1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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)?

    2. Relevant equations

    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=(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
  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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted