# Can the normal force between a blocks accelerating together be zero?

## Homework Statement

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?

## The Attempt at a Solution

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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?
you made a mistake somewhere doing this.

The easiest way to do it is to start by calculating the acceleration of all the blocks by using F = m*a for all of them together

tiny-tim
Homework Helper
Hi mistymoon_38! Hint: Call the acceleration A.

So the net force on the three blocks as a whole is F = 3mA.

Then what is the net force on the first block, to produce an acceleration of A? Hey! We have the same problem for my Physics Recitation.

I disagree with your block 1 which might lead to your error. I said that The Force Applied is going to the right and The Force (2 on 1) is going to the left.

This means that F(1 on 2) = -F(2 on 1) so it should not equal zero but be the same.