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Pushing force

  1. May 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Two boxes have masses m1 and m2, and m2 is greater than m1. The boxes are being pushed across a frictionless horizontal surface. As the drawing shows, there are two possible arrangements, and the pushing force is the same in each. In which arrangement, (a) or (b), does the force that the left box applies to the right bo have greater magnitude, or (c) is the magnitude the same in both cases?

    image.png



    2. Relevant equations
    F=ma


    3. The attempt at a solution
    I can't see why the answer isn't (c) here. If the pushing force is the same in both cases and the area of contact is the same between the boxes. Can someone explain why it's different please?
     
  2. jcsd
  3. May 26, 2014 #2

    ehild

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    You can write some equations how the accelerations and forces are related...

    ehild
     
  4. May 26, 2014 #3
    F = (m1+m2)*a
     
  5. May 26, 2014 #4

    ehild

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    Write the equations separately for both boxes.

    ehild
     
  6. May 26, 2014 #5
    I'm confused. Newtons third law would say that for every force acting on an object there is an equal and opposite reaction force. But according to this, how can a force ever move an object anyware? Since it's always counteracted by an opposite force. I know this is a stupid question btw.
     
  7. May 26, 2014 #6

    ehild

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    That opposite force acts on the other object. In this problem, the left box exerts f force on the right box. According to Newton 3, the right box exerts -f force to the left box.
    In case a:

    What forces act on the left box?
    What force acts on the right box?

    ehild
     
  8. May 26, 2014 #7
    On left box: pushing force and reaction force of the right box (equal in magnitude to pushing force)
    On right box: pushing force only

    Is that correct?
     
  9. May 26, 2014 #8

    Doc Al

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    No. If the force of the right box on the left equaled the pushing force, then the net force on the left box would be zero and it couldn't accelerate.

    No. The only force acting on the right box is the contact force from the left box. But that force does not equal the pushing force.
     
  10. May 26, 2014 #9

    ehild

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    The pushing force F acts on the left box only. The right box is pushed by the left box.

    ehild
     
  11. May 26, 2014 #10
    How come they don't equal each other? The pushing force must be transmitted through the first box. And since the boxes are together, the force must be transmitted and given to the second box also. Where else would the force be lost on the way?

    How can we determine how big force is on the right box?
     
  12. May 27, 2014 #11

    ehild

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    The force is not "transmitted". The force is not conserved. The force causes acceleration. If more forces act to a body, they add up. The resultant force determines the acceleration.
    The boxes touch each other and exert force to each other. The left box feels the pushing force and the force from the other box. The right box feels the force from the left box. These forces of interaction are opposite and equal in magnitude, according to Newton's Third Law.

    ehild
     
  13. May 27, 2014 #12

    Nathanael

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    Let's say it's your hand is what is pushing on the boxes. The force on the boxes (from your hand) will be equal to the force on your hand (from the boxes). The equal and opposite force doesn't happen on the same object, it happens between the objects, (one force on the box, one (equal and opposite) force on your hand). Does this make sense?

    The key is to notice that there are two (pairs) of forces in the situation.
    The first is the force on the left box (and the equal/opposite force on whatever is pushing the boxes).
    The other is the "pressing" force between the two boxes. Box 1 pushes on Box 2, and Box 2 pushes on Box 1 (equally in the opposite direction).

    (Assume you know the pushing force and the masses:)
    The boxes accelerate together, (right?) and you know the acceleration of the boxes (pretend they're a single object and use F=Ma).

    Now stop pretending they're a single object and just look at one (either one, it doesn't matter.) If you know the acceleration of (either) one and you know the mass of (either) one, then you therefore know what the net force has to be, right? (Net force = Mass times Acceleration)

    So now the only unknown is how hard the right box pushes on the left box, and how hard the left box pushes on the right box (which will be the same, since they're equal and opposite).

    Do you understand how to solve for that ("pressing") force?
     
  14. May 27, 2014 #13
    I understand that when one object pushes another, theres one force from object 1 acting on object 2 and one force from object 2 acting on object 1. But still, why don't they then cancel each other out if they are equal in magnitude?

    I understand when pushing the boxes, the force one pushes with has to overwin the masses of both m1 and m2 (since they are together). So Pushing force Fp = (m1+m2)*a. Let's say Fp = 10 N, m1 = 1kg and m2 = 2kg. This gives a = Fp/m 10/(1+2) = 10/3 = 3.3 m/s^2.

    The acceleration must be the same for both boxes, since they are moving together.

    Then, Pushing force on object 1 is a*m1 = 3.3*1 = 3.3 N and "pressing" force on object 2 is a*m2 = 3.3*2 = 6.6 N. The force from box 1 to box 2 is the pressing force.

    In case b) we have a*m1 = 6.6 N and a*m2 = 3.3 N. So then the force acting on the second box is smaller.

    Thanks! If I could also get explained my first question, why an object pushing another doesn't cancel the forces out, that would be great. In that case please consider only two objects, not 3 like in the case with the boxes
     
  15. May 27, 2014 #14

    Doc Al

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    As has been explained, force is not transmitted.

    Start by drawing free body diagrams for each box.

    Understand Newton's 3rd law. If A exerts a force on B, then B exerts and equal and opposite force on A. There are always two bodies (A and B) involved when forces are exerted; your job is to identify those bodies.

    For each force in this problem, identify what two objects are interacting to exert the force.
     
  16. May 27, 2014 #15

    Doc Al

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    Equal and opposite forces only 'cancel' (add to zero) if they act on the same body. Here they act on different bodies.
     
  17. May 27, 2014 #16
    Can you please also read my previous post where I explained how I reasoned about the example with the pushing force of the two boxes and say if it's correct?

    I could also say another scenario where the pushing force accelerates both objects to a = 1 m/s^2. Let's say m1=1kg and m2=2kg.

    Then the pushing force originally is F = (m1+m2)*a = 3*1 = 3 N.

    The pushing force on object 1 ONLY is then a*m1 = 1 N and on object 2 it is 2 N. However, object 2 pushes back on object 1 with 2 N (counter force). So object 1 gets 1 N from the left and 2 N from the right. Vector sum of this is -1N. Why doesn't it then starts accelerating backwards?
     
  18. May 27, 2014 #17

    Doc Al

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    You are confusing pushing force with net force.

    The pushing force is F and it acts on the first box only. It does equal F = (m1+m2)*a.

    The net force on box 1 must equal m1*a. But that's not the pushing force.
     
  19. May 27, 2014 #18
    image.png

    Is this free body diagram correct?

    Fp is the original pushing force, coming from object "mp" acting on "m1". -Fp is reacting force on that, coming from object "m1" acting on "mp".

    F2 is the pressing force coming from object "m1" acting on "m2". -F2 is the reaction force on that, coming from object "m2" acting on "m1".
     
  20. May 27, 2014 #19

    Doc Al

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    The diagrams for m1 and m2 look good. (You can take the minus sign off of -F2. The direction of the arrow will determine the sign.)

    There may be other forces acting on "mp", but we don't care. We are only analyzing m1 & m2.
     
  21. May 27, 2014 #20

    Nathanael

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    The red is where you made your mistake. The green part, however, is correct.

    It wouldn't be the pushing force on object 1 that equals 3.3N, it would be the NET force. The pushing force was already declared to be 10N, the NET force however is the sum of all the forces on it. In this case it would be F[itex]_{"push"}[/itex] - F[itex]_{"press"}[/itex] = F[itex]_{net}[/itex] and so 10N - F[itex]_{"press"}[/itex] = 3.3N and therefore F[itex]_{"press"}[/itex]=6.6N
    This agrees with your other answer (which it has to, since it's the same force in the opposite direction)

    The reason it was so simple to get the right answer for object 2 is that the pressing force is the only force on that object, and therefore it's the net force on that object, and so what you did in the green part of the quote would be correct.
    (a*m2 = net force = pressing force)


    As someone else said, the forces act on different objects. Forces only cancel out if they're on the same object.
    If you're looking at the 2 boxes as a single object, then the pressing forces do indeed cancel out, and the only relevant force becomes the "pushing" force.
    Once you look at them as separate objects, though, the "pressing" forces become relevant.
     
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