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2 blocks connected by massless string VS. pulley system

  1. Jun 24, 2015 #1
    (Note: The title would better describe the subject if written like this:

    "2 blocks on a horizontal frictionless surface connected by a massless string vs. pulley system")


    Hello, I have a problem regarding pulling two blocks of different masses connected by a string horizontally with opposing forces and evaluating the tension. Then, I would like to find out if this problem is set in such a way that it is analogous to two blocks on a pulley.

    1. The problem statement, all variables and given/known data



    If you pull to the left with a force such that if Block A were an isolated mass, it would accelerate 9.8m/s^2 ..and you pull to the right of Block B with such a force that if Block B were an isolated mass, it would accelerate 9.8m/s^2.

    Note: Let mass of block 1 be A and mass of block 2 be B. So m1 = A and m2 = B.

    Question 1: In what direction would the system accelerate?

    Question 2: What is the net force of the system?

    Question 3: What is the acceleration of the system?

    Question 4: What would be the tension in the massless cable connected to:
    1. the internal portion of A (T1 on to m1)
    2. the internal portion of B (T1 on to m2)

    Question 5:
    If you were to take the internal cable of this system and drape it over a pull, with Block A hanging down on the left side of the pulley and Block B hanging down on the right side of the pulley, would your values for tension change?

    2. Relevant equations

    F=ma

    3. The attempt at a solution



    Answer 1
    Then the block B would require more force:

    FA = [A(a) = 10kg(9.8m/s^2) = 98N] < FB = [B(a) = 100kg(9.8m/s^2) = 980N]

    Answer 2
    So the net force on the system would be Fnet = FA + FB = B(a) - A(a) = 980N - 98N = 882N

    Answer 3
    The acceleration would be F/(A+B) = 882N/110kg = 8.0181m/s^2

    Answer 4

    For Tensions,

    First establish net force of each block:

    Fnet_A = A(a) = 10kg(8.018m/s^2) = 80.18N
    Fnet_B = B(a) = 100kg(8.018m/s^2) = 801.81N,

    Then Tension in strings are:

    Tension of left most string on Block A

    FA = 10kg(9.8m/s^2) = 98N = T1/A

    Tension of internal string on Block A

    FA = T2/A - T1/A ==> T2/A = F1 + T1/A = A(a2) + A(a1) = A(a2+a1) = 10kg(9.8+8.018) = 178.18N

    Tension of internal string on Block B

    FB = T3/B - T2/B ==> T2/B = T3/B - F2 = B(a1) - B(a2) = B(a1 - a2) = 100kg(9.8 - 8.018) = 178.2N

    Tension of right most string on Block B

    FB = 100kg(9.8m/s^2) = 980N = T3/B

    Question 5

    I think that if you were to drape the internal string over a pulley, the tensions in the string would remain exactly the same because gravity, at 9.8m/s^2, would be pulling on each box just as the blocks are being pulled from opposite directions horizontally.

    Thank you
     

    Attached Files:

    Last edited: Jun 24, 2015
  2. jcsd
  3. Jun 24, 2015 #2

    haruspex

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    That all looks good to me.
     
  4. Jun 24, 2015 #3
    Great! Thank you. So it would be true that if you put the string on a pulley with the same masses, the system of blocks would accelerate downward at 8.018m/s^2 on the side of the pulley with the bigger block?
     
  5. Jun 24, 2015 #4

    haruspex

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    Yes.
     
  6. Jun 26, 2015 #5
    Thank you
     
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