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    brijmohan123456's Avatar
    brijmohan123456 Posts: 41, Reputation: 1
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    #1

    Jan 6, 2013, 11:36 PM
    How third law works like this?
    Newton's third law says that to every action there is a equal and opposite reaction then please consider the my following two doubts:
    1.suppose a box of mass 2 kg is put on a table and then the floor of table goes down with a acceleration of 0.1meter/sec^2 here want to ask that as every action there is a equal and opposite reaction so when a force will be applied on table by box then an equal and opposite reaction will also be applied by table on box.So why does only table showed an acceleration.please clear my doubt.

    2.everyday in physivs we come across the problems of this kind " spring is going over a horizontally clamped pulley of negligible mass and supports a block of mass M etc and many other things"here I mainly want to ask that as the spring will be pulled down by force Mg due to mass M of block the same reaction will be given upward to spring as a tension,thus due to this there should not be any elongation to spring but it happens.Why is it so?
    please consider my doubt seriously and answer it ,the question may appear big but it is more big headache for me as my exams are coming. So please help
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Jan 7, 2013, 06:47 AM
    For 1, please clarify the situiation. You put a box on a table and then the table starts to sink, I assume with the box still on it, right? So the box feels a reaction force equal to its weight minus its downward acceleration = this force is in the positive (upward) direction:



    The table surface feels a downward force due to the weight of the box minus the box's downward acceleration:



    Hence and are equal and opposite.

    For (2), if a mass hangs from a spring then the spring si stretched if the other end of the spring is constrained. Once the spring is stretched to its equilibrium the tension T is equal to the weight of the mass hanging from it, and the point at the end of the spring where the mass is attached feels a downward force due to the mass's weight and also an upward force due to T. Because these forces are equal when the system is in equilibrium the forces cancel out and so the spring experiences 0 acceleration - but remember this is true only when the spring is in equilibrium (ut has reached its final elongation). Hence at this point the spring is stretched but not accelerating. Note that 0 acceleration does not imply 0 displacement. Actually prior to reaching equilibrium the tension T is less than the mass's weight (recall that the tension in the spring is equal to T=kx, so if x is small the tension is less than the mass's weight). So prior to reaching equilibrium the point at the end of the spring does experience a net non-zero force, and hence accelerates downward, thus increasing x, until such time as T equals mg.

    Hope this helps.

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