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Acceleration of 2 boxes sliding down ramp - need help please!

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

    Please refer to the diagram. I have having trouble with B: "Determine the acceleration of the system. Explain your method, and identify and laws you are applying"

    Ok so for A:, I drew two FBD's. For package A there is Fg, Fnormal, Ffriction, and F from box B - Correct?

    Box B - Fg, Fnormal, Ffriction, and FfromboxA - Correct?

    So now B is where I am having trouble. I am not sure how to find the acceleration of the SYSTEM ?


    seub1d.png


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 11, 2011 #2

    Doc Al

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    Staff: Mentor

    OK. (Hopefully you drew the directions correctly.)

    Newton's 2nd law. (Analyze force components parallel and perpendicular to the incline.)
     
  4. Oct 11, 2011 #3
    Well, box A in front has a higher coefficient of friction, and so it will always be moving slower than box B. So the motion of box B is affected by the motion of box A, and vice versa (i.e. box B is pushing on box A because box A won't get out of its way).

    So the motion of the system is really determined by the motion of box A or B, since they move together. When you draw the FBD of box A, first assume that box B isn't there, and include all relevant forces, then add in the force box B exerts on A, and that should describe the motion of the system.
     
  5. Oct 11, 2011 #4
    Ok I see what you are saying.

    I am just having trouble figuring out the acceleration. What method do I use to figure this out?
     
  6. Oct 11, 2011 #5
    The acceleration of either box A or B is the acceleration of the system - they are both moving with the same acceleration.
     
  7. Oct 11, 2011 #6
    Yes I get that... since they are moving together, they have the same acceleration.

    BUT - how do I figure out the actual acceleration?
     
  8. Oct 11, 2011 #7
    You need to write your equations of motion for each box. After doing this what do you have in the inclined direction? Remember Newton's third law...
     
  9. Oct 11, 2011 #8
    so my first step would be to choose a cordinate system, and then delevope an equation?
     
  10. Oct 11, 2011 #9
    Yep. This is the case for tackling almost any problem. So what do you get?
     
  11. Oct 11, 2011 #10
    Give me a sec... We just started doing these (FBD's to equations) and IM not 100% comfortable yet doing them.

    Ill post it in a sec here.
     
  12. Oct 11, 2011 #11
    Ok drew it out, with coordinate system, and a table. Not sure if this is correct. Can yuou look over my table?

    t6v15e.jpg
     
  13. Oct 11, 2011 #12
    Draw them out as separate and draw Free body diagrams then equations for each block. Your table looks ok for the entire system but you should label 'Fgravity' as 'Fparallel' or something like that indicating that it acts parallel to the ramp. The value of 'Fgravity' is just 'mg' and it acts vertical. On the other hand 'Fparallel' and 'Fperpendicuarl' are the components of 'Fgravity' in the x and y directions.
     
  14. Oct 11, 2011 #13
    By the way you can solve for the acceleration of the blocks by solving it as a system vs writing equations for each block. But then you will still have to write and equation for one of the blocks to find the force 'F' that acts between them...
     
  15. Oct 11, 2011 #14
    So, if I go the system route, is my table correct?

    To add to my table the ma of x would be what? and what would the ma of y be?
     
  16. Oct 11, 2011 #15
    Can anyone confirm my table is correct?

    what is the ma of x and the ma of y

    Still dont know how to find those :(

    Thanks so much guys!
     
  17. Oct 12, 2011 #16

    SammyS

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    How are you able to find the frictional force on the system?
     
  18. Oct 12, 2011 #17
    Its just the sum of the friction forces for both blocks. I was not 100% sure how the system approach would work but it seems that if you do it as a system it is the same as if you wrote an equation for each block and added them together canceling out the interaction force 'F' between them.
     
  19. Oct 12, 2011 #18
    Now for C

    Now package A and package B are sliding down a ramp together, so they have the exact same acceleration. The acceleration of the system is 1.8 m/s^2 and the mass of package B is 10kg

    So, is it simply F = ma

    F = (10kg)(1.8) = 18 N ?
     
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