3 blocks no friction forces question

AI Thread Summary
The discussion revolves around calculating the forces acting on three blocks with varying masses. The initial calculations led to confusion regarding the forces exerted between the blocks, particularly between the 4 kg and 8 kg blocks. It was clarified that the force exerted by the 8 kg block on the 4 kg block is 13.14 N, with an equal and opposite force acting on the 8 kg block. The correct approach involves considering the net forces acting on each block to resolve the discrepancies in the calculations. Ultimately, understanding the interaction forces between the blocks is crucial for solving the problem accurately.
cmwilli
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Homework Statement


[PLAIN]http://img834.imageshack.us/img834/6920/captureqt.jpg


Homework Equations



f=ma

The Attempt at a Solution



Fsum=Fl + Fr=(m1+m2+m3)a
26+-8=(2+8+4)a
a=18/14=1.28571

f=ma
f=4(1.28571)=5.14268N

I know this is incorrect. The force of 4kg block pushes on the 8kg the same as the 8kg pushes on the 4kg, but the way I calculated it doesn't work with that but I know of no other way.
 
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You are right that the resultant force acting on the 4 kg block is F=4a=5.143 N. See the figure: What forces act on this block?

ehild
 
The homework systems seems to disagree.

Call the blocks from left to right A,B,C

The way the teacher explained it in class, the force of C acting on B is the same as the force of B acting on C. But the way I calculated it Fc=4a but Fb=8a
 
Look at the 4 kg mass. There are two forces acting on it: the force from the 8kg mass (F8) ponting to the right, and Fr=8 N, pointing to the left. The resultant force points to the right, and it is F8-Fr=4*a. So F8=4a+Fr.

ehild
 
Is the correct answer then 13.14268N?
 
Yes, the 8 kg mass exerts 13.14 N on the 4 kg one, and a force of the same magnitude and opposite direction is exerted by the 4 kg mass on the 8 kg one.

ehild
 
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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