Two crates, of mass 75 kg and 110 kg, are in contact and at rest on a horizontal surface. A 620-N force is exerted on the 75-kg crate. If the coefficient of kinetic friction is 0.15, calculate (a) the acceleration of the system, and (b) the force that each crate exerts on the other.

(c) Repeat with the crates reversed.

please help me... I don't get this problem:(

https://www.jiskha.com/questions/1808191/Two-crates-of-mass-75-kg-and-110-kg-are-in-contact-and-at-rest-on-a-horizontal

No problem, I'm here to help you step-by-step. Let's solve the problem together.

(a) To calculate the acceleration of the system, we'll use Newton's second law of motion. The net force acting on the system is the force applied minus the frictional force.

The force applied on the system is 620 N, and the frictional force is the product of the coefficient of kinetic friction (0.15) and the normal force. The normal force is the weight of the system, which is the sum of the weights of the two crates.

The weight of each crate can be calculated using the formula: weight = mass x gravity, where gravity is approximately 9.8 m/s^2.

For the 75 kg crate:
Weight1 = 75 kg x 9.8 m/s^2 = 735 N

For the 110 kg crate:
Weight2 = 110 kg x 9.8 m/s^2 = 1078 N

The total weight of the system is:
Total weight = Weight1 + Weight2 = 735 N + 1078 N = 1813 N

Now, we can calculate the frictional force:
Frictional force = coefficient of kinetic friction x normal force
Frictional force = 0.15 x 1813 N = 271.95 N (approximately)

The net force acting on the system is:
Net force = Force applied - Frictional force
Net force = 620 N - 271.95 N = 348.05 N

Now, we can calculate the acceleration using Newton's second law:
Net force = mass x acceleration
348.05 N = (75 kg + 110 kg) x acceleration
348.05 N = 185 kg x acceleration

Solving for acceleration:
acceleration = 348.05 N / 185 kg ≈ 1.881 m/s^2

Therefore, the acceleration of the system is approximately 1.881 m/s^2.

(b) To calculate the force that each crate exerts on the other, we'll use Newton's third law of motion. According to this law, the force exerted by crate 1 on crate 2 is equal in magnitude and opposite in direction to the force exerted by crate 2 on crate 1.

Let's call the force exerted by crate 1 on crate 2 as F12, and the force exerted by crate 2 on crate 1 as F21.

Since the crates are in contact and at rest, the force F12 and F21 are equal in magnitude and opposite in direction to balance the system.

The force F12 = F21 can be calculated using Newton's third law:
F12 = F21 = mass1 x acceleration
F12 = F21 = 75 kg x 1.881 m/s^2 = 140.85 N (approximately)

Therefore, each crate exerts a force of approximately 140.85 N on the other.

(c) To repeat the calculations with the crates reversed, we'll swap the masses.

The mass of the first crate becomes 110 kg, and the mass of the second crate becomes 75 kg. We'll follow the same process as in (a) and (b) to calculate the acceleration and the force on each crate.

Let's go through the calculations again:

The weight of the first crate is:
Weight1 = 110 kg x 9.8 m/s^2 = 1078 N

The weight of the second crate is:
Weight2 = 75 kg x 9.8 m/s^2 = 735 N

The total weight of the system is:
Total weight = Weight1 + Weight2 = 1078 N + 735 N = 1813 N

The frictional force is:
Frictional force = coefficient of kinetic friction x normal force
Frictional force = 0.15 x 1813 N = 271.95 N (approximately)

The net force acting on the system is:
Net force = Force applied - Frictional force
Net force = 620 N - 271.95 N = 348.05 N

The acceleration of the system is:
Net force = mass x acceleration
348.05 N = (110 kg + 75 kg) x acceleration
348.05 N = 185 kg x acceleration

Solving for acceleration:
acceleration = 348.05 N / 185 kg ≈ 1.881 m/s^2

Therefore, the acceleration of the system is approximately 1.881 m/s^2.

The forces exerted by the crates on each other are:
F12 = F21 = mass1 x acceleration
F12 = F21 = 110 kg x 1.881 m/s^2 = 206.91 N (approximately)

Therefore, each crate exerts a force of approximately 206.91 N on the other when the crates are reversed.

I hope this helps you understand the problem better. Let me know if you have any further questions!

Sure, I can help you with this problem. Let's break it down step by step.

(a) To calculate the acceleration of the system, we first need to find the net force acting on the system. The net force is the force that is responsible for the acceleration. In this case, the net force is the difference between the applied force and the frictional force.

The frictional force can be calculated using the equation:

Frictional force = coefficient of kinetic friction * normal force

The normal force is the force exerted by the surface on the crates. Since the crates are on a horizontal surface, the normal force is equal to the weight of the crates.

Normal force = mass * gravity

Using the given masses of the crates and the acceleration due to gravity (approximately 9.8 m/s^2), you can calculate the normal forces on both crates.

Next, subtract the frictional force from the applied force to find the net force:

Net force = applied force - frictional force

Finally, use Newton's second law of motion to find the acceleration of the system:

Net force = mass * acceleration

So, by rearranging the equation, you can solve for the acceleration of the system.

(b) To calculate the force that each crate exerts on the other, you need to consider that the crates are in contact, and therefore there must be an equal and opposite force between them. Since the crates are not accelerating vertically, the vertical forces cancel each other out. Therefore, the horizontal force exerted by one crate on the other will be equal in magnitude but opposite in direction. To calculate this force, use Newton's third law of motion, which states that every action has an equal and opposite reaction.

In this case, the horizontal force exerted by the 75-kg crate on the 110-kg crate is equal to the horizontal force exerted by the 110-kg crate on the 75-kg crate.

(c) To repeat the calculations with the crates reversed, you would need to swap the mass values for the two crates. Calculate the acceleration and the forces using the same steps as mentioned above, but with the mass values reversed.

I hope this explanation helps you understand how to approach this problem! Let me know if you have any further questions.