Search: The figure shows three crates being pushed over a concrete floor by a horizontal force of F magnitude 440 N. The masses of the crates are m1 = 30.0 kg, m2 = 10.0 kg, and m3 = 20.0 kg. The coefficient of kinetic friction between the floor and each of the crates is 0.700. (a) What is the magnitude F32 of the force on crate 3 from crate 2? (b) If the crates then slide onto a polished floor, where the coefficient of kinetic friction is less than 0.700, is magnitude F32 more than, less than, or the same as it was when the coefficient was 0.700?

Since the Figure is not available to us, we need to know which of the crates receives the external 440 N force.

Problems of this sort can be solved by applying free-body-diagram equations of motion to each of the three crates, remembering that the forces between any two crates that are in contact are equal and opposite. This allows one to solve for all three unkowns: F32 = -F23, F12 = -F21, and the acceleration, a.

To find the magnitude F32 of the force on crate 3 from crate 2, we need to analyze the forces acting on each crate.

(a) On crate 1:
- The only force acting horizontally is the force of magnitude 440 N.
- The force of kinetic friction acts in the opposite direction and its magnitude can be calculated as: Fk1 = μ1 * N, where μ1 is the coefficient of kinetic friction between crate 1 and the floor, and N is the normal force.
- The normal force N can be calculated as: N = m1 * g, where m1 is the mass of crate 1 and g is the acceleration due to gravity (approximately 9.8 m/s^2).
- The force of kinetic friction on crate 1 is: Fk1 = μ1 * m1 * g.

(b) On crate 2:
- Crate 2 experiences two forces: the force from crate 1 and the force on crate 3 from crate 2.
- The force on crate 3 from crate 2 is F32.
- Since crate 2 is being pushed horizontally, the force from crate 1 on crate 2 is equal in magnitude but opposite in direction to the force on crate 1 from crate 2.
- Therefore, the force from crate 1 on crate 2 is -440 N.

(c) On crate 3:
- Similarly to crate 1, crate 3 experiences a force of kinetic friction: Fk3 = μ3 * m3 * g.

To find F32, we need to equate the net force acting on crate 2 in the horizontal direction to zero:
0 = -440 N + F32 - Fk3.

Now, substitute the expression for Fk3:
0 = -440 N + F32 - (μ3 * m3 * g).

Rearrange the equation to solve for F32:
F32 = -(-440 N + μ3 * m3 * g).

Now, we can substitute the given values:
F32 = -(-440 N + 0.700 * 10.0 kg * 9.8 m/s^2).

The expression simplifies to:
F32 = 440 N - 68.6 N.

Therefore, the magnitude of F32 is:
F32 = 371.4 N.

(b) If the crates then slide onto a polished floor with a coefficient of kinetic friction less than 0.700, the force of kinetic friction on crate 3, Fk3, will decrease. As F32 is dependent on Fk3, its magnitude will be less than when the coefficient was 0.700.

To solve this problem, we can use Newton's second law of motion and the concept of friction.

(a) To find the magnitude F32 of the force on crate 3 from crate 2, we need to consider the forces acting on crate 2.

1. Calculate the gravitational force on crate 2:
Fg2 = m2 * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).

2. Calculate the force due to friction acting on crate 2:
Fk2 = μ * Fnorm
where μ is the coefficient of kinetic friction, and Fnorm is the normal force.

3. Calculate the net force on crate 2:
Fnet2 = Fg2 - Fk2

4. Calculate the force on crate 3 from crate 2:
F32 = -Fnet2

Let's substitute the given values into the equations:
m2 = 10.0 kg
g = 9.8 m/s^2
μ = 0.700

1. Calculate the gravitational force on crate 2:
Fg2 = m2 * g = 10.0 kg * 9.8 m/s^2 = 98 N

2. Calculate the force due to friction acting on crate 2:
Fk2 = μ * Fnorm = μ * (m2 * g) = 0.700 * (10.0 kg * 9.8 m/s^2) = 68.6 N

3. Calculate the net force on crate 2:
Fnet2 = Fg2 - Fk2 = 98 N - 68.6 N = 29.4 N

4. Calculate the force on crate 3 from crate 2:
F32 = -Fnet2 = -29.4 N

Therefore, the magnitude of the force on crate 3 from crate 2 is 29.4 N.

(b) If the crates slide onto a polished floor with a coefficient of kinetic friction less than 0.700, the force of friction will be smaller. This means that the net force on crate 2 will be smaller than before. Since the force on crate 3 from crate 2 is directly related to the net force on crate 2, the magnitude F32 will be less than it was when the coefficient was 0.700.