3 blocks no friction forces question

[cmwilli]

Homework Statement

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

Homework Equations

f=ma

The Attempt at a Solution

F_sum=F_l + F_r=(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.

 

[ehild]

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

 

[cmwilli]

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 F_c=4a but F_b=8a

 

[ehild]

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

 

[cmwilli]

Is the correct answer then 13.14268N?

 

[ehild]

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