Two masses M1 = 3.00 kg and M2 = 7.50 kg are stacked on top of each other as shown in the figure. The static coefficient of friction between M1 and M2 is μs = 0.520. There is no friction between M2 and the surface below it

To solve this problem, we need to consider the forces acting on the system and apply Newton's laws of motion. Specifically, we need to analyze the vertical and horizontal forces.

Let's start by considering the vertical forces. Since M1 and M2 are stacked on top of each other, they experience the same magnitude of gravitational force, which is given by Fg = m * g, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2).

For M1, the downward force is Fg1 = M1 * g = 3.00 kg * 9.8 m/s^2 = 29.4 N.

For M2, the downward force is Fg2 = M2 * g = 7.50 kg * 9.8 m/s^2 = 73.5 N.

Since the two masses are in contact and there is no friction between M2 and the surface below it, the normal force acting on M1 is equal to Fg2, which is 73.5 N.

Now let's consider the horizontal forces. The only horizontal force in this system is the friction force between M1 and M2. The friction force can be calculated using the equation Ff = μ * Fn, where μ is the coefficient of friction and Fn is the normal force.

For M1 and M2, the friction force is Ff = μs * Fn = 0.520 * 73.5 N = 38.22 N.

Since there is no acceleration in the horizontal direction (the system is not moving), the friction force must be equal in magnitude to the static friction force.

In summary, the friction force acting between M1 and M2 is 38.22 N.