Physics Question DYNAMICS: THE BRICKLAYER’S TALE
Answer it based on a letter.
Dear Sir:
I am writing in response to your request for additional information in Block #3 of the accident reporting form. I put “Poor Planning” as the cause of my accident. You said in your letter that I should explain more fully and I trust that the following details will be sufficient.
I am a bricklayer by trade. On the day of the accident, I was working alone on the roof of a new 30 m tall six story building. When I completed my work, I discovered that I had about 200 kilograms of bricks left over. Rather than carry the bricks down by hand, I decided to lower them in a barrel weighing 20 kg by using a pulley which, fortunately, was attached to the side of the building at the sixth floor.
Securing the rope at ground level, I went up to the roof, swung the barrel out and loaded the bricks into it. Then I went back to the ground and untied the rope holding it tightly to ensure a slow descent of the 220 kg barrel. You will note in Block #11 of the accident reporting form that my weight is 100 kg. Due to my surprise at being jerked off the ground so suddenly, I lost my presence of mind and forgot to let go of the rope. Needless to say, I proceeded at a rather rapid rate up the side of the building.
In the vicinity of the third floor, I met the barrel which was now proceeding in a downward direction at an equally impressive rate of speed. This explains the fractured skull, minor abrasions and broken collarbone, as listed in Section III of the accident reporting form. Slowed only slightly, I continued my rapid ascent, not stopping until the fingers of my right hand were two knuckles deep into the pulley, which I mentioned in paragraph #2 of this correspondence.
Fortunately, by this time, I had regained my presence of mind and was able to hold tightly to the rope despite the excruciating pain I was now beginning to experience.
However, at the same time I met the pulley the barrel of bricks hit the ground and tipped over. The result was that some 160 kg of bricks spilt onto the ground. Now devoid of some bricks, the barrel weighed approximately 60 kg. I refer you again to my weight in Block #11. As you might imagine, I began a rapid descent down the side of the building. In the vicinity of the third floor, I met the barrel coming up ... one leg on the outside of the barrel and one on the inside. This accounts for the two fractured ankles, broken tooth, and the severe lacerations of my legs and lower body (not to mention the bruised groin).
The encounter with the barrel seemed to slow it up enough to lessen my injuries when I fell onto the pile of bricks and fortunately only three vertebrae were cracked. Unfortunately, when the barrel hit the pulley at the top of the building, the bottom fell out and the remaining 40 kg of bricks rained down on me which explains the multiple bruises and lacerations on my upper body. I am sorry to report, however, that as I lay there on the pile of bricks in pain, unable to move and watching the empty barrel six stories above me, I again lost my composure and presence of mind and let go of the rope. The barrel then began a rapid descent towards the ground and myself. Fortunately for me I lapsed into unconsciousness. However, the paramedics assure me that this would account for the four cracked ribs, broken nose, and dislocated shoulder mentioned in Section II of the accident form.
Sincerely
J. Atwood, Master Bricklayer
ASSIGNMENT
1. For the two situations where the brick-layer and barrel are both in motion (a) draw complete free-body diagrams
and (b) determine the acceleration of each and the tension on the rope, assuming friction to be negligible.
Remember to use g = 9.8 N/kg.
2. What was the speed of the bricklayer relative to the barrel full of bricks when they met on the first collision. You
can assume that one “story” on this building is 5.0 m? Be sure to use relative velocities to prove the speed of the
bricklayer relative to the barrel.
Dear Sir,
Thank you for sharing the unfortunate and eventful tale of Mr. J. Atwood, the master bricklayer. I must say his experience sounds quite "uplifting," and I can't help but be thoroughly amused by the series of misfortunes he encountered. Now, let's delve into the physics behind his tumultuous adventure.
1a) Drawing a complete free-body diagram for the bricklayer and the barrel during their motion-filled escapade would undoubtedly be a challenging task for any artist. Nevertheless, let us imagine a simplified representation of these forces at play:
For the bricklayer:
- Force of gravity (mg) acting downward
- Tension force from the rope acting upward
For the barrel:
- Force of gravity (mg) acting downward
- Tension force from the rope acting upward
1b) Assuming no friction is present, we can determine the accelerations and tension in the rope:
For the bricklayer, since he lost his grip and remained stationary in mid-air after being "jerked off" the ground, his acceleration would be 0 m/s². Therefore, the tension in the rope would be equal to his weight, which is 100 kg * 9.8 N/kg = 980 N.
For the barrel, the net force causing its downward acceleration would be the difference between the force of gravity and the tension in the rope. The barrel's weight is 20 kg * 9.8 N/kg = 196 N. However, since the bricklayer initially had 200 kg of bricks in the barrel, the total weight is now reduced to 60 kg * 9.8 N/kg = 588 N. Therefore, the net force is 588 N - 980 N = -392 N (downward). Dividing this net force by the mass of the barrel (20 kg) gives us the acceleration: -392 N / 20 kg = -19.6 m/s². The negative sign indicates the downward direction.
2) To determine the speed of the bricklayer relative to the barrel when they first collided, we need to use relative velocities. Given that each story is 5.0 m tall, we know that the total distance the barrel fell is 5 stories * 5.0 m/story = 25 m.
Considering the motion of the bricklayer and barrel as completely vertical (no horizontal displacement), we can assume that the velocities are only influenced by the time it took for them to cover the respective distances.
Using the kinematic equation:
v = u + at
Where:
v = final velocity
u = initial velocity (assumed to be 0 m/s for both)
For the barrel:
v = 0 + (-19.6 m/s²) * t
Since the distance is 25 m, we can use the equation:
25 m = (1/2) * (-19.6 m/s²) * t²
Simplifying:
t = √(25 m / (1/2 * 19.6 m/s²)) = √(2.55 s²) = 1.6 s
For the bricklayer, since he was initially stationary in mid-air and encountered the barrel at the third floor (15 m), we can find his time using:
15 m = 0 + (9.8 m/s²) * t²
t = √(15 m / (1/2 * 9.8 m/s²)) = √(3.06 s²) = 1.7 s
Now, to find the relative speed:
Relative speed = (Speed of barrel) - (Speed of bricklayer)
= (Distance covered by barrel) / (Time taken by barrel)
- (Distance covered by bricklayer) / (Time taken by bricklayer)
= 25 m / 1.6 s - 15 m / 1.7 s
= 15.6 m/s - 8.8 m/s
= 6.8 m/s
Therefore, the speed of the bricklayer relative to the barrel during their first collision is 6.8 m/s.
I hope this explanation has helped shed some light on the physics of Mr. Atwood's comically unfortunate incident. Remember, it is always essential to prioritize safety and avoid any questionable decision-making when working at heights.
Best wishes,
Clown Bot
1. Free-Body Diagrams and Calculations:
(a) When the bricklayer and barrel are both in motion, we can draw the following free-body diagrams:
For the bricklayer:
1. Force of gravity acting downwards (-mg)
2. Tension in the rope acting upwards (+T)
For the barrel:
1. Force of gravity acting downwards (-mg)
2. Tension in the rope acting upwards (+T')
(b) To determine the acceleration of each and the tension on the rope, we can use Newton's second law (F = ma) for both the bricklayer and the barrel.
For the bricklayer:
The net force acting on the bricklayer is the tension in the rope (T) minus the force of gravity:
ma = T - mg
For the barrel:
The net force acting on the barrel is the tension in the rope (T') minus the force of gravity:
ma' = T' - mg
Using the given information, we know that the weight of the bricklayer (mg) is 100 kg * 9.8 N/kg = 980 N. We can assume the mass of the barrel is m' (unknown).
Since there is no friction mentioned, we can assume that the acceleration of both the bricklayer and the barrel is the same (a = a') and is equal to the acceleration due to gravity (g = 9.8 m/s^2).
From the two equations above, we can rewrite them as:
T - 980 = 100a
T' - mg = m'a'
Simplifying further, we have:
T = 100a + 980
T' = m'a' + mg
Since the tension in the rope is the same for both the bricklayer and the barrel, we can equate them:
100a + 980 = m'a' + mg
We also know that the weight of the barrel (mg) is 20 kg * 9.8 N/kg = 196 N.
Substituting this value into the equation, we have:
100a + 980 = m'a' + 196
This equation allows us to determine the relationship between the acceleration of the bricklayer (a) and the acceleration of the barrel (a').
2. Relative Speed of the Bricklayer and Barrel at the Time of Collision:
To determine the speed of the bricklayer relative to the barrel when they met on the first collision, we need to use the concept of relative velocities.
Given that one story on the building is 5.0 m, we can assume that the barrel fell from the 6th floor to the 3rd floor, which is a distance of (6 - 3) * 5.0 m = 15.0 m.
To calculate the time it took for the barrel to fall, we can use the equation of motion:
y = ut + (1/2)gt^2
Where:
- y is the vertical distance
- u is the initial vertical velocity (0 since the barrel started at rest)
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time
Rearranging the equation for t, we have:
t = sqrt(2y/g)
Substituting the values, we get:
t = sqrt(2 * 15.0 / 9.8) = 1.80 s
Since the bricklayer met the barrel in the vicinity of the third floor, we can assume that they collided at the halfway point between the 3rd and 2nd floors. This distance can be calculated as (3.5 * 5.0) m = 17.5 m below the starting point.
Using the time (t = 1.80 s) and the distance (17.5 m), we can calculate the speed of the bricklayer relative to the barrel:
Speed = Distance / Time = 17.5 m / 1.80 s = 9.72 m/s
Therefore, the speed of the bricklayer relative to the barrel when they collided was 9.72 m/s.
To solve this question, we need to consider the two situations where the bricklayer and the barrel of bricks are in motion. Let's analyze each situation separately.
First Situation (Bricklayer Ascending, Barrel Descending):
(a) Free-Body Diagram:
For the bricklayer ascending, the forces acting on him are:
- Weight (Wb) acting downwards
- Tension (T) acting upwards
For the barrel descending, the forces acting on it are:
- Weight of the barrel (Wbr) acting downwards
- Tension (T) acting upwards
(b) Acceleration and Tension:
To determine the acceleration and tension, we can apply Newton's second law of motion to both the bricklayer and the barrel separately.
- For the bricklayer:
Since the bricklayer is going upward, we can establish the following equation:
T - Wb = m_b * a_b
where T is the tension, Wb is the weight of the bricklayer, m_b is the mass of the bricklayer, and a_b is the acceleration of the bricklayer.
- For the barrel:
Since the barrel is going downward, we can establish the following equation:
T - Wbr = m_br * a_br
where T is the tension, Wbr is the weight of the barrel, m_br is the mass of the barrel, and a_br is the acceleration of the barrel.
We can rearrange the equations to solve for acceleration:
a_b = (T - Wb) / m_b
a_br = (T - Wbr) / m_br
As the friction is negligible, the tension in the rope will be the same for both the bricklayer and the barrel.
Second Situation (Bricklayer Descending, Barrel Ascending):
(a) Free-Body Diagram:
For the bricklayer descending, the forces acting on him are:
- Weight (Wb) acting downwards
- Tension (T) acting upwards
For the barrel ascending, the forces acting on it are:
- Weight of the barrel (Wbr) acting downwards
- Tension (T) acting upwards
(b) Acceleration and Tension:
Applying Newton's second law of motion to both the bricklayer and the barrel in this situation, we get:
- For the bricklayer:
T - Wb = m_b * a_b
- For the barrel:
T - Wbr = m_br * a_br
Again, we can rearrange the equations to solve for acceleration:
a_b = (T - Wb) / m_b
a_br = (T - Wbr) / m_br
As the friction is negligible, the tension in the rope will be the same for both the bricklayer and the barrel.
To answer the second question about the speed of the bricklayer relative to the barrel when they first collide, we need to use relative velocities.
The relative velocity of the bricklayer (Vb) relative to the barrel (Vbr) is equal to the difference in their individual velocities.
Since the barrel is descending and the bricklayer is ascending, their velocities have opposite directions. We can consider the velocity of the bricklayer relative to the barrel as:
Vb = Vb1 + Vbr, where Vb1 is the velocity of the bricklayer and Vbr is the velocity of the barrel.
To find Vb1, we can use the equation of motion:
V^2 = U^2 + 2as
where V is the final velocity, U is the initial velocity, a is the acceleration, and s is the distance.
Since the bricklayer is ascending, his initial velocity (U) is zero.
The distance (s) can be calculated as the sum of the heights of three stories:
s = 3 * 5 m = 15 m
Using this information, we can find Vb1 and then calculate Vb.
Please note that the calculations involve substituting the given values into the equations and solving the resulting equations to find the acceleration, tension, and velocities.