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The weight of two blocks

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data
    say we have a block resting on a incline of a wedge (the incline is 40 degrees). the wedge is on a scale(mass of block is .2 kg and mass of wedge is .8kg).if the block is at rest (due to to static friction) the scale would read 9.8N. but lets say we eliminate the friction on the wedge and the block starts to slide down the incline. the friction between the wedge and the scale is large enough to keep the wedge still on the scale.what would the scale read now?



    2. Relevant equations



    3. The attempt at a solution
    i somehow accidentally got the right answer to this (which is 8.99N) but i still dont wuite understand what happens when the block starts to slide. i know that the block would start pushing against the wedge as it goes down but i dont see which force is doing the pushing?
    can someone help me out with this
     
  2. jcsd
  3. Nov 19, 2008 #2

    PhanthomJay

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    It's the vertical component of the normal force between the block and the wedge that does the pushing. The scale reads the vertical normal force acting on it. you need to draw a free body diagram of the top block to look at what the normal force is between the block and wedge, then look at the vertical component of that force.
     
  4. Nov 19, 2008 #3
    thanks for the help
    one more thing how come the vertical component of the normal force doesnt affect the wedge when the blaock is standing still?
     
  5. Nov 19, 2008 #4

    PhanthomJay

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    Sure it does. The scale records a weight of 9.8N, which is the sum of the weights of both the wedge and block. There must be friction force on the block when it's still, and that vertical component of friction contributes also, along with the normal force vertical contribution, to the scale reading. But also in that at rest situation, you needn't consider the block's normal or friction force directly; just look at the system itself, it is in equilibruium, so the scale force must read the total weight of the block and wedge. In the other case where the block is sliding down the plane, the scale reading is less because the top block is not in equilibrium in the vertical direction.
     
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