A taut string connects a 5-kg crate to a 12-kg crate (see figure below). The coefficient of static friction between the smaller crate and the floor is 0.573; the coefficient of static friction between the larger crate and the floor is 0.443. Find the minimum horizontal force required to start the crates in motion.

Crate 1:

Wc = m*g = 5kg * 9.8N/kg = 49 N. = Wt.
of crate.
Fs = u * Wc = 0.573 * 49 = 28.08 N. =
Force of static friction.

Crate 2:
Wc = m*g = 12kg * 9.8N/kg = 117.6 N.
Fs = 0.443 * 117.6 = 52.1 N.

Fap-Fs1-Fs2 = m*a.
Fap-28.08-52.1 = m*0 = 0
Fap-80.2 = 0
Fap = 80.2 N. = Force applied.

Well, it seems like the crates are feeling a bit clingy! Alright, let's get them moving.

To find the minimum force required to start the crates in motion, we need to consider the static friction acting on each crate. The maximum static friction force can be calculated using the formula:

Maximum static friction force (fs_max) = coefficient of static friction (μs) × normal force (N)

For the smaller crate, the normal force acting on it is equal to its weight (m1 × g), where m1 is the mass of the smaller crate (5 kg) and g is the acceleration due to gravity. So, the maximum static friction force for the smaller crate (fs1_max) is:

fs1_max = μs1 × (m1 × g)

Similarly, for the larger crate, the normal force acting on it is equal to its weight (m2 × g), where m2 is the mass of the larger crate (12 kg). So, the maximum static friction force for the larger crate (fs2_max) is:

fs2_max = μs2 × (m2 × g)

Now, since the force is being applied horizontally to the smaller crate, it will also provide the friction force required to move the larger crate. In other words, the force applied will be equal to the maximum static friction force for the larger crate:

Force applied = fs2_max = μs2 × (m2 × g)

Plugging in the values, we get:

Force applied = 0.443 × (12 kg × 9.8 m/s²)

Calculating that, we find:

Force applied ≈ 51.95 N

So, the minimum horizontal force required to start the crates in motion is approximately 51.95 Newtons. Time to give those crates a little push and let the fun begin!

To find the minimum horizontal force required to start the crates in motion, we need to consider the forces acting on each crate.

Let's analyze the forces acting on the smaller crate first:

1. Weight (Gravity): The weight of the smaller crate is given by the formula: weight = mass * acceleration due to gravity = 5 kg * 9.8 m/s^2 = 49 N.
2. Normal Force: The normal force is the force exerted by the floor on the crate perpendicular to the surface. It is equal in magnitude but opposite in direction to the weight of the crate when it is at rest. So, the normal force exerted on the smaller crate is 49 N.

Next, let's analyze the forces acting on the larger crate:

1. Weight (Gravity): The weight of the larger crate is given by the formula: weight = mass * acceleration due to gravity = 12 kg * 9.8 m/s^2 = 117.6 N.
2. Normal Force: The normal force is equal in magnitude but opposite in direction to the weight of the crate when it is at rest. So, the normal force exerted on the larger crate is 117.6 N.

Now, we need to determine the maximum static friction force that can be applied between the crate and the floor:

1. For the smaller crate, the maximum static friction force is given by the formula: static friction force = coefficient of static friction * normal force = 0.573 * 49 N = 28.077 N.
2. For the larger crate, the maximum static friction force is given by the formula: static friction force = coefficient of static friction * normal force = 0.443 * 117.6 N = 52.0608 N.

Since both crates are connected by a taut string, the tension force in the string is the same for both crates. Let's denote it as T.

Considering the smaller crate, the horizontal forces acting on it are:

1. Tension force: T.
2. Static friction force: In order for the crate to start moving, the static friction force needs to be overcome. The maximum static friction force is 28.077 N.

Therefore, the net horizontal force acting on the smaller crate is:

net force = T - static friction force.

Considering the larger crate, the horizontal forces acting on it are:

1. Tension force: T.
2. Static friction force: In order for the crate to start moving, the static friction force needs to be overcome. The maximum static friction force is 52.0608 N.

Therefore, the net horizontal force acting on the larger crate is:

net force = T - static friction force.

Since both crates are connected by the same string and we want to find the minimum horizontal force required to start the crates in motion, the tension force must be the same for both crates.

Therefore, we can equate the net forces for both crates:

T - 28.077 N = T - 52.0608 N.

Simplifying the equation, we find:

-28.077 N = -52.0608 N.

This equation is not possible, and it implies that the horizontal force required to start the crates in motion is 0 N, which means that they will not start moving under these conditions.

To overcome this situation, we would need a larger force applied horizontally to overcome the static friction forces between the crates and the floor.

To find the minimum horizontal force required to start the crates in motion, we need to analyze the forces acting on the crates.

Let's denote the smaller crate as crate A (5 kg) and the larger crate as crate B (12 kg).

First, let's calculate the maximum static friction for each crate. The maximum static friction is given by multiplying the coefficient of static friction with the normal force.

For crate A:
Maximum static friction force on crate A = coefficient of static friction (μA) * normal force on crate A

Similarly, for crate B:
Maximum static friction force on crate B = coefficient of static friction (μB) * normal force on crate B

The normal force acting on each crate is equal to their respective weights since they are on a horizontal surface.

For crate A:
Normal force on crate A = weight of crate A = m1 * g

For crate B:
Normal force on crate B = weight of crate B = m2 * g

where m1 = mass of crate A = 5 kg, m2 = mass of crate B = 12 kg, and g = acceleration due to gravity = 9.8 m/s^2.

Now, let's calculate the maximum static friction forces for each crate.

For crate A:
Maximum static friction force on crate A = μA * m1 * g

For crate B:
Maximum static friction force on crate B = μB * m2 * g

The maximum static friction force for each crate represents the maximum force needed to overcome static friction and start the motion.

Since the two crates are connected by a taut string, the force applied to crate A will be applied to crate B as well. Therefore, the minimum horizontal force required to start the crates in motion will be equal to the maximum static friction force on crate B.

Minimum horizontal force required = Maximum static friction force on crate B

So, the minimum horizontal force required to start the crates in motion is given by:

Minimum horizontal force required = μB * m2 * g

Substituting the given values:

μB = 0.443, m2 = 12 kg, g = 9.8 m/s^2

Minimum horizontal force required = 0.443 * 12 kg * 9.8 m/s^2

Calculating this value will give us the answer.