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Newtons Laws Multiple Objects

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Three crates are arranged. The masses of the crates are m1= 29.8kg, m2= 15.3kg, and m3=27.3kg. The applied force of 910 N is directed horizontally, and the coefficient of friction is u= .277. what is the acceleration of the three masses and the magnitude of the contact force between m1 and m2, and m2 and m3

    2. Relevant equations



    3. The attempt at a solution

    Weight of m1= 292.04
    normal force on m2 = 150
    FF on m1 = 80.9
    FF on m2 = 150
    Ff on m3= 74.1
     
  2. jcsd
  3. Nov 15, 2009 #2

    rl.bhat

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    FF on m2 is wrong.
    Net force = applied force - FF.
    Acceleration = Net force/ mass
     
  4. Nov 15, 2009 #3
    Thanks! Thats what I tried mathematically but resulted with the wrong answer becuase I didn't multiply M2 by the resistance..
     
  5. Nov 15, 2009 #4
    And what about the contact forces?
     
  6. Nov 15, 2009 #5

    rl.bhat

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    If T1 is the contact force between M1 and M2, them
    T1 - FF(M1) = M1a.
    Similarly you can find T2.
     
  7. Nov 15, 2009 #6
    ok so FF of M1= 80.89
    FF of M2= 41.5
    and the acceleration would equal 9.85m/s/s

    so T1- 122.4(29.8)= (29.8)(9.85) ? That seems way to ghih.
     
  8. Nov 15, 2009 #7

    rl.bhat

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    From where did you get 122.4?
    It should be
    T1 - 80.89 = 29.8*9.85.
    In the formula FF(M1) is the frictional force on mass 1.
     
  9. Nov 15, 2009 #8
    so T1 would equal 374.42 The question asks what is the contact force between m1 and m2. 374.42 does not answer any of these questions. I have a feeling it has something to do with the friction source but im not sure.. Thanks!!
     
  10. Nov 15, 2009 #9
    Additionally, I have another question regarding a different problem.
    Three crates are connected with ropes and arranged. The masses of the crates are m1= 20kg m2= 16.2kg and m3= 26.6kg. The crates accelerate to the right at 4.05m/s/sand the fiction is 0.279.

    what is the magnitude of the tension force on the rope between m1 and m2 = 293.22
    T= 2MA
    T= 2(16.2)(4.05
    T= 293.22N

    What is the magnitude of the tension force in rope 2 between m2 and m3 ( m2====m2=====m1)
    ???????????????????????????????????????????????
     
  11. Nov 15, 2009 #10

    rl.bhat

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    In the first problem it is not clear to which mass the force is applied. Direction of the force decides the contact force.
    In the second case how did you get T = 2MA?
    Show the arrangement of the masses and the direction of the force clearly.
     
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