Physics - Unit V Challenge Problem
No matter how bad your day is going, here's proof that things could be A LOT worse.
This is a bricklayer's accident report that was printed in the newsletter of the English
equivalent of the Workers' Compensation Board. So here, thanks to John Sedgwick, is this
Bricklayer's report.
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 asked for a
fuller explanation and I trust 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 six-story
building. When I completed my work, I found I had some bricks left over which when
weighed later were found to weigh 240 lbs. Rather than carry the bricks down by hand, I
decided to lower them in a barrel by using a pulley which 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 down and untied the rope, holding it
tightly to insure a slow descent of the 240 lbs of bricks. You will note on the accident
reporting form that my weight is 135 lbs.
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 rapid rate up the side of
the building.(Q1) In the vicinity of the third floor, I met the barrel which was now
proceeding downward at an equally impressive speed. (Q2) This explains the fractured
skull, minor abrasions and the broken collarbone, as listed in Section 3, 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, in spite of the excruciating pain I was now beginning to
experience.
At approximately the same time, however, the barrel of bricks hit the ground-and the bottom
fell out of the barrel. Now devoid of the weight of the bricks, the barrel weighed
approximately 50 lbs. I refer you again to my weight. As you might imagine, I began a rapid
descent down the side of the building. (Q3) In the vicinity of the third floor, I met the
barrel coming up. (Q4) This accounts for the two fractured ankles, broken tooth and severe
lacerations of my legs and lower body.
Here my luck began to change slightly. The encounter with the barrel seemed to slow me
enough to lessen my injuries when I fell into the pile of bricks and fortunately only three
vertebrae were cracked. I am sorry to report, however, 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. (Q5, 6, 7)In order to do these problems, you need to know that 1 lb ≈ 4.5 N. Convert the weights
involved to newtons. Assume that the 6-floor building is 20 meters high.
1. Draw a force diagram detailing the forces acting on the man and on the barrel of bricks
(290 lbs). Determine the acceleration of the system..
2. How fast was the barrel traveling when it struck the man 1/2 way up the building? What
was the barrel's velocity relative to the man?
3. Draw a new force diagram for the barrel and man after the bricks have fallen out.
Determine the new acceleration of the system.
4. How fast was the barrel traveling when it struck the man a 2nd time? What was its
velocity relative to the man?
5. Once the man let go of the rope, assume that the barrel was essentially in free fall.
(Assume negligible air resistance). How long did it take the barrel to reach the ground?
6. How fast was it going when it hit the man?
7. How fast would it have been going if it had been full of bricks (290 lbs, not 50 lbs)?
Explain your answer.
hdjjdjd
To solve the physics problems related to the bricklayer's accident, we need to analyze the forces and calculate the acceleration and velocities involved.
1. To determine the acceleration of the system, we need to draw a force diagram detailing the forces acting on the man and the barrel of bricks. The forces acting on the man are his weight (135 lbs or 135 * 4.5 N) and the tension in the rope. The forces acting on the barrel are its weight (240 lbs or 240 * 4.5 N) and the tension in the rope.
Since the man and the barrel are both experiencing the same tension force, their accelerations will be the same. By using Newton's second law (F = ma), we can equate the sum of forces acting on each object to their respective masses and accelerations. Since the mass of the man is not given, we'll use the weight-force relationship (weight = mass * gravity) to convert his weight to kilograms (mass in kg = weight in N / g).
2. To find the velocity of the barrel when it strikes the man halfway up the building, we need to determine the distance it has traveled and the time it took. Here, we can use the kinematic equation v^2 = u^2 + 2a s, where v is the final velocity, u is the initial velocity (which is 0 since the barrel was initially at rest), a is the acceleration, and s is the distance traveled. Rearranging the equation, we get v = sqrt(2as).
The velocity of the barrel relative to the man can be calculated by subtracting the man's velocity from the barrel's velocity.
3. After the bricks have fallen out, the force diagram for the barrel and the man will change. The only force acting on the man now is his weight. The force acting on the empty barrel is only its weight. By applying Newton's second law as we did in problem 1, we can calculate the new acceleration of the system.
4. To find the velocity of the barrel when it strikes the man the second time, we can use the same kinematic equation as in problem 2. However, this time the initial velocity of the barrel will not be zero because it's falling from a certain height. The velocity of the barrel relative to the man can also be calculated in the same way as in problem 2.
5. When the man let go of the rope, the barrel essentially enters free fall. We need to calculate the time it takes for the barrel to reach the ground. For an object in free fall, the distance traveled can be calculated using the equation s = ut + (1/2)gt^2, where s is the distance, u is the initial velocity (0 in this case), g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Solve the equation for t to find the time.
6. To determine how fast the barrel was going when it hit the man, we can use the equation v = u + gt, where v is the final velocity (the velocity just before the impact), u is the initial velocity (0), g is the acceleration due to gravity, and t is the time calculated in problem 5.
7. To find the velocity the barrel would have had if it had been full of bricks, we can use the same equation as in problem 6, but substitute the weight of the full barrel (290 lbs or 290 * 4.5 N) for its mass in the equation. This will give us the velocity just before the impact.
It's important to note that these calculations assume idealized conditions, neglecting air resistance and other factors.