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B Net force on two books

  1. May 15, 2017 #1
    This is an example given in my textbook:

    "On a frictionless horizontal surface, you push with a force ##\vec{F}## on a book of mass ##m_1## that in turn pushes on a book with a mass ##m_2##. What force does the second book exert on the first?"

    First they calculate the acceleration of book 2. This is how they do it:

    "The total mass of the two blocks is ##m_1 + m_2##, and the net force applied to the combination is ##\vec{F}##.

    I want to make sure that I understand correctly why the net force is ##\vec{F}##. Here is my explanation:

    I originally push book 1 with a force of ##\vec{F}##. Because of this, book 1 starts to accelerate in the direction of my push and thus exerts a force, say ##\vec{F_{12}}## on book 2, and by Newton's third law, book 2 then exerts a force, say ##\vec{F_{21}}## on book 1. Thus the net force of the system is ##\vec{F} + \vec{F_{12}} + \vec{F_{21}} = \vec{F} + \vec{F_{12}} - \vec{F_{12}} = \vec{F}##. Is my reasoning correct?
     
    Last edited: May 15, 2017
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  3. May 15, 2017 #2

    Nugatory

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    That explanation works. You've just demonstrated the general principle that all the internal forces have to cancel; if A exerts a force on B then B must exert an equal and opposite force on A so forces between A and B cannot contribute to the net force on the system made up of A and B. Do a few more of these problems and you'll find yourself effortlessly recognizing and ignoring the internal forces.

    Of course you will still need the internal forces when it comes time to apply ##F=ma## to each individual object.
     
  4. May 15, 2017 #3

    BvU

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    Did someone tell you the force ##\vec F## is parallel to the surface ? Because you sure didn't tell us !

    Yes. Gravity and the normal force from the surface cancel each other.
     
  5. May 15, 2017 #4
    Sorry, ##\vec{F}## is parallel to the surface. Let me ask a follow up question now. Suppose the normal force didn't exist in this problem, so that there was nothing to cancel out gravity. This would not affect any of the calculations that I made above because gravity is acting perpendicularly to the forces in question. Right?
     
  6. May 15, 2017 #5

    BvU

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    Somewhat artificial: the net force would have a component downwards (the books would start accelerating that way too....).

    And: the books would not stay together without glue or velcro :smile:
     
  7. May 15, 2017 #6
    Ah, that makes sense - if ##(F_x, 0)## is the force I apply with my hand, then ##(F_x, -g)## would be the net force if the normal force did not exist - it's just that gravity wouldn't have any affect on the horizontal distance traveled by the books.

    Is that last point right?
     
  8. May 15, 2017 #7

    BvU

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    $$
    (F_x, -{\bf m_1}g)$$on book 1 (which is on top of -- or underneath ? -- book 2). And it has no way to exert a force on book 2, so that experiences ##(0, -m_2g)## only
     
  9. May 15, 2017 #8
    It is beside book 2. Let me ask a question is more general: I know that if acceleration is constant, then the component of acceleration in one direction has no effect on the motion in a perpendicular direction. To me, this would remain true even if acceleration was not constant. Am I right?
     
  10. May 15, 2017 #9

    BvU

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    Next time post the complete problem statement; can avoid endless misunderstandings :rolleyes:
    As long as they are independent, yes. Comes in handy when projectile trajectories are considered.
     
  11. May 15, 2017 #10
    Would Galileo agree with this?
     
  12. May 16, 2017 #11

    BvU

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    We can't ask him any more. But I'm sure he would have appreciated a nice (sk)etching of the situation !

    In my defence: that the books were beside each other instead of piled up wasn't revealed at that stage of the thread ... :wink:
     
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