A force of magnitude 7.50 N pushes three boxes with masses m_1 = 1.30 kg,m_2 = 3.20 kg, and m_3 = 4.90 kg .The mass is side by side.

1-Find the magnitude of the contact force between boxes 1 and 2, and
2-Find the magnitude of the contact force between boxes 2 and 3.

*the position of the box is like this:-

7.50N(push)-> |m1|m2|m3|

Ok, I assume you didn't mean side to side, but they are in a front to back line, 1 in front, 3 in back, and you are pushing on 3.

find the net acceleration
F=ma
7.5=(summasses)a solve for a.

Now
1) box one is being pushed, so force=m1*a

2) box 2 and 1 are being pushed, so
f=(M1+m2)a

Well, it seems like these boxes are having a friendly push and shove session! Let's calculate their contact forces.

1. To find the contact force between boxes 1 and 2, we need to know the acceleration of the system. If we assume they all move together, the total mass of the system is:

m_total = m1 + m2 + m3
m_total = 1.30 kg + 3.20 kg + 4.90 kg
m_total ≈ 9.40 kg

Using Newton's second law, F = m * a, we can calculate the acceleration:

F = m_total * a
7.50 N = 9.40 kg * a
a ≈ 0.80 m/s^2

Since boxes 1 and 2 move together, the force between them is just the force exerted on the system minus the force on box 3:

F_contact12 = F_system - F3
F_contact12 = 7.50 N - F3

2. To find the contact force between boxes 2 and 3, we assume they move together as well. So the force between them is the force on box 3:

F_contact23 = F3

Can I just say, these boxes are really close-knit, right? The contact force is basically like a friendly pat on the back between them!

To find the magnitudes of the contact forces between the boxes, we can follow these steps:

1. Determine the net force on each box using Newton's second law:
F_net = ma
where F_net is the net force, m is the mass of the box, and a is the acceleration.

2. Calculate the acceleration of each box using the given force:
a = F_net / m
In this case, since the force is the same for all three boxes, the acceleration will also be the same for all three.

3. Use Newton's third law to determine the contact forces between the boxes. According to this law, the forces between two objects in contact are equal in magnitude and opposite in direction:
F_contact1-2 = -F_contact2-1 (force between boxes 1 and 2)
F_contact2-3 = -F_contact3-2 (force between boxes 2 and 3)

Let's calculate the magnitudes of the contact forces step by step:

Given:
Force (F) = 7.50 N
Mass of box 1 (m1) = 1.30 kg
Mass of box 2 (m2) = 3.20 kg
Mass of box 3 (m3) = 4.90 kg

1. Determine the acceleration (a):
Using Newton's second law, we have:
a = F / (m1 + m2 + m3)
a = 7.50 N / (1.30 kg + 3.20 kg + 4.90 kg)
a ≈ 0.708 m/s²

2. Calculate the net force on each box using the acceleration:
F_net1 = m1 * a = 1.30 kg * 0.708 m/s²
F_net1 ≈ 0.919 N

F_net2 = m2 * a = 3.20 kg * 0.708 m/s²
F_net2 ≈ 2.27 N

F_net3 = m3 * a = 4.90 kg * 0.708 m/s²
F_net3 ≈ 3.47 N

3. Determine the magnitudes of the contact forces:
Using Newton's third law, we have:
F_contact1-2 = -F_contact2-1
F_contact2-3 = -F_contact3-2

Therefore,
F_contact1-2 = F_net2 = 2.27 N
F_contact2-3 = F_net3 = 3.47 N

To summarize the results:
1. The magnitude of the contact force between boxes 1 and 2 is approximately 2.27 N.
2. The magnitude of the contact force between boxes 2 and 3 is approximately 3.47 N.

To find the magnitude of the contact force between boxes 1 and 2, and between boxes 2 and 3, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

1) To find the magnitude of the contact force between boxes 1 and 2:

Step 1: Calculate the total mass of boxes 1 and 2, since they are side by side.
Total mass (m_total) = m1 + m2

Given:
m1 = 1.30 kg
m2 = 3.20 kg

m_total = 1.30 kg + 3.20 kg = 4.50 kg

Step 2: Calculate the acceleration of the system.
Using Newton's second law, we know that the net force acting on the system is equal to the product of the total mass and acceleration.
Net force (F_net) = m_total * a

Given:
F_net = 7.50 N
m_total = 4.50 kg

Substitute the known values into the equation and solve for the acceleration.
7.50 N = 4.50 kg * a

a = 7.50 N / 4.50 kg = 1.67 m/s^2

Step 3: Calculate the contact force between boxes 1 and 2.
Now, since boxes 1 and 2 are side by side and have the same acceleration, the contact force between them will be equal. Let's call this force F_12.

Using Newton's second law again, we can write:
F_12 = m1 * a

Given:
m1 = 1.30 kg
a = 1.67 m/s^2

Substituting the values, we get:
F_12 = 1.30 kg * 1.67 m/s^2 = 2.171 N

2) To find the magnitude of the contact force between boxes 2 and 3:

Step 1: Calculate the total mass of boxes 2 and 3, since they are side by side.
Total mass (m_total) = m2 + m3

Given:
m2 = 3.20 kg
m3 = 4.90 kg

m_total = 3.20 kg + 4.90 kg = 8.10 kg

Step 2: Calculate the acceleration of the system.
Using Newton's second law, we know that the net force acting on the system is equal to the product of the total mass and acceleration.
Net force (F_net) = m_total * a

Given:
F_net = 7.50 N
m_total = 8.10 kg

Substitute the known values into the equation and solve for the acceleration.
7.50 N = 8.10 kg * a

a = 7.50 N / 8.10 kg = 0.926 m/s^2

Step 3: Calculate the contact force between boxes 2 and 3.
Now, since boxes 2 and 3 are side by side and have the same acceleration, the contact force between them will be equal. Let's call this force F_23.

Using Newton's second law again, we can write:
F_23 = m2 * a

Given:
m2 = 3.20 kg
a = 0.926 m/s^2

Substituting the values, we get:
F_23 = 3.20 kg * 0.926 m/s^2 = 2.963 N

Therefore,
1) The magnitude of the contact force between boxes 1 and 2 is 2.171 N.
2) The magnitude of the contact force between boxes 2 and 3 is 2.963 N.