Forces misconception hammer and nail

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SUMMARY

The discussion centers on the mechanics of how a hammer drives a nail, specifically addressing the confusion surrounding Newton's Third Law of Motion. Participants clarify that while the hammer exerts a force on the nail, the nail simultaneously exerts an equal and opposite force on the hammer. This interaction does not prevent the nail from accelerating because the forces act on different objects, and external forces, such as the hand's force and gravity, play a crucial role in the system's overall motion. The conversation emphasizes the importance of understanding net forces and the distinction between internal and external forces in physical interactions.

PREREQUISITES
  • Newton's Laws of Motion
  • Concept of net force and acceleration
  • Understanding of internal vs. external forces
  • Basic principles of momentum conservation
NEXT STEPS
  • Study Newton's Laws of Motion in detail
  • Explore the concept of net force and how it affects acceleration
  • Learn about momentum conservation and its applications in physics
  • Investigate real-world examples of force interactions, such as collisions and recoil
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the fundamental principles of mechanics and force interactions in physical systems.

  • #31
Wait wait wait, Newton's law says in your example that if you give me 5 dollars i have to give you 5 dollars back. If I push on a wall the wall has to push me back with the same force. If a mack truck hits a bike the mack truck gives a force to the bike and the bike gives that same force back to the truck. That's what it says
 
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  • #32
zoobyshoe said:
It's baby simple: If I give you $5, you experience a change of +$5 and I experience a change of -$5. The changes are equal and opposite! The action and reaction are automatically equal and opposite!

It does not mean: If I give you +$5 you must give me +$5. That's not what Newton III is saying at all. Newton III is saying if I give you exactly $5 I automatically suffer a loss of exactly $5. Equal, but OPPOSITE. You: +$5, me: -$5.

Newton's Third Law guarantees that if you take x amount of force from me, I lose exactly x amount of force. If I lose x amount of momentum to you, you gain exactly x amount of momentum. No more, no less. It is the foundation of all the conservation laws. You experience a net force of +$5, so to speak, and I simultaneously experience a net force of -$5.
I do not like this analogy at all! There are two big problems here.

First, in this type of financial transaction the total money is conserved. There is no conservation of force. There is no sense in which one body takes force from another body which is therefore lost from the first body.

Second, the amount of money is a scalar quantity, not a vector quantity like force is, and that makes a big difference. An object experiencing an external force to the right is not gaining force and an object experiencing an external force to the left is not losing force.

eulerddx4, I do not recommend that you study this analogy.
 
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  • #33
The problem with the analogy is that it confuses "force" with "momentum". It is really trying to be an analogy about conservation of momentum. That's a valid lesson-- we can either say that the 3rd law must be true because momentum is conserved, or we can say that momentum is conserved because of the 3rd law. Getting back to the gun, it means we can impart a momentum into the bullet only if we are willing to get that same momentum imparted into our arm (in the opposite direction, so a "negative" momentum), in the form of recoil (though we can pass that momentum along, via our body, into the ground).
 
  • #34
Yes, as an analogy with momentum it is not too bad. Momentum is conserved so if A gives momentum to B then A loses momentum. The scalar/vector bit can be glossed over for 1D "vectors".
 
  • #35
OK Dale and Ken, if we all agree what I said applies to momentum I'm happy.

I realize I'm on shaky ground suggesting something like "conservation of force". It seemed to make sense when I wrote it if I thought of force in terms of Newtons. It seemed that any agent of force must lose as many Newtons as it succeeds in imparting to something else. I made the assumption that kg*m/s2 are transferable from one body to another just as kg*m/s are. If body A exerts force to accelerate B, B now has the force to accelerate something else. A, it seemed, must have lost some force. A must now not have the ability to produce as much acceleration as it did before. This seems logical. However all that went through my head without much examination and I probably should have stuck exclusively to the concept of momentum.
 
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  • #36
zoobyshoe said:
I made the assumption that kg*m/s2 are transferable from one body to another just as kg*m/s are. If body A exerts force to accelerate B, B now has the force to accelerate something else. A, it seemed, must have lost some force. A must now not have the ability to produce as much acceleration as it did before. This seems logical.
The problem is, that extra "s" in the denominator makes a big difference. Force is momentum imparted per time, so if a body acquires some momentum in time t1, it can impart that momentum into something else, but if it does so in a different time t2, then it will be imparting a different force than what it received. The forces must be the same for action/reaction pairs only, because only then must the times be the same-- it's the same interaction, so it has to take the same time, so involves both the same momentum imparted and the same force, except for the opposite direction.
 
  • #37
zoobyshoe said:
I realize I'm on shaky ground suggesting something like "conservation of force". It seemed to make sense when I wrote it if I thought of force in terms of Newtons. It seemed that any agent of force must lose as many Newtons as it succeeds in imparting to something else.
Your hand may exert a force of only one Newton over a second on a hammer, but the hammer can exert a force of a thousand Newtons over a millisecond on the nail. If force were conserved then the hammer could never have generated more than one Newton of force on the nail because that was all the force that was imparted to it. A hammer would be useless because it could not generate any more force than you could already generate with your hand.
 
  • #38
Ken G said:
The problem is, that extra "s" in the denominator makes a big difference. Force is momentum imparted per time, so if a body acquires some momentum in time t1, it can impart that momentum into something else, but if it does so in a different time t2, then it will be imparting a different force than what it received. The forces must be the same for action/reaction pairs only, because only then must the times be the same-- it's the same interaction, so it has to take the same time, so involves both the same momentum imparted and the same force, except for the opposite direction.
Yes. You're alluding to impulse, I believe: F (Δt)=Δρ. The longer the time over which a force acts the greater the change in momentum it produces. And I see what you mean in the whole final sentence.

DaleSpam said:
Your hand may exert a force of only one Newton over a second on a hammer, but the hammer can exert a force of a thousand Newtons over a millisecond on the nail. If force were conserved then the hammer could never have generated more than one Newton of force on the nail because that was all the force that was imparted to it. A hammer would be useless because it could not generate any more force than you could already generate with your hand.
This suggests the quantity that is conserved I was grasping for each time I said "force" is Impulse. If the hammer generates a thousand Newtons it is only because it acts on the nail over a proportionately shorter time than the one Newton arm acts on the hammer. That latter sentence is correct, right? The multiplication of the force has to come at the expense of the amount of time over which the larger force can be applied, just as with distance and force in the same case of leverage? The distance over which the magnified force of the short arm of a lever can work is inversely proportional to the distance over which the smaller force has to work on the long arm of a lever. F (Δt) = -F (Δt) as F (d) = -F (d). (d = distance)
 
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