Three rectangular blocks of equal mass m slide to the right along a frictionless horizontal surface accelerated by a force F applied to the left side of the left-most block. (a) Draw a free-body diagram for each block. (b) Determine the magnitudes and directions of all of the forces on each block in terms of m, F and/or g.
The Attempt at a Solution
As for vertical forces, each block has the force of gravity downwards and the normal force from the surface pointing upwards, which sum to 0. That part is fine.
As for the horizontal forces, on block 1 I have the applied force to the left, and the normal for from block 2 to the right. For block 2 I have the normal force from block 1 to the right and the normal force from block 3 to the left. And then for block 3 there I have the normal force from block 2 to the right. Using Newton's Law of equal and opp. pairs, I know that the normal force of 2 on 1=-(force on 1 on 2) and normal forces of 2 on 3=-(force 3 on 2).
By equating these forces and using Newton's Law, I get that the force of 2 on 3 is just F (applied force) and 3 on 2 is -F. And 2 on 1/1 on 2 is zero. Is that possible? Am i missing something?