Three boxes are lined up so that they are touching each other as shown in the figure below. Box A has a mass of 30.0 kg, box B has a mass of 45.0 kg, and box C has a mass of 75.0 kg. An external force (F) pushes on box A toward the right, and the force that box B exerts on box C is 190 N.What is the acceleration of the boxes? (Express your answer to three significant figures.) What is the magnitude of the external force F (Express your answer to three significant figures.)

Well, it seems like these boxes are related to each other like a dysfunctional family. But let's not judge – they need our help.

To determine the acceleration of the boxes, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.

First, let's figure out the net force acting on box A. Since there are no other external forces mentioned, the net force acting on box A is simply the force it experiences from box B. We know that force is equal to mass multiplied by acceleration (F = ma), so we can rearrange the equation to solve for acceleration:

acceleration = force / mass.

The force exerted by box B on box C is 190 N. But since this force is equal to the net force acting on box A, we can say that the acceleration of the boxes is 190 N divided by the mass of box A, which is 30.0 kg.

So the acceleration of the boxes is approximately: acceleration = 190 N / 30.0 kg = 6.33 m/s².

Now, let's move on to the magnitude of the external force, F. Since we know the acceleration, we can use the same equation as before to solve for the force:

force = mass × acceleration.

For this, we need to consider the total mass of box A with box B (since they are both being pushed by the external force, F). The total mass is the sum of their individual masses, which is 30.0 kg + 45.0 kg = 75.0 kg.

Now we can plug in the known values:

force = 75.0 kg × 6.33 m/s² = 474.75 N.

So, the magnitude of the external force, F, is approximately 474.75 N. Oh, the things we do for dysfunctional families.

Hope that helps!

To find the acceleration of the boxes, we can use Newton's second law of motion which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

Step 1: Calculate the net force on Box A:
The net force on Box A is the force pushing it to the right (which we'll denote as F) minus the force Box B exerts on it (which we'll denote as F_BA).
Net force on Box A = F - F_BA

Step 2: Calculate the net force on Box C:
The net force on Box C is the force Box B exerts on it (F_BA) minus the force used to push it to the right (which we'll denote as F_CA).
Net force on Box C = F_BA - F_CA

Step 3: Equate the net forces to the respective masses multiplied by the acceleration:
For Box A:
Net force on Box A = mass of Box A x acceleration of the boxes
F - F_BA = mass of Box A x acceleration

For Box C:
Net force on Box C = mass of Box C x acceleration of the boxes
F_BA - F_CA = mass of Box C x acceleration

Step 4: Substitute the given mass values:
Mass of Box A = 30.0 kg
Mass of Box C = 75.0 kg

Step 5: Solve the equations simultaneously to find the acceleration:
From the equation for Box A:
F - F_BA = 30.0 kg x acceleration ...(Equation 1)

From the equation for Box C:
F_BA - F_CA = 75.0 kg x acceleration ...(Equation 2)

Adding Equation 1 and Equation 2 together:
F - F_BA + F_BA - F_CA = 30.0 kg x acceleration + 75.0 kg x acceleration

F - F_CA = (30.0 kg + 75.0 kg) x acceleration

F - F_CA = 105.0 kg x acceleration ...(Equation 3)

Step 6: Substitute the given force value:
The force that Box B exerts on Box C (F_BA) is given as 190 N.

Substitute this value into Equation 3:
F - 190 N = 105.0 kg x acceleration

Step 7: Rearrange the equation to solve for the acceleration:
F = 190 N + 105.0 kg x acceleration

acceleration = (F - 190 N) / 105.0 kg

Now, to solve for the acceleration, we need the magnitude of the external force F, which brings us to the next part of the question.

The magnitude of the external force F can be found by isolating F in Equation 1:
F = F_BA + 30.0 kg x acceleration

Step 8: Substitute the given values to solve for F:
Given: F_BA = 190 N, acceleration = (F - 190 N) / 105.0 kg

F = 190 N + 30.0 kg x [(F - 190 N) / 105.0 kg]

Simplify:
F = 190 N + 0.2857(F - 190 N)

Solve for F:
F = 190 N + 0.2857F - 54.285 N

0.7143F = 244.285 N

F = 244.285 N / 0.7143

F ≈ 342.049 N (rounded to three significant figures)

Now that we have found the magnitude of the external force F, we can substitute it back into Equation 1 to find the acceleration (using the rearranged equation from Step 7):

acceleration = (F - 190 N) / 105.0 kg

Substitute F = 342.049 N:

acceleration = (342.049 N - 190 N) / 105.0 kg

acceleration ≈ 1.275 m/s² (rounded to three significant figures)

Therefore, the acceleration of the boxes is approximately 1.275 m/s², and the magnitude of the external force F is approximately 342.049 N.

F accelerates the three boxes equally

since C is half of the total system mass, F equals twice the force on C

Fa+b=-Fc

therefore, acceleration can be calculated by:
a=F/(mass of a+b)

External force
Fa=-Fb+c
The force of box a on boxes b+c is the external force, therefore external force = force of boxes b+c on a
Fb+c= (mass b+c)*acceleration from above