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B Normal Force

  1. Jan 8, 2017 #1
    If 3 blocks have some mass(a,b,c) are kept on one another (Vertically i mean)....Why block a doesnt experience the normal force exerted by the ground on block c? wont that force get transfered through b to a?
     
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  3. Jan 8, 2017 #2

    PeroK

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    If there were no gravity and the normal force was an active force (someone pushing block c), then some of that force would be transferred through block c to the next block. But, the normal force is simply opposing gravity, so there is no net force on block c.

    Of course, some of the force is transferred as the normal force between blocks c and b.

    And, there is nothing special about the blocks, the same sort of internal forces are acting within each block. The bottom of each block is supporting all the block and the ones above it, whereas the top of each block is supporting only the blocks above it.
     
    Last edited: Jan 8, 2017
  4. Jan 8, 2017 #3
    At a position within a block, the weight supported will be between the two. wHAT DOES THATMEAN?
     
  5. Jan 8, 2017 #4

    PeroK

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    I meant in the middle of a block, the weight supported will be the top half of that block plus the blocks above it. There is nothing special about the contact points between the blocks. The same balance of forces is at play throughout each block, eventualy down to inter-molecular forces.
     
  6. Jan 8, 2017 #5

    Vanadium 50

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    It does. When you remove block C, don't the rest of the blocks start tumbling down?
     
  7. Jan 8, 2017 #6
    Isnt that because as C is removed so gravity acts on them? Morever if the force acts on A,Why dont we consider it while drawing the FBD of block A?
     
  8. Jan 8, 2017 #7
    Have you drawn free body diagrams of each of the three blocks, showing the forces acting on each, or do you feel that you have advanced to the point beyond which you no longer need to use free body diagrams? If you have drawn the free body diagrams, please write down for us the force balance equation that applies to each of the blocks.
     
  9. Jan 8, 2017 #8
    Assuming the mass of A is M1
    Mass of B is m2 and mass of C is M3...As these blocks are kept vertically and are rest

    The force acting on the uppermost A would be Gravitaional M1g downwards and the Normal force (N1)exerted by B on A (Equal to m1g) upwards


    Forces acting on B would be Normal (N1) Exerted by A on B (Third Law) Equal to m1g and gravitational force downwards equal to m2g and the normal (N2) exerted by block C on B ( N2= mig + m2g) upwards

    Forces on C would be N2 downwards (Thid law) and m3g as well as N3 upwards(Between ground and C)
     
  10. Jan 8, 2017 #9
     
  11. Jan 8, 2017 #10
    Excellent. Now you have 3 linear algebraic equations in the three unknowns N1, N2, and N3. Please solve for these three unknowns.
     
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