Refer to Concept Simulation 4.4 for background relating to this problem. The drawing shows a large cube (mass = 28.9 kg) being accelerated across a horizontal frictionless surface by a horizontal force P. A small cube (mass = 4.4 kg) is in contact with the front surface of the large cube and will slide downward unless P is sufficiently large. The coefficient of static friction between the cubes is 0.710. What is the smallest magnitude that P can have in order to keep the small cube from sliding downward?

smallest magnitud is P=33.3 N

Explanation:

We are analyze the situation as an external force is applied and there is a friction force. We have a problem with Newton's second law.

F = ma

As the two blocks go together they must have the same acceleration, so we can calculate this for the entire system

P = (m1 + m2) a

a = P / (m1 + m2)

In this case there is no friction force because the small block does not touch the ground.

In order to calculate the friction force, we must analyze each system component separately.

The large block on the X axis has an applied P force and as it moves feels a force from the small block. In the Y axis has the weight (W1) and the reaction to normal (N1)

For the small block on the X axis, the force it feels is the thrust of the large block, note that this is an action and reaction force between the two blocks, it is the same definition we have of the normal one, so we can call this force (N)

Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

N = m2 a

fr - W2 = 0

fr = W2

The definition of friction force is

fr = μ N

Let's replace and calculate

μ (m2 a) = m2 g

μ (P / (m1 + m2)) = g

P = g / μ (m1 + m2)

Let's calculate the value of this force

P = 9.8 / 0.710 (28.9 + 4.4)

P = 13.80 (33.3)

P = 33.3 N

This is the minimum friction force that prevents the block from sliding down