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Two boxes

  1. Nov 21, 2011 #1
    The problem:

    You are lowering two boxes, one on top of the other, down the ramp shown in Figure 5.53 by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 15.0 cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.444, and the coefficient of static friction between the two boxes is 0.800. (a) What force do you need to exert to accomplish this? (b) What are the magnitude and direction of the friction force on the upper box?

    http://www.flickr.com/photos/53947633@N06/4988606991/


    Equations

    (2)w = mg
    (1)F = uk(m1+m2)g+us*m2g

    Solution:

    Well i added the two boxes masses to 80 kg and i got the angle to be 27.7 degrees
    (1)w = m(tot)*g = 784,8 N

    (2)F = uk(m1+m2)g +us*m2*g <==>

    F = 599.4 N

    The answer should be 57.1 N
     
  2. jcsd
  3. Nov 21, 2011 #2
    You are approaching the problem in the wrong way I think. Try resolving the forces in the problem into their components perpendicular and parallel to the plane, which would make life a lot easier for you.
     
  4. Nov 21, 2011 #3

    gneill

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    The crates are on a slope, so the forces will have components that are normal to the surface and parallel to the surface. Your force due to friction with the ramp, for example, depends upon the component of the total weight that is perpendicular to (normal to) the ramp.
     
  5. Nov 21, 2011 #4
    ƩFy = g(m1+m2)sinα

    ƩFx = μk(m1+m2)g*sinα

    F = (m1+m2)g(sinα-μkcosα)

    Is this right ?

    How can i find magnitude and direction of the friction force on the upper box ?
     
  6. Nov 21, 2011 #5
    The trig isn't correct, try drawing a free body diagram, it always helps.

    You have the frictional force, Fr, which is equal to μN, where N is the force normal to the plane, you have the force parallel to the plane due to the weight of the mass(es) and you have the force of the man pulling on the rope. The boxes are moving at a constant velocity, therefore forces up the plane must equal forces down the plane.
     
  7. Nov 21, 2011 #6
    I have plotted a free-body diagram, but how do i know which equation i should use ?

    ƩFx = Tcos27.7 (-fk)

    ƩFy = Tsin27.7 + n +(-w)
     
    Last edited: Nov 21, 2011
  8. Nov 21, 2011 #7
    Like I said, calculate the forces in the problem, from that, you should know which forces act up or down the plane, then equate them.
     
  9. Nov 21, 2011 #8

    gneill

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    attachment.php?attachmentid=41102&stc=1&d=1321888834.jpg

    I think his trig looks fine. Where do you see a problem?
     

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  10. Nov 21, 2011 #9
    Is
    ƩFy = g(m1+m2)sinα

    ƩFx = μk(m1+m2)g*sinα

    F = (m1+m2)g(sinα-μkcosα)

    the right way ?

    For the upper box:

    ƩFx = Tcos27.7 + (-fk) = 0 <==> Tcos27.7 = μkn

    ƩFy = Tsin27.7 + n +(-w) = 0 <==> n=w-Tsin27.7
     
  11. Nov 21, 2011 #10

    gneill

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    That looks fine. What answer does it give for F?
    For the upper box you won't know what the actual friction force is until you calculate the downslope component of the box's weight and compare it to the maximum static friction. This maximum is calculated just like for static friction: find the normal force and multiply by μs. What happens if the downslope component of the weight exceeds this maximum value?
     
  12. Nov 21, 2011 #11
    I got F to be 56,39 N

    fs = μn
    w1 = 470.88 N

    w*cosα*μs = 555,8 N tot (both boxes)

    w1*cosα*μs = 222.3 N (upper box)

    what am I doing wrong here ?
     
  13. Nov 21, 2011 #12

    gneill

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    That's just a little low. Are you carrying enough decimal places through your intermediate steps?
    Only the weight of the upper box is pressing down on the lower box. The lower box is only providing the surface. What's the weight of the top box? What's the normal component?
     
  14. Nov 21, 2011 #13
    F = (m1+m2)g(sinα-μkcosα) <==>

    F = (32+48)9.81*(sin27.7-0.4444cos27.7) = 56

    The upper box:

    313.92*cos27.7 * 0.800 = 222.3 N
     
  15. Nov 21, 2011 #14

    gneill

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    Keep a couple more decimal places in angle value: 27.759°
    Okay, that's the maximum value that static friction can be. Now, what's the downslope component of the upper box's weight?
     
  16. Nov 21, 2011 #15
    Im not sure

    But is it

    ƩFy = wCosα ?
     
  17. Nov 21, 2011 #16

    gneill

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    No, that's the normal component (that you just used to find the maximum friction). What's the component along the slope?
     
  18. Nov 21, 2011 #17
    Is it
    w*sinα <=>

    313.92*sin(27.7) = 145.9 ~146 N
     
  19. Nov 21, 2011 #18

    gneill

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    Yes it is.

    Again, you should keep additional digits in your angle value. This will prevent rounding errors interfering with your results.
     
  20. Nov 21, 2011 #19

    Forgive me if I'm being blatently dumb, but shouldn't that be cosine, and not sine?
     
  21. Nov 21, 2011 #20

    gneill

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    Yes, well, it looks like he made a typo in that particular line. Note that he wrote the correct expression for the force directly below:
    which was the point of the exercise.
     
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