Action/reaction pair Newton's law

AI Thread Summary
In discussing action/reaction pairs in Newton's law, the focus is on the forces involved when a man swims. The swimmer pushes against the water, creating a forward motion, while the water exerts an equal and opposite force back on the swimmer. This principle is similar to how an aircraft interacts with air, where the aircraft pushes against the air, resulting in drag. The exchange of momentum occurs as one object transfers momentum to another, leading to equal gains and losses. Understanding these interactions clarifies the mechanics of motion in fluid environments.
moomoocow
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hi
i have a question
if in a question, and they ask, action/reaction pair of the force
lets say a man is swimming..and they say name the action/reaction pair of the force, of the force that causes the man to accelerate..how would i reply?
 
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When someone pushes or pulls an object or pushes a fluid, the object or fluid resists with a force.

In order to swim, one must push water, or otherwise transfer momentum to the water in order to propel oneself. Then when one moves forward, the water in which one swims pushes back.

Similarly, when an aircraft flies, their air is pushing back (in reality, the aircraft pushes against the air and loses momentum at a more or less contant rate.)

The resistance force of the fluid is the drag on an object moving through the fluid.
 
oh ok i see thanks!
 
You can think of one object flinging a piece of momentum from itself to the other object, which absorbs that momentum piece to its own.

Thus, the first object loses momentum and the second gains momentum in equal measure.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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