how to write newtons 2nd law equation for horizontal forces and to solve them for acceleration and tension?

Question

Heres a crappy diagram of what the problem looks like:
<----[12kg]----[20kg]----[16kg...

Three boxes are tied together and are pulled forward on a frictionless surface by a rope with a tension of Ft=12N.

a) Write Newtons 2nd law equation for the horizontal forces on each box individually. (No numbers allowed)

b) Solve your three Newton's law equations for the acceleration of each box and the tension in each string. (Find numerical answers (with proper units.)

Any help? Im not sure what to do
Additional Details

The last box is 16kg since it cut the end of it off

Answer 1

a) Write Newtons 2nd law equation for the horizontal forces
Ft= F1 + F2 + F3
Ft= m1a1 +m2a2 + m3a3

Since they all move together
a1=a2=a3=a

we have

Ft=(m1+m2+m3)a

a=Ft/(m1+m2+m3)

(b)Solve your three Newton's law equations for the acceleration of each box and the tension in each string.

T(1)=Ft
T(1,2)=F2+F3
T(2,3)=F3
F1=m1a
F2=m2a
F3=m3a

I hope it helps

Answer 2

A rope exerts a consistent horizontal stress of 250 N to drag a 60-kg crate by the floor. the speed of the crate is talked about to develop from a million m/s to three m/s in a time of two seconds lower than the effect of this stress and the frictional stress between the floor and the crate. what's the magnitude of the frictional stress appearing on the crate? Use g = 10 m/s/s.

Answer 3

From left to right I'm calling this
Ma,Mb,Mc for the box masses,
Ta,Tb,Tc for the tesions on the three strings. (Ta=Ft)

Basically F=Ma so:

Box a
Ta-Tb=Ma*A

Box b
Tb-Tc=Mb*A

Box c
Tc=Ma*A

Now part B, this is real simple.
The total mass is 48kg, the Tension Ta=12N so
A=F/m=12/48=0.25m/s^2

We now know enought to solve for every thing.
Tc=Mc*A=4N

Tb=Mb*A+Tc=%N+4N=9N