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Newton's Third Law and Unbalanced Forces

  1. Aug 11, 2013 #1
    Hello,

    I have seen several explanations on the Internet about how to resolve Newton's Third Law and Unbalanced Forces, but none made sense to me in a way that I really understood it.

    Scenario 1:

    There is a block on the ground, the ground has friction with say static friction force 10 N. I push the block with 20 N. It moves because there is now an unbalanced force on on the block. The net force on the block is 10 N in the direction I am pushing it.

    Force relationship 1: When I push on the block with 20 N, the block is pushing on me with 20 N.

    Force relationship 2: Is the block is pushing on the floor with 20 N and the floor is pushing back on the block with 10 N of friction?

    In this case for the second relationship, there is an opposite force of the floor pushing on the block, but it is not equal, correct?

    How can Newton's Third Law account for unbalanced forces?


    Scenario 2:

    Maybe a variation on the same type of problem, just need to know how best to think about it and explain it.

    One car going fast approaching another car going slow in opposite directions head on. When they collide, I understand that there is an equal and opposite force. Meaning if both had infinite force meters that collided head on, at the moment of impact they would say the same number.

    However, one car would probably knock the other out of the way.

    Is the best explanation that despite the equal and opposite forces, there is a clear difference in momentum?


    Scenario 3:

    A projectile is traveling at a constant velocity and hits someone. It will clearly do damage to the person. However, if it is traveling at a constant velocity then there is no acceleration. If there is no acceleration then there is no Force.

    Why does it do damage?

    Is it because when the projectile strikes something, the equal and opposite forces makes it decelerate over a short span of time? Would that deceleration be used in the F=ma equation to figure out how much force it hits the person with? Does it relate to impulse?

    Thank you so much for clearing up these misconceptions in my head, and I hope you won't mind the follow up questions I am sure to have.
     
  2. jcsd
  3. Aug 11, 2013 #2

    Dale

    Staff: Mentor

    Hi nate99, welcome to PF
    No, the block pushes on the floor with 10 N of friction force, equal and opposite to the 10 N of friction force of the floor pushing on the block.

    The change of momentum for one car is equal and opposite to the change in momentum for the other car (according to Newtons 3rd law). But if their beginning momenta are unequal then the one with the larger initial momentum will have less "relative" change for the same "absolute" change.

    That is poorly worded, hopefully you get the idea, but don't hesitate to ask.

    A projectile which hits something and does damage definitely decelerates, usually very rapidly.
     
  4. Aug 11, 2013 #3
    Yes! Ok, I think I finally understand that.

    Scenario 1

    So let me see if I can explain this back correctly. There are forces happening between the box and the floor due to friction. In this case it is 10 N. 10 N the floor is pushing on the box, 10 N the box is pushing on the floor. The maximum amount of Newtons would be what ever the Static Coefficient of Friction x Normal force of the box at first, and then the Kinetic coefficient of Friction X Normal force to keep it moving. However, the additional force that is past that amount is being exerted by me on the box (an additional 10 N) and of course is also being exerted back on me by the box as well.

    I guess my followup question could be where does the additional 10 N come from, but I think I understand that. It is part of the 20 N total that I exert on the box (and the box back on me) that comes from the chemical energy in my cells being converted into mechanical energy to do work (I'm sure that is incorrectly worded).

    So in reality while the forces are UNBALANCED from the box's perspective causing acceleration, the system as a whole (me and what is going on in my body, box, floor, surroundings), the forces are all balanced.

    Do I have this correct?


    In Scenario 3, so while the projectile is moving at a constant velocity, it can be said there is no force acting on it. However, if it hits someone it will have impact from force due to the fast deceleration. Correct?

    Thanks for the help, I think I understand that a lot better now.
     
  5. Aug 11, 2013 #4
    Another question about balanced and unbalanced forces...

    So the more I think about balanced and unbalanced forces, the more confused I get.

    I think I now get the unbalanced part IF I re-explained that first scenario correctly....If not then I need to ask more followup questions.

    Anyway, how about balanced forces that are not stationary?

    I've learned that when forces are balanced on an object, there is no acceleration, and therefore no net force. This is easy to understand with stationary objects. This is much harder for me to truly understand when it comes to non stationary objects.

    Scenario 1: Horizontal

    Just like the scenario above, I'm pushing a box that is touching the ground that has friction. Let's say that I've got it moving and the force that I am exerting is right at the amount necessary to keep the box moving (coefficient of kinetic friction X normal force of box?) at a constant velocity.

    Question 1: So in respect to the box, is the force I am exerting on it the same amount but opposite direction to the force the ground is exerting on the box due to friction? Or is it slightly above that?

    If the forces are equal on the box, it's hard for me to wrap my head around why the box is still moving then.

    Question 2 (dependent on question 1 answer): And when I think about it, I'm still exerting a constant force (F=ma), right? Even though it is not accelerating?

    Question 3 (also dependent on question 1 answer): If, with respect to the box, I am exerting a force slightly above the force due to kinetic friction, then the box is slightly accelerating, right? Does that then mean the forces are not balanced, and given enough time the box will slowly approach infinite velocity?

    Question 4: Maybe that was a bad example. Are there better examples of balanced forces on non-stationary objects? I know a common example is something floating in space, may not be stationary but has balanced forces (no forces?).


    Scenario 2: Vertical

    Terminal velocity is something I have always found confusing. I understand it as the velocity something is falling at where there is no more change in velocity, like the maximum it can reach. I saw a video of a skydiver and the screen showed the downwards velocity and you can see at some point it just tops out. I believe the force due to gravity is equal to the drag force...but let me know if i am wrong...

    Question 5: Maybe i'm thinking about this a weird way, but if there is no change in velocity, then acceleration = 0...then what happened to the F = m a? Is that now F = m (0)? What about the drag force, which is dependent on velocity I believe? Wouldn't it accelerate the object upwards now that gravity force is F = m (0)?

    Question 6: It's still hard for me to wrap my head around the drag force = force due to gravity on the object, having no net force, and the object is still moving downwards at a constant velocity...i can't help but just keep wondering "Why is it moving down still?"
     
  6. Aug 11, 2013 #5

    Dale

    Staff: Mentor

    Yes.

    If you take any isolated system (i.e. no external forces, just internal forces) then every force on one part of the system will be balanced by an equal and opposite force on another part of the system, so there is no net acceleration, just relative movement of different parts.

    Yes. You can tell that there is a force on the bullet because the bullet deforms on hitting the target. Typically it deforms a lot.
     
  7. Aug 11, 2013 #6

    Dale

    Staff: Mentor

    OK, but this is a lot of questions. I probably won't get to all of them. I will do what I can, but you may want to think a bit and really focus on the key questions.

    It is equal. If it were greater then it would accelerate.

    Yes. Remember it isn't F=ma, it is ƩF=ma. I.e. it is the net force which equals mass times acceleration. This is an important distinction.

    Yes.

    The answer to this question requires special relativity, let's table it for now.

    I like the example of an object falling through the atmosphere at terminal velocity. EDIT: I see you thought of that example too :smile:

    Again, it is not F=ma, it is ƩF=ma. Here we have two forces, weight (mg) and drag (Fd) so ƩF=Fd+mg=ma=0m=0 or Fd=-mg.

    I guess I did get through all of them. It keeps moving because there is no net force to decelerate it. In the absence of a net force objects continue moving at a constant velocity in a straight line.
     
  8. Aug 11, 2013 #7
    Excellent! I'm going to mull over those for a little while. Thanks!
     
  9. Aug 11, 2013 #8
    If you push the block, you are also bracing your feet against the floor and pushing on the floor.
    Let the subscript Y be you, G be the ground, B be the block.
    ##F_{YB} = 20##N
    ##F_{BY} = -20##N
    ##F_{BG} = 10##N
    ##F_{GB} = -10##N
    Since the block has unbalanced forces, it moves in the positive direction. Assume you aren't moving. Then your forces must be balanced, and so ##F_{BY}+F_{GY} = 0##
    ##F_{GY} = 20##N
    ##F_{YG} = -20##N
    Therefore, the ground must also accelerate in the negative direction by ##F_{BG}+F_{YG}=-10\text{N}/m_{G}##

    Does that make sense?
     
  10. Aug 12, 2013 #9
    Khashishi,

    Thanks for that explanation!

    I had to draw it out and use arrows and quantities.

    So in the way that this diagram would be set up, when I look at all of the forces in the same way that you did, I would get

    The box is accelerating in the positive direction. (10 N)

    The ground is accelerating in the negative direction. (-10 N)

    I am not moving. (0 N)

    I think I understand it, at least the box and the ground. The box and the ground, or maybe I just think of it as the exact surface of the ground in contact with the box, accelerate forward and backwards respectively.

    I'm quite clear about the box accelerating in the positive direction from my reference point in this scenario. I guess I'm still a little unsure about thinking of the ground as accelerating in the negative direction in regards to ME. That chunk of ground under the box would still be in the same position relative to me, and I did not move. Am I not thinking about it in the right way?

    Thanks again. I can tell once I wrap my head around this last part, I'll really understand it.
     
  11. Aug 12, 2013 #10

    rcgldr

    User Avatar
    Homework Helper

    The total momentum of the earth, you, and the box is conserved, assuming an appropriate frame of reference is chosen, such as the center of mass of the earth, you, and the box. If the change in momentum of the box is equal in magnitude but opposite of the change in momentum of the earth, then you do not move.
     
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