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Homework Help: Two-Mass Pulley System

  1. Jul 15, 2017 #1
    1. The problem statement, all variables and given/known data
    In the figure shown if the system is released from rest, acceleration of block B will be
    (a). g/2
    (b). g/3
    (c). 2g/3
    (d). g/7
    2. Relevant equations
    ma= F[net]
    taking the direction of motion as positive

    3. The attempt at a solution

    So my attempt was to first find tension in the string connected to Mass 3M from
    3Ma= 3Mg-T
    T= 3M(g-a)

    now the tension in the second string connected to the M say T1 is half of T ( am i correct?)
    T1= T/2= 3Mg/2 - 3Ma/2

    now equation of motion for the mass M

    Ma= T1-Mg
    Ma= M(3g/2 -3a/2 -g)

    a = 3g/2 - g - 3a/2
    a + 3a/2 = 3g/2 -g
    5a/2= g/2
    ∴ a= g/5

    where did i go wrong , because i looked up the answer and it is g/7
    please help...
  2. jcsd
  3. Jul 15, 2017 #2
    Acceleration of block B is not 'a' i.e it is not equal to that of block A .

    But they have a simple relationship .
  4. Jul 15, 2017 #3
    can you please tell me the relationship between the two accelerations?
  5. Jul 15, 2017 #4
    If the lower pulley goes down by a distance 'x' how much does block B go down ?
  6. Jul 15, 2017 #5
    since the string isn't directly connected to the pulley so they won't travel the same distance, the string goes along the circumference of the pulley, this is as far as i can think sorry i'm really bad at this type of thinking
  7. Jul 15, 2017 #6
    No problem .

    Suppose you hold block B i.e you do not allow it to move .Now let lower pulley go down by distance 'x' , how much string length of string whose one end is fixed at ground and other end connects block B , goes slack/loose ?

    Hint: Carefully look at both the left and right parts of string going over lower pulley .
  8. Jul 15, 2017 #7
    its half the circumference (πr) so if the pulley moves up say x distance the mass B will move πr +x ?
  9. Jul 15, 2017 #8

    You do not need to worry about the part that lies on the pulley circumference . Why ?
    Because before pulley moves down , πr length is over the pulley and after pulley moves down , same length πr is on the pulley circumference , so net change in the length of string as far as part that lies on the string is zero .

    So forget about the part of string that lies on pulley circumference .

    Just look at left and right parts of string .Pick a pen and paper and make a rough sketch . Without moving block B , move lower pulley down by a distance 'x' .

    How much string length of left part gets loosened ?

    How much string length of right part gets loosened ?

    What is the total length of string that gets slack/loosened ?
  10. Jul 15, 2017 #9
    hahah i get it now 2x

    so acceleration of block B will be two times the accelaration of the pulley

    i have done the calculation using this result and now the answer is g/7 ...... thanks for your help , i really appreciate it
  11. Jul 15, 2017 #10
    Well done !
  12. Jul 15, 2017 #11


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    Is this what you get for the acceleration of block B, or is it the acceleration of block A?
  13. Jul 15, 2017 #12
    It's for block B
  14. Jul 15, 2017 #13


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    Maybe I'm making a mistake, but I get that block A has the acceleration of g/7.
    Do you agree with the following equations?

    3Mg - TA = 3MaA

    TB - Mg = MaB

    TA = 2TB

    aA = aB / 2
  15. Jul 15, 2017 #14
    Are you taking the acceleration of the block A and B as the same?
  16. Jul 15, 2017 #15


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    No. See the last equation I wrote.
  17. Jul 15, 2017 #16
    The acceleration of block B will be two times that of block A while tension of Block B will be half that of block A
  18. Jul 15, 2017 #17


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    Yes, that agrees with the 3rd and 4th equations that I wrote.
  19. Jul 15, 2017 #18
    I have done the math again ... And the result is

    Acceleration for block A is g/7 and for B is 2g/7


    I made a mistake in a my previous math
  20. Jul 15, 2017 #19


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    OK, good. That's what I get, too.
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