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Regarding a block and an inclined plane

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Person pushing a block with mass m on an inclined plane with angle [theta]. The friction coefficient is u. If the force applied is constant and parallel to the slope of the plane, show that the acceleration ax of the block in the direction of motion can be expressed as :
    ax=(P/m)-g(u*cos[theta]+sin[theta])


    2. Relevant equations
    Force = P
    Mass = m
    Gravity = g
    Friction coefficient = u
    Force = mass*acceleration


    3. The attempt at a solution
    First I drew the free body diagram, but I'm seriously lost coming up with equations. My book has nothing but vague allusions to formulae (with no examples whatsoever), and the internet is rife with blocks sliding down planes, but barren regarding blocks being pushed up them.

    Furthermore, the problem does not give information concerning the distance the block is pushed, nor does it include that measure in the "final answer" I'm supposed to solve for, but I thought distance was essential to measuring acceleration.

    Anyways, I'm lost here, folks. My physics acumen is lacking, to say the least, and the fact that my book asks these little "concept" questions while seeming deliberately wordy and mystifying only serves to frustrate me further.

    Help please.
     
  2. jcsd
  3. Oct 28, 2009 #2

    sylas

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    Science Advisor

    The distance doesn't matter here. You have three forces acting on the block. One is the force of gravity. One is the force from the plane. One is the force of the person pushing. All of these add up to a single total force that gives a certain acceleration to the block.

    What do you know about the co-efficient of friction? Have you got any equations for that?

    Cheers -- sylas
     
  4. Oct 28, 2009 #3
    Ok, I've got the force of gravity equaling m*(-g)*sin[theta]. Friction doesn't apply to that one, so no "u".
    Then, pushing force = u*mg*cos[theta].
    Finally, force of the plane = P*(u*m).

    If I add them together (?) I get P(u*m)+(m*g)(ucos[theta]-sin[theta])

    I seem to have the last part correct, only everything before it is off. There is undoubtedly an error in the equations I just listed, but they're all I have. I guess an explanation of a little theory here might help me. Why am I subtracting the g(ucos - sin) part, and how do I get there? How do I get rid of the u in the first part?

    I think I'm using equations for work instead of equations for forces, which might explain the extra variables.
     
  5. Oct 28, 2009 #4
    you seem to have mixed up the pushing force and the force of the plane.
    The pushing force is just P.
    The size of the force of the plane (I would call it friction force) is indeed u*mg*cos(theta). It's working against the direction of movement, so it needs a minus sign.
    If you add them and use F=ma you get the right answer
     
  6. Oct 28, 2009 #5
    Thank you, sylas and willem! I understand it now.
     
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