What is the total force exerted by block 2 on block 3 ?

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The discussion centers on calculating the total force exerted by block 2 on block 3 in a system of four blocks on a frictionless table. Initially, a participant calculated the force as 67.5 N, which was incorrect. They later considered the combined weight of the two blocks above block 2, calculating it as 196 N, but still faced issues with the horizontal force component. The conversation emphasizes the importance of analyzing both horizontal and vertical forces to determine the net force accurately. Ultimately, the correct approach involves understanding the vector nature of the forces acting on the blocks.
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Four blocks EACH of mass m = 10.0 kg are arranged as shown in the picture, on top of a frictionless table. A hand touching block 1 applies a force of Fh1 = 90.0 N to the right. The coefficient of friction between the blocks is sufficient to keep the blocks from moving with respect to each other.

What is the total force exerted by block 2 on block 3 ?

I got 67.5 N which was wrong

F=90N-22.5=67.5N since a=9/4ms-2
 

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Start by drawing a free-body diagram showing all the forces. Pay particular attention to blocks 2 and 3, but don't ignore the others.
 
I now think block 2 would exert a force to block 3 equal to the combined weight of the 2 books above it which is f=2(mg) = 2*10*9.8 = 196 N

but i still get it wrong
 
That's part of it, but what about the horizontal direction?
 
i got it now its sqrt(of horizontal direction ^2 + vertical direction ^2)
 
Assuming you are referring to the forces on the block, that is indeed the magnitude of the net force. Now what is the direction?
 
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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