3. Think about this pulling scenario: A 60 N force pulls three boxes across a surface with no friction. T1 is the tension force the rope on the left pulls with at both ends, and T2 is the force the rope on the right pulls with on both ends. (a) Draw diagrams of the three boxes and the forces they feel. (b) Find how much all three accelerate (will they have equal or unequal accelerations?)

To answer this question, let's break it down into two parts: (a) drawing the diagrams of the boxes and the forces they experience, and (b) finding their accelerations.

(a) Drawing the diagrams:
In this scenario, we have three boxes being pulled by two ropes. Let's label the boxes A, B, and C from left to right. We'll also label the tension forces acting on them as T1 and T2.

----------- -----------
A B C

T1 T1 T2

The leftmost box A is being pulled by the rope on the left with a tension force T1. The middle box B is also being pulled by the same rope on the left with a tension force T1. Finally, the rightmost box C is being pulled by the rope on the right with a tension force T2.

(b) Finding the accelerations:
Since there is no friction on the surface, the only external forces acting on the boxes are the tension forces T1 and T2. The net force acting on an object is equal to the product of its mass and acceleration (Newton's second law).

We know that the force pulling each box is 60 N, but we need to determine the individual masses of the boxes or the ratio of the masses to calculate their individual accelerations.

For simplicity, let's assume that the masses of the three boxes are equal, so their individual accelerations will also be equal. In this case, when we sum the forces acting on the system, we have:

T1 + T2 = (mass of each box) * (acceleration of each box)

Since the tension forces T1 and T2 are acting in the opposite direction to the motion, we can say:

T1 + T2 = 3 * m * a, where m is the mass of each box and a is the acceleration.

Since we have T1 = T2, we can simplify the equation:

2T1 = 3 * m * a

Now, we know that the tension force T1 is 60 N. Substituting this value into the equation, we have:

2(60) = 3 * m * a

Simplifying further:

120 = 3 * m * a

40 = m * a

Therefore, the acceleration of each box is 40 m/s². Hence, all three boxes will have equal accelerations in this scenario.

Note: If the masses of the boxes were different, their accelerations would also differ.

(a) Diagrams of the three boxes and the forces they feel:

Box 1
_______
| |
| m1 |
|_______|

↗️ T1 ⬅️️⬅️️⬅️️️️️ ⬅️️⬅️️⬅️️️️️️ F1
_____________________________________________________

Box 2
_______
| |
| m2 |
|_______|

↗️ T1 ⬅️️⬅️️⬅️️️️️ ⬅️️⬅️️⬅️️️️️️ F2
_____________________________________________________

Box 3
_______
| |
| m3 |
|_______|

↗️ T1 ⬅️️⬅️️⬅️️️️️ ⬅️️⬅️️⬅️️️️️️ F3
_____________________________________________________

Here, T1 is the tension force acting on all three boxes from the left rope, and F1, F2, and F3 are the external forces acting on each box.

(b) To find how much all three boxes accelerate, we can use Newton's second law of motion:

ΣF = m * a

Since there is no friction and the tension force T1 acts on all three boxes, the net force ΣF on each box will be the same. Therefore, using the equation above:

F1 = m1 * a₁
F2 = m2 * a₂
F3 = m3 * a₃

Since ΣF = F1 = F2 = F3, we can write:

m1 * a₁ = m2 * a₂ = m3 * a₃

Since the masses of the boxes are different, their accelerations will be unequal.