# Newton's Third Law: Action and Reaction Explained

• thomas_shvekher
I avoid the use of reaction because in Newtonian physics where action and reaction occur simultaneously, there is no justification of calling one force action and the other reaction. I call the forces drawn in FBDs as action forces because they are external to the system and "act" on it. More generally, I give a name to the whatever force, e.g. "Fred", and then talk about the Third Law counterpart of "Fred".In summary, the force on the system is non zero and depends on how inclusive your system is.

#### thomas_shvekher

Homework Statement
I understand that every action force has a reaction force of equal magnitude and opposite direction. In the drawing below, I undertsand that box 1 exerts a force on box 2, and box 2 exerts a force of the same magnitude but opposite direction on box 1 (it is an internal force). That being said, let's say that the person pushing on the two boxes pushes with a force of 25N, and the boxes expericne 10N of friction - what are the opposite and equal reaction forces to the forces listed above (what force is opposite and equal to the 25N of the applied force, and what force is opposite and equal to the 10N of friction force?).
Relevant Equations
Newton's Third Law.

The net force is non zero on the system I.e. the boxes are accelerating. I don’t think you can tell without the masses of the boxes, and friction force acting in each.

MatinSAR, topsquark and thomas_shvekher
erobz said:
The net force is non zero on the system
Depends on how inclusive your system is.

topsquark
hmmm27 said:
Depends on how inclusive your system is.
Can you explain.

hmmm27 said:
Depends on how inclusive your system is.
Am I misunderstanding the problem?

topsquark
The problem looks pretty straightforwards (except maybe if you wanted numerical values for the BoxA<>BoxB force interaction, the value for friction [and mass] needs be specified for each box).

But, regardless of acceleration, the forces are still balanced when all the players are accounted for ; not just the person and the boxes.

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topsquark and erobz
hmmm27 said:
The problem looks pretty straightforwards (except maybe if you wanted numerical values for the BoxA<>BoxB interaction, more detail needs be shown regarding friction).
This is what I was thinking, I guess that's not actually what is being asked.

topsquark
Regradless of whether the person and the boxes are accelerating or not, if the person exerts 25 N on box 1 to the right, Box 1 exerts on the person 25 N to the left. If the ground exerts 10 N to the left on the boxes, the boxes exert 10 N on the ground to the right.

If the system starts moving from rest, how does it accelerate? Note that friction between the feet of the person and the ground provides the accelerating force to the right and must be greater than 10 N. If it's equal to 10 N, the system will be moving at constant velocity.

PeroK, MatinSAR, topsquark and 2 others
I think I skim read a bit too much (and I was secretly hoping they were going to solve for the Newtons Third Law pair between the boxes - it feels a tad slow recently?)!

Unfortunately, the problem is basically reduces to figuring out what to call the reaction force that acts opposite to the whatever the force in question is?

hmmm27 and topsquark
erobz said:
I think I skim read a bit too much (and I was secretly hoping they were going to solve for the Newtons Third Law pair between the boxes - it feels a tad slow recently?)!
That can be done. If the coefficient of kinetic friction between ground and boxes is ##\mu_k##, and the force exerted by the person is ##F_0##, we can write the Second Law fro the two-block system and find the common acceleration first. $$F_{net}=F_0-\mu_k(m_1+m_2)g=(m_1+m_2)a\implies a=\frac{F_0}{m_1+m_2}-\mu_k g.$$The net force on leading block 1 is $$F_{net}=m_1a=m_1\left(\frac{F_0}{m_1+m_2}-\mu_k g\right)=F_{21}-\mu_k m_1g \implies F_{21}=\frac{m_1}{m_1+m_2}F_0.$$The net force on trailing block 2 is $$F_{net}=m_2a=m_2\left(\frac{F_0}{m_1+m_2}-\mu_k g\right)=F_0-F_{12}-\mu_k m_2g \implies F_{12}=\frac{m_1}{m_1+m_2}F_0.$$ Note that the forces between the boxes are independent of the coefficient of kinetic friction. They have the same value even when the boxes slide on a frictionless surface.
erobz said:
Unfortunately, the problem is basically reduces to figuring out what to call the reaction force that acts opposite to the whatever the force in question is?
I avoid the use of reaction because in Newtonian physics where action and reaction occur simultaneously, there is no justification of calling one force action and the other reaction. I call the forces drawn in FBDs as action forces because they are external to the system and "act" on it. More generally, I give a name to the whatever force, e.g. "Fred", and then talk about the Third Law counterpart of "Fred".

jbriggs444, topsquark and erobz
kuruman said:
Note that the forces between the boxes are independent of the coefficient of kinetic friction. They have the same value even when the boxes slide on a frictionless surface.
. . . by which you mean "are still equal to each other" not "same for each CoF".
kuruman said:
More generally, I give a name to the whatever force, e.g. "Fred", and then talk about the Third Law counterpart of "Fred".
Derf.

topsquark
hmmm27 said:
. . . by which you mean "are still equal to each other" not "same for each CoF".
I mean the same for any CoF. As you see from the derivation in post #10, $$F_{21}=F_{12}=\frac{m_1}{m_1+m_2}F_0.$$ The CoF, whatever its value, affects only the acceleration not the Third Law pairs.

topsquark
kuruman said:
That can be done. If the coefficient of kinetic friction between ground and boxes is ##\mu_k##, and the force exerted by the person is ##F_0##, we can write the Second Law fro the two-block system and find the common acceleration first. $$F_{net}=F_0-\mu_k(m_1+m_2)g=(m_1+m_2)a\implies a=\frac{F_0}{m_1+m_2}-\mu_k g.$$The net force on leading block 1 is $$F_{net}=m_1a=m_1\left(\frac{F_0}{m_1+m_2}-\mu_k g\right)=F_{21}-\mu_k m_1g \implies F_{21}=\frac{m_1}{m_1+m_2}F_0.$$The net force on trailing block 2 is $$F_{net}=m_2a=m_2\left(\frac{F_0}{m_1+m_2}-\mu_k g\right)=F_0-F_{12}-\mu_k m_2g \implies F_{12}=\frac{m_1}{m_1+m_2}F_0.$$ Note that the forces between the boxes are independent of the coefficient of kinetic friction. They have the same value even when the boxes slide on a frictionless surface.

I avoid the use of reaction because in Newtonian physics where action and reaction occur simultaneously, there is no justification of calling one force action and the other reaction. I call the forces drawn in FBDs as action forces because they are external to the system and "act" on it. More generally, I give a name to the whatever force, e.g. "Fred", and then talk about the Third Law counterpart of "Fred".
Seems like you might have labled your blocks opposite of the diagram in the OP?

Interesting though, I didn’t see that they would only depend on mass and pushing force, not friction (I didn’t solve it).

Thanks!

topsquark and kuruman
erobz said:
Seems like you might have labled your blocks opposite of the diagram in the OP?
Good catch. Thanks.

topsquark and erobz
kuruman said:
That can be done. If the coefficient of kinetic friction between ground and boxes is ##\mu_k##, and the force exerted by the person is ##F_0##, we can write the Second Law fro the two-block system and find the common acceleration first. $$F_{net}=F_0-\mu_k(m_1+m_2)g=(m_1+m_2)a$$ $$\implies a=\frac{F_0}{m_1+m_2}-\mu_k g$$
Awesome !

The next two formulae are giving me problems...more cogitation required.

(Note that - on my screen, at least - all of them are cut off somewhere around the \implies's, which I didn't realize for awhile : tossed a couple extra \$'s into the Quote to make it legible for myself.)

thomas_shvekher said:
Homework Statement: I understand that every action force has a reaction force of equal magnitude and opposite direction. In the drawing below, I undertsand that box 1 exerts a force on box 2, and box 2 exerts a force of the same magnitude but opposite direction on box 1 (it is an internal force). That being said, let's say that the person pushing on the two boxes pushes with a force of 25N, and the boxes expericne 10N of friction - what are the opposite and equal reaction forces to the forces listed above (what force is opposite and equal to the 25N of the applied force, and what force is opposite and equal to the 10N of friction force?).
Relevant Equations: Newton's Third Law.

View attachment 324585
By my reckoning there are five pairs of third law forces in that scenario. Can you label them?

topsquark