Question

During another immunity challenge, a box (m = 26 kg) is positioned on an inclined surface (angle = 33°). A rope is attached to the box and the rope is placed through a pulley and another box(m = 32 kg) filled with puzzle pieces and is suspended from the other end of the rope (see figure below). Several tribemates are positioned on the inclined surface holding onto the first box andwaiting for the rest of their tribe to position themselves below the box with the puzzle pieces in order to catch it and complete the rest of the challenge. When the tribemates on the inclinedsurface release the first box from rest, how long does it take the other tribemates to catch the box of puzzle pieces if the box falls from a height of 4.2 m one the system is released and thecoefficient of kinetic friction between the first box and the inclined surface is 0.25? What is the tension in the rope connecting the two boxes?

Answer 1

Here's a diagram of the situation with all the forces acting on the boxes.

To calculate the tension in the rope, we need to calculate the sum of forces pulling on the rope.

Let's call Fg1 the force of gravity of the first box (on the incline) and Fg2 the force of gravity on the second.

Remember that F = mg, where m is mass and g is gravitational acceleration (9.81 m/s^2)

Fg2 = 32*9.81 = 313.92 N

Fg1 = 26*9.81 = 255.06 N

Since the first box is on an incline, the gravitational force can be broken up into two components: the component parallel (tangent) to the incline, and the other perpendicular (normal) to the incline. Let's call those forces FT and FN respectively.

Let's rearrange the force arrows to form a right triangle:

Since we know Fg1 and an angle measure in this triangle, we can calculate FN and FT using trigonometry.

FN = Fg1*cos(33) = 255.06*cos(33) = 213.911 N

FT = Fg1*sin(33) = 255.06*sin(33) = 138.916 N

Now that we have FN, we can calculate the force of friction, let's call that Ff.

Ff = μ*FN, where μ is the coefficient of friction.

Ff = 0.25*213.911 = 53.478 N

In relation to the whole pulley system, the boxes pull each other toward their respective directions. For example, the first box pulls the second box to the left of the pulley, and the second box pulls the first to the right. Imagine the boxes on opposite sides of a tug-of-war. The box that pulls with the most force will win the tug-of-war and pull the entire pulley system in its direction.

The second box is tugging with a force of Fg2, its gravitational force.

The first box is tugging with FT+Ff, its tangent gravitational force and the force of friction.

The total tension in the rope is the sum of all of these tugging forces.

Total tension = Fg2 + FT + Ff = 313.92+138.916+53.478

Total tension = 506.314 N (answer to second part)

Now, the second box tugs with a force of 313.92 N, and the first box tugs with a force of 138.916+53.478 = 192.394 N, and since the tugging force of the second box is greater, the whole pulley system will move in that direction. The resultant force is the tugging force of the second box minus that of the first box. Let's call this FR.

FR = 313.92 - 192.394 = 121.526 N

The pulley now pulls the mass of both of these boxes combined with this force. Let's call this m.

m = 26kg + 32kg = 58 kg

Remember again that F = m*a. Let's substitute FR for F, then calculate a.

121.526 = 58a; a = 2.0953 m/s^2

So the pulley system accelerates at this rate, so therefore the second box falls at this rate as well.

We also have that it falls a distance d = 4.2m before it's caught, and the pulley system starts at rest (up until the second box is released), so the initial velocity vi = 0.

We need to calculate the time it takes for the box to fall this far, t.

Given the variables we know and need, we can use the following equation:

d = vi*t + (a/2)*t^2

We can substitute for d, vi, and a, then solve for t.

4.2 = 0 + (2.0953/2)*t^2

Solving for t,

t = 2.0022 s (answer to first part)