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Three Blocks On Two Pulleys

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-05-Newtons-Laws/blocks_on_pulleys/12.gif [Broken]

    In the above picture, the masses are M1 = 10 kg, M2 = 15 kg, and M3 = 25 kg. The pulleys and strings are massless, there is no friction, and gravity points downward. The whole system is held fixed, then released at rest. What is the acceleration of the 25 kg block? Enter a positive answer if it goes up, a negative answer if it goes down, or 0 if the acceleration is zero.

    Hint 1) Draw a free-body diagram for each of the three masses. Try to convince yourself that the tension in the strings supporting M1 and M2 are equal, and that tension is half of the tension in the string supporting M3.

    Hint 2) Let's define the directions of all accelerations to be positive if they are up, negative if they are down. Consider the accelerations of the three blocks (a1, a2, and a3). The basic result is that a3 = -(a1 + a2)/2. How can we convince ourselves of that? This is not easy to derive, but we can make it plausible. Consider what happens if M1 = M2. Then we know that those two blocks will accelerate together just as though they were a single mass. They are essentially one half of a simple Atwood's machine, the other half of which is Block 3. Under those conditions, a1 = a2 and they are both equal to -a3. The negative sign arises because if Block 3 goes down, then the other two blocks must go up (and vice versa). This is exactly what our formula gives. In the general case, the formula is still valid. Using this information plus the hint about the tensions plus writing F = ma for each mass, we end up with 3 equations and 3 unknowns (1 tension and 2 accelerations). Solve these to find the answer. At this point the algebra gets a little messy, and it is better to plug the masses into the equations right away rather than keeping things algebraic.

    2. Relevant equations
    F = ma for each block

    3. The attempt at a solution
    I set up free body diagrams for each block and found:
    (FT1 relates to the tension force for block 1 etc...)
    FT1 = 10(g + a1)
    FT2 = 15(g + a2)
    FT3 = 25(g + a3)

    I agree that FT1 = FT2 but I don't know why they are half the tension in FT3 or that their accelerations are a3 = -(a1 + a2)/2. Could someone please explain this a bit better than they did?

    Also, just by using the hints/formulas they gave me I solved for a3 to be .2087 but this isn't the right answer. So I must be doing something wrong.

    All help would be greatly appreciated
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 24, 2010 #2


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    Gold Member

    The pulley connecting them also obeys F=ma. It is accelerating but massless so the net force on it must be zero.

    Also since the other pulley is fixed the acceleration of that pulley will be equal and opposite in sign to the acceleration of the third mass. Its height will be the average height of the first two masses (plus half the length of the string which is constant) so then to is its velocity and acceleration. Hence a3 = - (a1+a2)/2.

    I just worked the problem and got something different from you. Double check your work.
  4. Sep 24, 2010 #3
    Thanks! got the right answer. The real key for me was applying F = MA to that knot and a3 = -(a1+a2)/2

    I was able to prove that the tension is 1/2 of that in F3, but I'm still having trouble proving the acceleration one. Could you help me get started in proving that as well (if you don't mind could you call the pulley connecting the three masses pulley 1 and the other pulley 2? I'm getting a bit confused by which pulley you were referencing)? I really like to understand all of the equations I use.

    Thanks again!
  5. Sep 25, 2010 #4
    If you don't mind I'd call positive a downward acceleration.

    By the superposition of effect on [tex] M_1, M_2[/tex]
    we write that
    [tex] a_2= g\frac{M_1-M_2}{M_1+M_2}[/tex]

    and of course [tex] a_2 = - a_1[/tex]

    So we have that
    [tex] F_2 = M_2(g+a_2)[/tex]
    [tex] F_1 = M_1(g-a_2)[/tex]

    The force pulling down the left pulley is
    since [tex] F_{12} = F_1 + F_2 [/tex]

    [tex] F_{12} = g( M_1+M_2 - \frac{(M_1-M_2)^2}{M_1+M_2} ) = g \frac{4M_1M_2}{M_1+M_2}[/tex]

    That also gives you the "virtual mass" of the moving system of M1 and M2.
    Their mass is:

    Now if you make the same reasoning on the principal pulley we got the nice expression:

    [tex] a_3 = g \frac{M_3-\frac{4M_1M_2}{M_1+M_2}}{M_3+\frac{4M_1M_2}{M_1+M_2}} [/tex]
    or, simpifying

    [tex] a_3 = g \frac{M_1M_3+M_2M_3-4M_1M_2}{M_1M_3+M_2M_3+4M_1M_2} [/tex]

    Plug in the masses and got the acceleration of 3.

    Play with the final formula.
    What if M1 = 0 ?
    What if M1=M2 ?
    What if M3 = 2*M1 = 2*M2 ?
    Imagine the pulley moving while masses go up and down.
    Try make real experiments as long as it is possible with short time and money.
    Have fun.
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