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Tension in of string between two boxes with friction.

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Two boxes are connected by a massless string. The first box's mass is 16kg and second box's mass is 21kg. You pull with a total force of 185N on the 16kg box. The coefficient of friction is 0.5. Calculate the tension in the string between the two boxes.

    2. Relevant equations

    Ffr=μFn
    Fnet = ma


    3. The attempt at a solution

    Thinking of the acceleration of the whole system (including friction)

    Fnet = Fa - Ffr = ma
    divide by m to get the acceleration of the whole system.

    a = (Fa - Ffr)/m

    m = 16+21 = 37kg

    Fa = 185N

    Ffr = μFn = μmg =(0.5)(37)(9.8)

    a = [185-(0.5)(37)(9.8)]/37 = 0.1m/s2

    Using this we should be able to calculate the Fnet of the 21kg box on the end of the string.

    Fnet=ma = (21)(0.1) = 2.1N

    Is the Fnet of the 21kg box equal to the tension of the string? I cant seem to get my head around this part. For me it doesnt make sense that even though I'm pulling on the whole system at 185N, only 2.1N is the tension in the string pulling on the 21kg box. Should I have taken the force of friction into account in calculating the Fnet of the 21kg box? Should I be adding or subtracting 185N from something?
     
  2. jcsd
  3. Oct 15, 2012 #2

    tiny-tim

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    welcome to pf!

    hi psychochef! welcome to pf! :smile:
    yes
    if the tension in the string was the same as the force with which you're pulling the front box, then (even without friction) the front box wouldn't move, would it? :wink:
     
  4. Oct 15, 2012 #3
    Awesome! It's good to know that I'm on the right track. So assuming I need to consider the force of Friction on the 2nd box (21kg), here's the calculations I get.

    Fnet = Fa-Ffr-T = ma
    solve for T:

    T= Fa-Ffr-ma

    Now should I use 185N as my Fa? I know that on the first box the applied force is the 185N pull, so knowing that the massless string should conserve that force should I put 185N as my applied force?

    If so, heres what I get:

    185-μFn-21(0.1)=T

    185-(0.5)(21)(9.8)-(21)(0.1)= 80N

    Does that look about right?
    Thanks!
     
  5. Oct 16, 2012 #4

    tiny-tim

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    hi psychochef! :smile:

    (just got up :zzz:)
    no, this is wrong

    draw a free body diagram for the second box …

    there are only two horizontal forces on it: tension and friction

    (the applied force is only applied to the first box … the question says so … it has nothing to do with the second box)

    try again :smile:
     
  6. Oct 16, 2012 #5
    I got it!

    T-Ffr = ma

    I solved for T, got 105N, that seems much better than 2.1 or 80N.
    Thanks a bunch!
     
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