Force between blocks pushed from one side
Two blocks, m1 = 2.0 kg and m2 = 3.0 kg are put in contact on a frictionless surface. A horizontal force F = 5.0 N is applied to one of them (see Fig.1). (1) Find the force Fa between the two blocks. (2) Find the force Fb if the force F is applied to m2 in the opposite direction.
Fig.1 is drawn in such a way that it suggests that the force F is applied from the outside, but we know that this force can be considered as concentrated in the center of gravity of this block. We draw the Free Body Diagrams for these blocks. For the block m1, it is on the left hand side of Fig.2, and for the block m2 on the right hand side of this Figure.
The forces are applied to the CG (center of gravity) of mass m1 and m2. In the explanations and calculations we will use bold face notation for vector quantities. It is easer to handle with text editors as opposed to handwriting, where arrows are more convenient to use. The N1 force is a reaction force to the gravity force m1g as described by Newton’s third Law of Motion. Analogically, N2 and m2g are equal and opposite as this law requires.
For the m1 block (the left part of Fig.2) the direction of force F is obvious from the text of the problem (see Fig.1). But why is the force Fa directed to the left? Because this is the force exerted on block m1 by block m2. Block m1 “pushes” block m2 so block m2 creates an “opposite” force according to Newton’s Third Law of Motion. Stop and think it over for a moment. This is a very crucial point of all problems involving Newton’s Laws. Once you understand the idea of drawing the Free Body Diagram you will solve all problems involving mechanics much faster and more efficiently.
Now it’s time to write equations based on Newton’s Laws. In the vertical direction
m1g = N1
so we can forget about these forces in further analysis.
In the horizontal direction the resultant force exerted on m1 is F - Fa and this is the force accelerating block m1. Therefore we can write
F - Fa = m1a (1)
The FBD for m2 shows that the only the horizontal force acting on it is the one exerted by block m1. This force has a magnitude of the Fa from the FBD on the left side if Fig.2, but the direction is opposite. This is the force accelerating block m2. As both blocks are in contact they must have the same acceleration a. So, for the second block the equation of motion is
Fa = m2 a (2)
We can drop out the vector notation from these two equations as the directions are well defined on the FBD’s for both blocks.
From the (2) we have
a = Fa/m2
and substituting this acceleration into (1) we find, after a little elementary algebra,
Fa = F m2 / (m1 +m2)
And this is the answer to question (1) from the problem.
If the force F is exerted from right to left, as in part (2) of the problem, the analogical reasoning will lead to the answer
FB = F m1 / (m1 + m2)
Substituting the values given in the problem we get
Fa =3.0 N and Fb = 1.2 N
You can wonder why the force between the blocks is larger when you push from the left. This is because in that situation the block which is a kind of transmitter of force must push the larger mass (m2) than in the second situation, when the larger block is pushing the smaller one.