Quick thing How did they get this?

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The discussion focuses on calculating the contact force between two masses, mA and mB, in a system where an applied force is given. The acceleration is determined using the formula a = F/(3m), where F is the applied force and m is the mass of each block. To find the contact force, the problem simplifies by considering the force needed to accelerate the two remaining masses, which is derived from the overall acceleration. The contact force is essentially the force exerted by the first mass on the other two masses to achieve the same acceleration. Understanding this relationship clarifies how the contact force is calculated in the context of the problem.
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https://www.physicsforums.com/showthread.php?t=259689

In that thread how did the person figure out the contact force between mA and mB? I know they got the acceleration by doing F/mA+mB. But I don't get how they got the contact force, because all the blocks are equal mass. Something like F-a*(10+10/2)? What are they multiplying the F-a by?

Thanks.
 
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In the problem applied force is given, masses are given. Find the acceleration.
a = F/3m.
For the time being, forget the first mass.
How much force is to be applied to the remaining two masses to have the acceleration a?
That is the contact force acted by the first mass on the remaining two masses.
 
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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