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Newtons law of motion

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    The problem statement, all variables and given/known data
    the blocks shown in the figure are connected by flexible, inextensible cords passing over frictionless pulleys. at A, the coefficient fiction is 0.30 while at B, it is 0.40. compute the magnitude and direction of the friction forces and each block.

    2.jpg

    2. Relevant equations
    for block A=
    T-mgsin53.13-Fmax=0
    Fmax=0.3(300cos53.13)

    for block B=
    T-mgsin36.87-Fmax=0
    Fmax=0.40(200cos36.87)

    m(1)g-T=m(2)
    3. The attempt at a solution

    for A:
    Fmax=(o.30)(300cos53.13)
    Fmax=54

    T= 300(9.81)sin53.13
    T=2354.4

    for B :
    Fmax=0.4(200cos36.87)
    Fmax=54
    T=200(9.81)sin36.87
    T=1177.20

    m1g-T=m2a
    a=(2943-T)/200
    a=244.3

    AM i RiGHT ? please help thnx .. im too confuse
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 26, 2011 #2
    Hello!
    Are the blocks to remain static? Does this question keep them stationary?
    Daniel
     
  4. Sep 26, 2011 #3
    Then this
    is impossible, since a = 0.
    So try again, on that point alone.
     
  5. Sep 26, 2011 #4
    sorry it is kenetic..
     
    Last edited: Sep 26, 2011
  6. Sep 26, 2011 #5
    for A:

    Fmax=μk(m1)gcos53.13
    54=μk(m1)gcos53.13
    μk=0.03

    T-300(9.81)sin53.13-μk(300)9.81cos53.13=(300)a eq(1)
    200g-T=200a eq(2)

    subtitute (1) and (2) to find a

    a=-3.79

    for block B:

    Fmax=μk(m2)gcos36.87
    64=μk(m2)gcos36.87
    μk=0.04



    T-200(9.81)sin36.87-μk(200)9.81cos36.87=(200)a eq(1)
    300g-T=300a eq(2)

    subtitute (1) and (2)

    to find a

    a=0.2


    AM i DOiNG RiGHT ?? please comment .. im lack of knowledge ..
     
    Last edited: Sep 26, 2011
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