Calculating Contact Forces Between Three Blocks on a Frictionless Surface

AI Thread Summary
Three blocks of equal mass (10 kg each) are in contact on a frictionless surface with a force of 96 N applied to block A. The acceleration of the system is calculated to be 3.2 m/s². The contact force between block A and block B is determined to be 64 N, while the contact force between block B and block C is 32 N. The discussion emphasizes the importance of applying Newton's second law correctly, focusing on the forces acting on each individual block rather than the forces between them. The final clarification highlights the relationship between the forces exerted by the blocks on each other, confirming that the forces are equal and opposite.
robvba
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Homework Statement


Three blocks on a frictionless horizontal surface are in contact with each other. A force F is applied to block A (mass mA ). If mA=mB=mC=10.0 kg and F = 96.0 N, determine the force of contact that each block exerts on its neighbor.


Homework Equations





The Attempt at a Solution



a=96/(10+10+10)=3.2N

contact force between mA and mB: 96-3.2*10=64
contact force between mB and mC: 3.2*10=32 or 64-3.2*10=32

did i solve this right?

thank you
 
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robvba said:
a=96/(10+10+10)=3.2N

contact force between mA and mB: 96-3.2*10=64
contact force between mB and mC: 3.2*10=32 or 64-3.2*10=32

Hi robvba! :smile:

Correct result … and it looks like you got it the right way, but I'm not sure.

This is a Newton's second law problem … you really shouldn't talk about the "force between" two bodies … each equation should only concern the forces on a particular body. :smile:
 
tiny-tim said:
Hi robvba! :smile:

Correct result … and it looks like you got it the right way, but I'm not sure.

This is a Newton's second law problem … you really shouldn't talk about the "force between" two bodies … each equation should only concern the forces on a particular body. :smile:

thank you. you're right. it's force of contact that each block exerts on it's neighbor. which would mean that block mA exerts a force of 64N on block mB while block mB exerts an equal and opposite force back; so mA = -mB. is that a fair statement?
 
robvba said:
thank you. you're right. it's force of contact that each block exerts on it's neighbor. which would mean that block mA exerts a force of 64N on block mB while block mB exerts an equal and opposite force back; so mA = -mB. is that a fair statement?

erm … I don't think you mean mA = -mB, do you?

I was thinking more of mass times acceleration = sum of forces …

you know the acceleration of block A, so the sum of forces on block A is … , and so the reaction force from block B on block A is … ? :wink:
 
tiny-tim said:
erm … I don't think you mean mA = -mB, do you?

I was thinking more of mass times acceleration = sum of forces …

you know the acceleration of block A, so the sum of forces on block A is … , and so the reaction force from block B on block A is … ? :wink:

right. i meant Fba = - Fab (Force of A on B = negative Force of B on A)...?
 
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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