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Confusion over an Inclined Plane Question

  1. Oct 12, 2007 #1
    1. The problem statement, all variables and given/known data

    A box of textbooks of mass 25.0 kg rests on a loading ramp that makes an angle "a" with the horizontal. The coefficient of kinetic friction (muk) is 0.230 and the coefficient of static friction (mus) is 0.340.
    a) As the angle "a" is increased, find the minimum angle at which the box starts to slip.
    b) At this angle, find the acceleration once the box has begun to move.
    c) At this angle, how fast will the box be moving after it has slid a distance 5.30 along the loading ramp?

    2. Relevant equations/Attempts

    I have solved a), with an answer of 19 degrees, and my answer for b) does not seem to be right although I can't see where I went wrong. I cannot complete c) till b) is finished.

    for b):

    F= ma
    mgsin"a" - (muk)mgsincos"a" = ma
    therefore:
    a = g*sin"a" - (muk)*g*cos"a"
    and I get a = 1.059 which is wrong.

    Any insight is appreciated.
     
  2. jcsd
  3. Oct 12, 2007 #2

    Doc Al

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    Good. (I assume the "sincos" was a typo.)
    Try using a more accurate value for the angle.
     
  4. Oct 12, 2007 #3
    - Thank you. sincos was an error. It was meant to be just cos.

    The questions asks me to express myself to two significant digits, and when I do whether I input 19 degrees or 18.77803322 degrees, I get the same, "1.0". Do you know what it could be?

    Thanks, Ty.
     
  5. Oct 12, 2007 #4

    Doc Al

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    Beats me. Your answer is correct.
     
  6. Oct 12, 2007 #5
    Bugger of a question, ain't it?

    Is there anyway I can do c) without using acceleration? If I do, I am going to assume that a = 1.0 since I, myself, also can't see anything wrong with it. But you'd know better. What equation can I then use.

    Thanks, Ty.
     
  7. Oct 12, 2007 #6

    Doc Al

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    You can use energy methods, but they end up using the same values as you calculated for acceleration--so don't expect anything different.

    I would use the more accurate value for acceleration and just round off your answer at the end.

    (Unfortunately, these online systems can be flaky. Don't think this is the first time that such puzzlers have been reported.)
     
  8. Oct 12, 2007 #7
    That's why I like the good old fashioned "write-it out" technique.

    If I was going to find d then, can I use:
    vf^2 = vi^2 +2ad
    or would it be wrong to assume I started from rest?
     
  9. Oct 12, 2007 #8

    Doc Al

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    Yes.
    That would not be wrong--assume it starts from rest.
     
  10. Oct 12, 2007 #9
    Thank you, Doc Al. Weirdly enough it accepted my answer for d, even though I used 1.0 as the acceleration.

    Would you mind if I asked you another question?
     
  11. Oct 12, 2007 #10

    Doc Al

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    Ask away.
     
  12. Oct 12, 2007 #11
    Thank you again.

    < http://people.hws.edu/faux/Phys1/practice2.pdf >

    Question number 5:
    5. The 4.00 kg block in the figure to the right (see link) is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are
    extended as shown in the diagram and the tension in the upper string is 80.0 N.
    a) What is the tension in the lower cord?
    b) How many revolutions per minute does the system make?
    c) Find the number of revolutions per minute at which the lower cord just goes slack.


    I have T figured out to 30, but I am stuck on finding b) and c). I know I need to find a, to get v to get T but I don't know how to get a.

    I have tried:
    F=ma
    79 - 30 - 39.2 = 4*a
    a=2.45
    then, a = v^2/R
    v= 1.355 544 171
    then, T = (2*pi*R)/v
    T= 3.4576 381 723 rev/s
    = 208.582 903 4 rev/min
    which I know is too high. What am I doing wrong?

    When I do c)
    F=ma
    79 - 0 - 39.2 = 4*a
    a=9.95
    then, a = v^2/R
    v= 2.731 757 676
    then, T = (2*pi*R)/v
    T= 1.725 039 165 rev/s
    = 103.502 349 9 rev/min
    This seems to low compared to b).

    By the way, I have used R=0.75. I did calculate that right, right?
     
  13. Oct 12, 2007 #12

    Doc Al

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    The acceleration you need is the centripetal acceleration, which in this case is horizontal. What horizontal forces act on the mass?

    Forces are vectors--direction counts. Treat horizontal and vertical components separately. Once you straighten this out you can redo parts b and c.
    Right.
     
  14. Oct 12, 2007 #13
    To find centripetal acceleration is a = v^2/r, but I do not have v. I do know that v and a are at perpendicular to one another though. Right?
    I know centripetal force (F= (mv^2)/R) is a horizontal force acting on the mass. By the way, what about centrifugal "force"; does it play a role?


    I thought my signs where correct; may I ask how I was mistaken? No insult meant, but you probably right. As for the treating horizontal and vertical components separately, how can I set it up? I don't know what I am going to resolving for, but I am going to play around with it now.
     
  15. Oct 12, 2007 #14

    Doc Al

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    Since the mass is whirling in a circle, you know it's centripetally accelerating. What forces are producing the centripetal acceleration? (That's why I asked: What horizontal forces act on the mass?)

    Once you find the net force towards the center, you'll set that equal to the formula for centripetal force.

    No. Centrifugal force is a "fictitious" force that arises when analyzing things from a rotating reference frame. Not relevant here. (You could use it to solve the problem. Don't.)

    FYI: "Centripetal" means "towards the center" and "centrifugal" means "away from the center".


    The problem is not just signs. There are three forces acting on the mass: the two tensions and gravity. Find the net horizontal force acting on the mass. (Add up the horizontal components of each force.)
     
  16. Oct 13, 2007 #15
    I understand now! I took what you told me, about my signs and horizontal forces, and I got my answer of 44.5 revolutions per minute, and then for part c) I and got 29.9 rev/ min. I appreciate all the help you gave me, man. Thanks, Ty.
     
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