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Homework Help: Need help really quick.

  1. Oct 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A bos of mass m2= 3.5 kg rests on a frictionless horizontal shelf is attached by strings to boxes of masses m1= 1.5 kg and m3= 2.5 kg, which hang freely. Both pulleys are frictionless and massless. The system is initially held at rest. After it is released, find (a) the acceleration of each of the boxes, and (b) the tension in each string.

    If you are confused about the graph, imagine the the m2 is in the middle with 2 strings on 2 sides, at 2 ends of the shelf there are 2 pulleys and m1 hangs on the left of the shelf and m3 hangs on the right of the shelf.

    2. Relevant equations


    3. The attempt at a solution

    I tried to get the Tension at rest but it not gonna do anything with the final answer.
  2. jcsd
  3. Oct 21, 2008 #2
    Guys, I need some help really quick
  4. Oct 21, 2008 #3
    Ok, this is rather long.

    First you need to find the acceleration of the masses. The acceleration of all the masses is the same. So, for now look at the three masses as one big box (added them up to together). Then draw the forces pulling them each way. (There should be two).

    Then use F= ma

    subtract the smaller force from the larger force, because the force is net force, and solve:

    so it should be set up like this:

    winner force- loser force = mass(of entire big blog) * a
  5. Oct 21, 2008 #4


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    You are going to have to show some work. You will need to identify the forces acting on each block, and apply Newton's 2nd law to each. Are you familiar with free body diagrams?
  6. Oct 21, 2008 #5

    Draw the diagram

    draw a free body diagram for each box (remember the tensions in each string may be different).

    If the strings are taut and unstretchable, what can you say about the accelerations of each body?

    Now you should have lots of equations! Solve!
  7. Oct 21, 2008 #6
    How do I find the acceleration ? Does it involve tension here ??
  8. Oct 21, 2008 #7
    I did draw FBD for each box and creates some 2nd law equation:


    What do you guys thinks ?
  9. Oct 21, 2008 #8
    I posted how to find the acceleration. forget about the tension until you find the acceleration. (you need acceleration to find the tension)
  10. Oct 21, 2008 #9
    I mean how do you find the force since we don't know a ?
  11. Oct 21, 2008 #10
    you can't
  12. Oct 21, 2008 #11
    so how can I use your formula:
    winner force- loser force = mass(of entire big blog) * a

    to find a ?
  13. Oct 21, 2008 #12
    find a, then find tension. It's the only way.
  14. Oct 21, 2008 #13
    I know that so how do you find a ? Which formula do you use ?
  15. Oct 21, 2008 #14
    winner force- loser force = mass(of entire big blog) * a

    look at my first post
  16. Oct 21, 2008 #15
    That's what I thought you wrote. So I can find the mass. How about the winner force and loser force ??
  17. Oct 21, 2008 #16
    the winner force and the loser force come from the boxs hanging down. Since they are hanging, they have weight, so: mg
  18. Oct 21, 2008 #17
    so we ignore the tension there ?

    Now I get it. Sorry I am a little too dumb tonight.
  19. Oct 21, 2008 #18
    it's alrgiht. Tell me when you get the accleration, and then I'll help with the tension. (there are two)
  20. Oct 21, 2008 #19
    I got 1.31 m/s^2. Is that right ?

    Now what's next ?

    Do we use 2nd law again as:

    T1-m1g= m1a ?

    m3g-T3=m3a ?
  21. Oct 21, 2008 #20
    sorry, I actually have not done the calculations. but the set up for the next part looks right!
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