my lesson, I performed this little exercise. Please check my work.
    I figured that the problem above was just a specific instance of the general case below (where A1=A2=pi/2). Both masses are on frictionless surfaces. They are attached by a rope which runs along a frictionless pulley. See the attached diagram. I'll show my work briefly (since I'm sure you know much more than I do, this should be very easy to follow).
    I figured out that the problem becomes very simple if I treat the rope as if it were straight (and rotate the other forces accordingly). Then I don't have to worry about vector notation using two different coordinate systems. Anyway,
    T - m1*g*sin(A1) = m1*a
    a = (1/m1)(T-m1*g*sin(A1)) = (1/m2)(m2*g*sin(A2) - T)
    T = (m1*m2*g)/(m1 + m2)*(sin(A1) + sin(A2))
    When A1=A2=pi/2
    T = 2(m1*m2*g)/(m1 + m2)
    Thanks again!

    .......
    I was so confident in this result, but it doesn't appear to be right either. Using my work above, I'm still getting 2400J for the work on the smaller mass in the original problem. What am I doing wrong?
    W=FgD=m2*g*1/2*(Fnet)/m2*t^2 = g/2*t^2*Fnet
    For Fnet I have Fnet = T - m2*g = 2(m1*m2*g)/(m1 + m2) - m2*g
    W = 1/2*(tg)^2*m2*[2m1/(m1+m2)-1]
    From there it's just plugging in numbers. Where did I go wrong?
     

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