Calculating Int. Forces: Why the Difference?

AI Thread Summary
When two boxes of different masses are pushed with the same force, the internal forces between them differ based on which box is being pushed. If the lighter box (m1) is pushed, it exerts a greater force on the heavier box (m2) compared to when m2 is pushed. This is due to the acceleration being uniform for both boxes, requiring the same total force. The internal forces reflect the mass distribution and the direction of the applied force. Understanding these dynamics clarifies the concept of internal forces in a system of connected masses.
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I know that you can calculate that if two boxes next to each other of different masses are pushed on end 1 and on end 2 with the same force have different internal forces... But I can't think of the physical meaning why?
 
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I have absolutely no idea what you mean by "different internal forces". Perhaps it would help if you told us what you mean by "end 1" and "end 2".


I might guess that you mean you have two boxes, of masses m1[/sup] and m2 abutting and by "pushed on end 1 and end 2" you mean on the one hand pushing on box of mass m1 so that both move or on the other pushing on the box of mass m2 so that they both move.

The obvious points are: Either way, since you are accelerating the same mass, m1+ m2, in the same way, you must apply the same force. However, is m1 < m2, the force that box m2 must apply to it, it you are pushing on box m2, is less than the force box m1 must apply to box m2 if you are pushing on box m1.
 
yikes... i should brush up on my english! thank you for trying to understand my broken english and helping me with my problems!
 
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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