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I need help with my problem

  1. Nov 21, 2007 #1

    dur

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    I need help with my problem!!!!

    a box slides down a frictionless plane that is sloped 20 degree above the horizontal. Find the acceleration of the box.
    can any one help me to solve this problem
     
    Last edited by a moderator: Nov 21, 2007
  2. jcsd
  3. Nov 21, 2007 #2

    dur

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    thank you
     
  4. Nov 21, 2007 #3
    This problem is quite generic, uses Newton's laws. Do you have any formulas in your mind that could help you solve this problem? You need to show effort so the PF members can assist you. Forum rules.
     
  5. Nov 22, 2007 #4

    dur

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    thank you so much Siracuse. The formulas I have on my book a=f/m
    but how can i solve ?? and I have no force and mass.
    thank you
     
  6. Nov 22, 2007 #5
    Think about it...a mass at rest in a plane has forces acting in it. You have to assume there is a gravitational force, Earth's gravitational force to be more specific. There is also another force acting on the mass, called normal force.

    Did you learn all of Newton's laws yet?
     
  7. Nov 22, 2007 #6

    dur

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    Thank you Siracuse. because the box plane that is sloped 20 degree above the horizontal I think, I have to use cos.
    cos(20)=.940
    .940(-9.8m/s)=-9.212
    I dont know if that's correct.
    yes I learn all of Newton's laws.
     
  8. Nov 22, 2007 #7

    Shooting Star

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    Tell us why you chose cos and not tan or sin, and why the minus sign in (-9.8)?
     
  9. Nov 22, 2007 #8
    Use f=ma

    So mgxsin 20-mgcos20xu=ma

    gxsin20-gcos20u=a

    but since its frictionless I would just use gxsin20=a or F=ma, no friction. Try that
     
  10. Nov 23, 2007 #9

    dur

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    Hello , shooting star and Silently loud. Thank you so much>>> both.
    I use cos, because the box sloped 20 degree horizontal.
    Silently I have no force how can I use F=ma?? I dont understand that.
    Thank you
     
  11. Nov 23, 2007 #10

    Shooting Star

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    Hi Dur,

    No, you should use sin, because mg*sin 20 is the component of the weight along the plane. (Draw a freebody diagram of the forces acting on the mass.)

    From F=ma, you get, mg*sin 20 = ma => a = g*sin 20.

    Don't use -ve sign for g.
     
    Last edited: Nov 24, 2007
  12. Nov 24, 2007 #11

    dur

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    Hi Shooting star,

    sin(20)=.342
    .3429(-9.8)=3.352

    I dont know if I have correct answeer.
    This problem confused me so bad:confused:
     
  13. Nov 24, 2007 #12
    that seems correct. This is a pretty standard problem that we call "Inclined Plane" problem. Try also looking at this webpage:
    http://www.lon-capa.org/~mmp/applist/si/plane.htm

    I can provide you with one other explanation, not sure how much you will like it:
    Basically, you know that gravity will act on the box. Gravity acts vertically downward. What you need to do in this problem though is to rotate your coordinate system from x,y coordinate system, to a coordinate system "perpendicular" and "parallel" direction of the plane. Then you can say that the box cannot move at all in the Perpendicular direction (because the plane is there!) but it can freely move in the parallel direction, up and down the plane. So now the problem becomes to find the parallel component of the force of gravity to the plane. This ends up being MgSin(20). At this point, the problem is EXACTLY equivalent to a problem where I simply push some box on the ground with a force MgSin(20). This, again, made possible by the fact that we rotated the entire world and "flattened" our plane by rotating the entire coordinate system. In this new rotated world, the force of gravity does not act exactly downward, but a little to the right(or left) as well. Hope that helps a little
     
  14. Nov 24, 2007 #13

    Shooting Star

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    Hi dur,

    Why are you using a negative sign for g?

    The correct ans is 3.352 m/s/s down the plane, which you have got. But I'm not very sure whether you understand how. I think you should read up a bit on forces and how to split them up into components. Then all this will become very easy for you.
     
    Last edited: Nov 24, 2007
  15. Nov 25, 2007 #14

    dur

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    Hi , and thanks alot.
    Shooting star, I usa -9.8 becuase the box slope down.
    Mephisto thanks you for the link you gave me, and I think slove the problem like above on replay #11.
    than you
     
  16. Nov 25, 2007 #15

    Shooting Star

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    Please use +9.8 because the box moves down.
     
  17. Nov 26, 2007 #16

    dur

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    Hi Shooting star,
    Howcom we use (-9.8) when the object go down?
    I'm a college student, I study college physics (seventh edition) so we use - 9.0 on chapter 3 when any object go down. Just check and tell me.
    Thank you
     
  18. Nov 26, 2007 #17

    Shooting Star

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    If you consider the y-axis to be positive in the upward direction, then the vector g is indeed -9.8 j.

    But in this problem, the downward motion along the plane was being considered positive. This is totally a matter of convention. You got a value of 3.352 m/s^2 down the plane, which is the same as -3.352 m/s^2 up the plane.
     
  19. Nov 26, 2007 #18

    dur

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    Hi Shooting star, you are correct. We solve the problem today.
    thank you so much.:smile:
     
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