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Confusion regarding the concept of acceleration along a component

  1. Aug 9, 2013 #1
    Suppose we have a wedge with angle θ on which there is a block. We are pushing the wedge with the least force required to keep the block stationary. If the acceleration of wedge is A then the acceleration of the block along the inclined plane will be gsinθ+Acosθ=0. That means A is negative.

    How is this possible physically that we apply a force and the body is accelerating negatively in that direction?
     
    Last edited: Aug 9, 2013
  2. jcsd
  3. Aug 9, 2013 #2

    PhysicoRaj

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    A is negative, relative to the block ( as you are calculating the acceleration of the block). As you apply the force on the wedge, the block tends to slide down,(as far as I have inferred, the sloping surface of the wedge faces the source of force) which means it accelerates in the opposite direction. Or, standing on the block, you can say the wedge is accelerating in the direction opposite to the block.
     
  4. Aug 9, 2013 #3
    thank you for replying. please reread my post. i edited it. made my question more clear.
     
  5. Aug 9, 2013 #4

    Nugatory

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    It's not. You've either
    1) been inconsistent in your convention for deciding which is the positive direction: left to right or right to left. Are you seeing the wedge accelerating right to left, while your x-axis has increasing values of x to the right? If so, both the force and the acceleration are negative (in the direction of decreasing x).
    2) used the force that the block exerts on the plane, when you should have used the equal and opposite force that the plane exerts on the block.
     
  6. Aug 9, 2013 #5
    i have taken the x axis parallel to the incline of the wedge.
     
  7. Aug 11, 2013 #6

    PhysicoRaj

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    Yeah. I had considered that.
    O K then...Check the equation again. Acosθ and gsinθ are in opposite direction. If one is towards the positive X-Axis direction then the other is in the negative.
    The resultant acceleration must be zero.
    That is,
    Acosθ+(-gsinθ)=0
    Acosθ=gsinθ
     
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