Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: A simple pulley problem

  1. Feb 29, 2004 #1
    I'm working on this rather simple pulley problem, and can't quite figure it out. It doesn't even involve acceleration. Could anyone help me out? Thanks.

    Here's the problem:
    A crate is pulled up using frictionless pulleys in the manner shown in the figure. The angle is 45 degrees. The masses are, for the small pulley, m1=3.5 kg, for the traveling pulley, M2=6.3 kg, and for the crate, MC=37.4 kg. What is the tension with which the operator must pull on the cable (assume the cable is of neglible mass) in order to raise the crate at a constant speed?

    Here's the illustration.
    Last edited: Feb 29, 2004
  2. jcsd
  3. Feb 29, 2004 #2


    User Avatar
    Science Advisor

    There are two things going on here. Imagine the person pulling on the line for a distance of one meter. The line from pulley m1 to the person lengthens by x meters so the total line from point of attachment to pulley m1 decreases by x meters. Since the "travelling pulley", m2, stays at the midpoint of the line, its height and the height of the mass increases by x/2 meter. Now apply "conservation of energy". The crate, MC= 37.4 kg, so weight 37.4g Newtons, has raised by x/2 m so its potential energy has increased by (37.4g)(x/2) Joules. The "travelling pulley", m2= 3.5 kg so 3.5 Newtons weight, has raised by x/ m so its potential energy has increased by (3.5g)(x/2)Joules. The stationary pulley, m1, does not move- its potential energy doesn't change. The potential energy for the system has increased by (37.4g)(x/2)+(3.5g)(x/2)= (40.9g)(x/2) Joules. That means that that amount of work has been done by the man pulling on the line. Since Work= Force*distance and he has moved the line for x meters, he must have done xF= (40.9g)(x/2) Joules of work and so F= 40.9g/2= 20.45g N.
    General rule: divide the weight lifted by the number of supporting lines in the pulley system: a total of 40.9g N weight was lifted by two lines: Force necessary is half the weight.

    Since pulley m1 changes the direction of pull, the angle θ is irrelevant.
  4. Mar 1, 2004 #3
    That works

    Thanks for the reply. I think I follow your reasoning, although you did it differently than I did (I was trying to analyze the system only in terms of Newton's second law). The answer is correct, though. :smile:
  5. Mar 2, 2004 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Re: That works

    I believe that the answer Halls gave is incorrect: he used the wrong M2 (which is 6.3 Kg, not 3.5 Kg).

    In any case, using Newton's second law is an easier way to solve this problem. For the system (moving pulley plus crate) to be lifted at a constant speed, it must be in equilibrium: The upward force (which is 2T) must equal the downward force (M2+MC)g. So, the tension required is (6.3 + 37.4)g/2 = 214 N.
  6. Mar 4, 2004 #5

    Thanks, Doc. The correct answer is indeed 214 N. I meant that the method Halls used to solve it was correct, although I went through the calculations and plugged in the numbers myself.

    It is easier to use Newton's 2nd Law. I had tried this the first time, but arrived at the wrong answer because I thought I needed to take into account the small pulley (now I see that I didn't have to since it's attached to the ceiling). Thanks for everyone's help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook