A cube of mass m1 = 7.7 kg is sitting on top of another cube of the same size and mass m2 = 0.6 kg while they are both in free-fall. Ignoring any air resistance, what is the magnitude of the normal force with which the bottom cube is acting on the top cube?

for this question. I used FN=mg. I multiplied 7.7 x9.8 to get 75N but I got it wrong.
Please help

To answer this question, we need to consider the forces acting on both cubes.

First, let's analyze the forces acting on the top cube:
1. Weight force (from gravity) acting on the top cube = m1 * g
Plugging in the given values, this is equal to 7.7 kg * 9.8 m/s^2 = 75.46 N.

Now, let's consider the forces acting on the bottom cube:
1. Weight force (from gravity) acting on the bottom cube = m2 * g
Plugging in the given values, this is equal to 0.6 kg * 9.8 m/s^2 = 5.88 N.

The normal force is the force exerted by a surface to support an object in contact with it. In this case, it is the force exerted by the bottom cube on the top cube.

From Newton's third law, we know that the magnitude of the normal force acting on the top cube is equal to the magnitude of the normal force acting on the bottom cube.

Therefore, the magnitude of the normal force with which the bottom cube is acting on the top cube is 5.88 N.

So, you made a mistake by using the weight force of the top cube to calculate the normal force when you should have used the weight force of the bottom cube.

They are equal to each other so zero force