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Force and motion problem

  1. Jun 26, 2015 #1
    1. The problem statement, all variables and given/known data
    In the figure here three ballot boxes are connected by cords, one of which wraps over a pulley having negligible friction on its axle and negligible mass. The masses are mA = 33 kg, mB = 45 kg, and mC = 14 kg. When the assembly is released from rest, (a) what is the tension in the cord connecting B and C, and (b) how far does A move in the first 0.300 s (assuming it does not reach the pulley)?

    http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c05/fig_5_E.gif
    2. Relevant equations
    F=ma

    3. The attempt at a solution
    http://s1294.photobucket.com/user/jhoversten/media/be59570a-8c6b-4612-aa51-52d1bd5d3a7d_zpsqc5pc7ce.jpg.html [Broken]

    Im not sure how to solve the simultaneous equations for T or a. my attempt at solving for T was incorrect.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jun 26, 2015 #2
    The first part of the question asks for the tension in the cord between blocks B and C. So, if you've run into trouble, you may want to put some of your initial assumptions under more scrutiny. Do you know for a fact that the tension in the cord is uniform throughout both segments or was that just a gut feeling?
     
  4. Jun 26, 2015 #3
    ah crap. yea the tension in the cord is not the same through out.
    I just understood the problem wrong.
     
  5. Jun 26, 2015 #4
    Live and learn :)
     
  6. Jun 26, 2015 #5
    im going to take another stab at this.
     
  7. Jun 26, 2015 #6
    So i need to find a for the whole system because its asking for tension after the system is released right?
     
  8. Jun 26, 2015 #7
    From what I've done, I am seeing that finding ##a## for the system will be beneficial, yes.
     
  9. Jun 26, 2015 #8
    So what is the difference between the tension between box B and A and the weight of B+A?
    T=ma
    T=(45+14)(-9.8)=-578.2N that is the same as the weight of box B and A? So then how do i find acceleration for the system?
     
  10. Jun 26, 2015 #9
    Have you written down Newton's 2nd law for the 3 masses?
     
  11. Jun 26, 2015 #10
    here what i have so far; FA=(33kg)(a), FB=572.8N - T, FC=mg + (14kg)(-a)
     
  12. Jun 26, 2015 #11
    FB= 572.8 + mg + TBtoC
    edit: i think mg and + TBtoC are the same = -137.2N
    edit again: nevermind they are different T in B=-5.4N
     
    Last edited: Jun 26, 2015
  13. Jun 26, 2015 #12
    how would I find the tension between C and B? if FC= mg + tension from B?
     
  14. Jun 26, 2015 #13
    no
     
  15. Jun 26, 2015 #14
    im having trouble finding the acceleration on the system because i am not getting the net forces correctly.
     
  16. Jun 26, 2015 #15
    FA=MA a
    FB=TB to A + TB to C + mg
    FC=TC to B + mg
    how does that look for newtons 2nd law for each box?
     
  17. Jun 26, 2015 #16
    Sorry, I had to step out for a while. When writing Newton's Second Law for a thing, just put ##m \vec{a} = \vec{F}^{(net)}## where the net force would include things like gravity and tension (not the total acceleration, ##\vec{a}## which appears on the left).
     
    Last edited: Jun 26, 2015
  18. Jun 26, 2015 #17
    This might be right depending on your notation! Are those tensions meant to be vectors or positive scalars?

    You may also find it more instructive to have ##m_{A,B,C}a## on the left side instead of ##F_{A,B,C}##.
    Edit: I jumped the gun a bit. Your equation for the forces on mass A isn't wrong, but go ahead and put the forces acting on it in the place of ##F_A##. The previous suggestion does apply to your second two equations though.
     
    Last edited: Jun 26, 2015
  19. Jun 26, 2015 #18
    the tensions are vectors, eg. TB to A = (45kg + 14kg)a -59a

    if ##m_{A,B,C}a## is on the left then what is on the right?
     
  20. Jun 26, 2015 #19
    Just so that I can be sure that you're sure, can you note your coordinate system (which directions are positive and negative) and write the three equations in scalar form?
     
  21. Jun 26, 2015 #20
    Sure, FnetA=33a, FnetB= -59a + 14a - 441, FnetC= -14a - 137.2
    standard coordinate system: up positive, down negative, right positive
     
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