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Force - boxes connected by strings

  1. Jul 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Three boxes are connected by strings of constant length. (A--B--C--> F) An applied force F acts horizontal in the positive x direction. mc=5kg, mb=2mc, ma=2mb. The coefficients of static and kinetic friction are 0.2 and 0.1 respectively and are the same for each box. The boxes are initially at rest.

    1) What is the magnitude of the maximum force that can be applied without causing the boxes to move?

    2) The magnitude of the applied force is varied until the boxes are moving to the right at a constant speed of 2m/s. What is the magnitude of the force when the boxes are moving at this constant speed?

    3) The magnitude of the applied force is varied until the boxes are moving to the right at a constant speed of 2m/s. What is the magnitude of the tension in the string connecting A and B?

    4) The applied force is adjusted so that the boxes accelerate to the right at a rate of 2m/s^2. What is the magnitude of the applied force?

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    mc=5kg
    mb=10kg
    ma=20kg
    mtotal=35kg
    us=.2
    uk=.1

    1) 20*9.8 + 10*9.8 + 5*9.8 = 343
    343*.2 = 68.6N

    2) and 3) I'm not really understanding. I know F=ma, but a isn't given; v=2m/s. And I don't even know where to begin with tension.

    4) F=35*2 = 70N

    Thanks for your help in advance.
     
  2. jcsd
  3. Jul 11, 2011 #2

    PhanthomJay

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    What is the value of a when the blocks are moving at a constant speed?
    Draw a Free Body Diagram of Block A, which is moving at constant speed.
    That's the NET force, not the applied force.
    Welcome to PF!:smile:
     
  4. Jul 11, 2011 #3
    "What is the value of a when the blocks are moving at a constant speed?"

    a at constant speed = 0m/s^2?

    "Draw a Free Body Diagram of Block A, which is moving at constant speed."

    fw=20 down, fn=20 up
    v=2m/s right, uk=.1 left

    that's about all i got.

    "That's the NET force, not the applied force."

    Fapp= 35*2 + 0.1*35 = 73.5?
     
  5. Jul 12, 2011 #4

    PhanthomJay

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    yes, so proceed as you did in part a, except change the friction coefficients.
    fw down is 20*9.8 N, fn up is 20*9.8 N. But what are the forces in the x direction? The velocity v = 2 m/s is not a force, and its value is not relevant in this problem. Only the fact that v is constant is relevant (a = 0).
    Almost correct, you again forgot to convert the total mass to total weight when calculating the friction force.
     
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