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Pulley system on rough surface.

  1. Oct 21, 2016 #1
    1. The problem statement, all variables and given/known data

    Question :- Find the acceleration of block of mass ##M##. The coefficient of friction between blocks is ##\mu_1## and between block and ground is ##\mu_2##.

    free body diagram at the end.

    Variables :-
    ##f_1## - friction between blocks.
    ##f_2## - friction between block and ground.
    ##N_2## - Normal reaction between blocks.
    ##N_1## - Normal reaction between block and ground.
    ##T## - Tension in the string.
    ##a## - acceleration of the blocks since string is in-extendable.
    ##M < m##

    2. Relevant equations

    Since small block will move down, and big one right.
    $$ma = mg - (f_1 + T) \qquad (1)$$
    $$T - (N_2 + f_2) = Ma \qquad (2)$$
    $$N = Mg + f_1 \qquad (3)$$

    If big block moves right then, pulleys will move right and small block will move right. Thus big block is non inertial frame. So we add the adequate pseudoforces to balance out unbalanced forces.

    ##N_2 = ma \qquad (4)##

    3. The attempt at a solution

    ##(1) + (2)##

    $$a = mg - f_1 - N_2 - f_2 $$

    $$a = mg - N_2 - ( f_2 + f_1 )$$

    $$a = mg - N_2 - ( \mu_1 N_2 + N_1 \mu_2 )$$

    substituting ##(3)## and ##(4)##

    $$a(m + M) = mg - ma-(ma\mu_1 + \mu_2(Mg + f_1))$$
    $$a(2m + M + \mu_1) = mg - \mu_2(Mg + ma\mu_1)$$
    $$a(m(2 + \mu_1 + \mu_1 \mu_2) + M) = g(m - \mu_2 M)$$
    $$a = {g(m - \mu_2 M)\over (m(2 + \mu_1 + \mu_1 \mu_2) + M)}$$

    But the given answer is ##a = {[2m - \mu_2(M + m)]g\over M + m[5 + 2(\mu_1 + \mu_2)]}##.
    I think the given solution is incorrect, because i can't find any error in my calculations but i am not sure.
    Please check my free body diagrams in the picture below and my equations. I think any error in those would be sufficient for me to proceed ahead.

    My kindest Thanks for your time and help.

    asdaasf.png
     
    Last edited: Oct 21, 2016
  2. jcsd
  3. Oct 21, 2016 #2

    TSny

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    Is it correct to assume that the vertical acceleration of m equals the horizontal acceleration of M?
     
  4. Oct 21, 2016 #3

    haruspex

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    Did you mean those the other way around? It's a bit inconsistent with the other suffixes.
    Their horizontal accelerations are the same, clearly, but what about the vertical acceleration of m?

    TSny beat me to it.
     
  5. Oct 21, 2016 #4

    TSny

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    I believe you are missing some of the tension forces T that act on the big block M.
     
  6. Oct 21, 2016 #5
    I think vertical acceleration will also be ##a##.

    Let total length of string be ##x## and displacement of smaller block be ##u## and bigger block be ##p## in some time ##t##.
    So,

    $$u + p = x$$
    $${(du)^2\over dt^2} + {(dp)^2\over dt^2} = {(dx)^2\over dt^2}$$
    $$\text{vertical acceleration of m} + \text{horizontal acceleration of M} = 0$$
    $$\text{vertical acceleration of m} = -a$$
    if we take magnitude, then ##\text{vertical acceleration of m} = a##

    May you please point my mistake ?
    Sorry for inconsistency. I wrote what i mean, did not mean it other way.
     
  7. Oct 21, 2016 #6

    haruspex

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    There are two horizontal strings that change length, but only one vertical.
     
  8. Oct 21, 2016 #7

    TSny

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    I don't follow this. The string can be broken into four sections as shown
    upload_2016-10-21_15-51-45.png
     
  9. Oct 21, 2016 #8
    So should it be like this :-

    $$L_1 + L_2 + L_3 + L_4 = x$$
    $${d(L_1)\over dt^2} + {d(L_2)\over dt^2} + {d(L_3)\over dt^2} + {d(L_4)\over dt^2} = {d(x)\over dt^2} $$
    $${d(L_1)\over dt^2} + 0 + 2{d(L_3)\over dt^2} = 0$$
    $$\text{vertical acceleration of m} + 2\text{horizontal acceleration of M} = 0$$
    $$\text{vertical acceleration of m} = |2\text{horizontal acceleration of M}| $$
    $$\text{vertical acceleration of m} = |2a| $$
     
  10. Oct 21, 2016 #9

    TSny

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    Yes, that looks good. (Your notation for second derivatives is a bit off.)
     
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