posted by Physics on .
Ok I got a question I asked before except there are other parts that I didn't ask so here we go
Three blocks on a frictionless horizontal surface are in contact with each other
A force F is applied to block 1 (mass m1).
Draw a free-body diagram for each block
ok I did this
the acceleration of the system in terms of m1, m2, m3
I just found the acceleration of each block
a of m1 = m1^-1 F
a of m2 = (m1 + m2)^-1 F
a of m3 = (m1 + m2 + m3)^-1 F
did I do that right?
the net force on each block
how do I do this???
the force of contact that each block exerts on its neighbor
How do I do this???
If you are pushing on block m1, I assume the other blocks are in front of it; otherwise m2 and m3 would be left behind and the solution would be trivial
(a1 = F/m1 and a2 = a3 = 0).
Therefore I would assume that
a1 = F/(m1 + m2 + m3)
The other blocks, being in front of m1, would accelerate at the same rate.
For the contact force, start with block m3, for which you know the acceleration AND the mass. use F3 = a3/m3 = a1/m3
Force F3 acts backwards on m2, and another force F2 acts forwards on it. Use that fact and a FBD to solve for F2.
what's FBD how do i solve for F2
FBD..free body diagram.
what's the contact force on m1
F - F2 = m1*a1
Get F2 from the FBD and Newton's 2nd law for block 2, as explained earlier.