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As per the book:

$\sum_{j=1}^N f_{j}^{ext}$+$\sum_{j=1}^N f_{j}^{int}$=$\sum_{j=1}^N dp_{j}/dt$

The first term, $\sum_{j=1}^N f_{j}^{int}$, is the sum of all internal forces acting on all the particles.

$\sum_{j=1}^N f_{j}^{int}$=$0$ (because according to newton's 3rd law, every particle will have the reaction force as a counteracting force within the system-which I understood)

The second term, $\sum_{j=1}^N f_{j}^{ext}$, is the sum of all external forces acting on all the particles. It is the total external force $F_{ext}$ acting on the system.:

$\sum_{j=1}^N f_{j}^{ext}$ ≡$ F_{ext}$

However, I didn't really get how can we say that the second term is equal to total external force. It would be appreciated if it could be explained with the help of some graph or figure by choosing small number of particle and showing that the sum of force of interaction of these particle is indeed the total external force applied to the system.

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    $\begingroup$ What is it that you don't understand ? $\endgroup$ – Mitchell Jan 1 '18 at 6:17
  • $\begingroup$ A bit vague... isn't it just Newton's second law? $\endgroup$ – QuIcKmAtHs Jan 1 '18 at 6:30
  • $\begingroup$ I didn't understand why $\sum_{j=1}^N f_{j}^{ext}$ ≡$ F_{ext}$ . Ok so lets say that I exert a force on a rigid body neglecting friction or gravity for a while. Maybe I exert a 10 N force on the body. How can it be said that the sum of all external forces on all the particles (which I don't even know how we can determine) is equal to the total external force I applied i.e 10N. Whoever came up with the thought must have had a reason to be able to ascertain it. If I'm wrong do correct me. $\endgroup$ – suiz Jan 1 '18 at 6:33
  • $\begingroup$ @XcoderX could explain in a bit more detailed way. $\endgroup$ – suiz Jan 1 '18 at 6:37
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    $\begingroup$ I mean, isn't this just the definition of the the net external force? -- that it is the sum of the external forces acting on all the objects in the system? Perhaps you're asking about why this is the force corresponding to the center of mass of the system? $\endgroup$ – march Jan 1 '18 at 23:01
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Since you seem happy about the internal forces let's ignore them and just set them all to zero so only consider external forces.

The total momentum $P$ is just the sum of all the individual momenta:

$$ P = \sum p_i $$

and we can differentiate both sides of this to get:

$$ \frac{dP}{dt} = \sum \frac{dp_i}{dt} $$

For any object, simple or composite, force is the rate of change of momentum - that is just Newton's second law. Now, the left side of the equation above is the rate of change of total momentum so that's the total force:

$$ \frac{dP}{dt} = F_\text{tot} $$

The right side is the rate of change of momenta of the individual particles so that's the force on the individual particles:

$$ \frac{dp_i}{dt} = f_i $$

So we end up with:

$$ F_\text{tot} = \sum f_i $$

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So what you are asking is how the sum of external forces acting on all the particles is equal to the sum of external force on the entire system. Right?

They are the same thing. When we consider a system of particle the external force refer to any force which is not exerted by particles within the system.

So lets take a rigid box as our system which I try to push. The body starts moving in a straight line. The external force here is the contact forces from my hand since my hands are not part of the system. Now my hands are in contact with only some of the particles so the external force is acting only on these particles. But since the entire system is moving there is some net force acting on all individual particles in the system even those who are not in contact with the external force. They receive the force from nearby particles. So when we add all the forces acting on individual particle the internal force which is any force exerted by the particles within the system on nearby particles cancel out because they have an action reaction counterpart. And the only thing that remains is the force from my hand which were exerted on only the few particles which came in contact with my hand. This force also has a reaction counterpart which is the reaction force the box exerts on my hand but since my hands are not part of the system those force are not considered. I hope that I haven't confused you even more.

I think what you should focus here is that external force refers to any force which is not coming from any particles within the system. That definition holds even when we consider the forces acting on individual particles

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