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.
Instead of posting the same question five times, try showing your work.
i don't understand where to start from!
1. To draw a force diagram detailing the forces acting on the man and the barrel of bricks, we need to consider the following forces:
For the man:
- Weight (Wman) = 135 lbs ≈ 135 * 4.5 N = 607.5 N (acting downward)
- Tension in the rope (T) = 607.5 N (acting upward)
For the barrel:
- Weight (Wbarrel) = 240 lbs ≈ 240 * 4.5 N = 1080 N (acting downward)
- Tension in the rope (T) = 1080 N (acting upward)
The acceleration (a) of the system can be calculated using Newton's second law:
ΣF = ma
For the man:
T - Wman = ma
607.5 N - 607.5 N = (135/9.8) * a
a = 0 m/s^2
- The man experiences no acceleration as he is stationary.
For the barrel:
T - Wbarrel = ma
1080 N - 1080 N = (240/9.8) * a
a = 0 m/s^2
- The barrel also experiences no acceleration as it is stationary.
2. To find the velocity of the barrel when it struck the man halfway up the building, we need to consider the distance covered and the time taken. Since there is no acceleration, we can use the equation:
v = d/t
Given that the 6-floor building is 20 meters high, and the man was halfway up (at the third floor) at the time of impact, the distance covered by the barrel is 20/2 = 10 meters.
We need to find the time it took for the barrel to cover this distance. Since there is no acceleration, we can use the equation of motion:
s = ut + (1/2)at^2
Since the initial velocity (u) is 0 m/s and the acceleration (a) is 0 m/s^2, the equation simplifies to:
s = (1/2)at^2
10 = (1/2) * 0 * t^2
10 = 0
Therefore, the time taken is 0 seconds.
Since the time is 0, the barrel did not reach the man, so we cannot determine its velocity relative to the man.
3. After the bricks have fallen out, the new force diagram for the barrel and the man would only include the weight of each object. Therefore:
For the man:
- Weight (Wman) = 135 lbs ≈ 607.5 N (acting downward)
For the barrel:
- Weight (Wbarrel) = 50 lbs ≈ 225 N (acting downward)
The new acceleration (a) of the system can be calculated using the same process as before:
For the man:
Wman = ma
607.5 N - 607.5 N = (135/9.8) * a
a = 0 m/s^2
For the barrel:
Wbarrel = ma
225 N - 225 N = (50/9.8) * a
a = 0 m/s^2
Once again, there is no acceleration as both the man and the barrel are stationary.
4. Since the barrel has no acceleration, its velocity remains the same as when it struck the man halfway up the building. As we calculated previously, the barrel did not reach the man, so the velocity relative to the man is also 0 m/s.
5. When the man let go of the rope and the barrel was essentially in free fall, we can calculate the time (t) it took for the barrel to reach the ground using the equation of motion:
s = ut + (1/2)at^2
Given that the height of the building is 20 meters and assuming negligible air resistance, the equation simplifies to:
s = (1/2) * 9.8 * t^2
20 = 4.9 * t^2
t^2 = 20 / 4.9
t ≈ 2.02 seconds
Therefore, it took approximately 2.02 seconds for the barrel to reach the ground.
6. When the barrel hit the man, it reached the ground and its velocity would be equal to the final velocity (vf) after falling a height of 20 meters. We can calculate this velocity using the equation of motion:
vf = u + at
Given that the initial velocity (u) is 0 m/s and the acceleration (a) due to gravity is 9.8 m/s^2, the equation simplifies to:
vf = 0 + (9.8 * 2.02)
Therefore, the velocity of the barrel when it hit the man is approximately 19.80 m/s.
7. If the barrel had been full of bricks (weighing 290 lbs), the force diagram and the acceleration would remain the same as in step 3. However, the weight of the barrel would be:
Wbarrel = 290 lbs ≈ 1305 N (acting downward)
The velocity of the barrel can be calculated using the same process as before:
vf = 0 + (9.8 * 2.02)
Therefore, the velocity of the barrel if it had been full of bricks would also be approximately 19.80 m/s. The presence or absence of bricks does not affect the velocity in free fall, as all objects fall at the same rate in the absence of air resistance.
To solve these physics problems, we need to apply the principles of Newton's laws of motion and the concepts of forces, weight, and acceleration. Let's go through each question step by step:
Q1: Draw a force diagram detailing the forces acting on the man and on the barrel of bricks (weight: 290 lbs). Determine the acceleration of the system.
First, let's convert the weight from pounds to Newtons. We know that 1 lb ≈ 4.5 N. Therefore, the weight of the bricks is approximately 290 lbs × 4.5 N/lb = 1305 N.
The forces acting on the man are his weight (135 lbs × 4.5 N/lb) and the tension in the rope from his pulling force. The forces acting on the barrel are its weight (1305 N) and the tension in the rope.
To determine the acceleration of the system, we need to find the net force acting on the system and apply Newton's second law (F = ma). The net force is the difference between the force of the man pulling up and the force of the weight of the bricks pulling down.
Let's assume that the upward force exerted by the man is greater than the weight of the bricks, which means the system will accelerate upwards.
Q2: How fast was the barrel traveling when it struck the man halfway up the building? What was the barrel's velocity relative to the man?
To find the speed of the barrel, we first need to know its initial speed before it met the man. We don't have this information provided in the problem. However, we can assume that the speed of the barrel is the same as the speed of the man before they collided. Therefore, the barrel's velocity relative to the man is zero.
Q3: Draw a new force diagram for the barrel and man after the bricks have fallen out. Determine the new acceleration of the system.
After the bricks have fallen out, the weight of the barrel decreases to 50 lbs × 4.5 N/lb = 225 N.
The forces acting on the man remain the same, but now the force diagram for the barrel only includes its weight.
To determine the new acceleration of the system, we need to recalculate the net force as the difference between the force of the man pulling up and the weight of the barrel.
Q4: How fast was the barrel traveling when it struck the man a second time? What was its velocity relative to the man?
Similarly to Q2, we don't have enough information to determine the speed of the barrel when it struck the man the second time. We can assume that the barrel's velocity relative to the man is zero as they collide again.
Q5: 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?
When the man let go of the rope, there is no upward force acting on the barrel. Therefore, the only force acting on the barrel is its weight, 225 N.
We can use the equation of motion s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity (zero), a is the acceleration (due to gravity, approximately 9.8 m/s^2), and t is the time taken.
Since the barrel is in free fall, the initial velocity (u) is zero, and the distance it travels is the height of the building (20 meters). Rearranging the equation, we have 20 = 0 + (1/2) × 9.8 × t^2. Solve for t.
Q6: How fast was it going when it hit the man?
To find the speed of the barrel when it hit the man, we can use the equation of motion v = u + at, where v is the final velocity, u is the initial velocity (which is zero in this case since the barrel starts from rest), a is the acceleration (due to gravity, approximately 9.8 m/s^2), and t is the time taken.
First, we need to calculate the time taken for the barrel to reach the ground (which was found in Q5). Then, we can use this value to calculate the final velocity.
Q7: How fast would it have been going if it had been full of bricks (290 lbs, not 50 lbs)? Explain your answer.
To determine the speed of the barrel if it had been full of bricks, we need to apply the same calculation method as in Q6, using the weight of the bricks (290 lbs × 4.5 N/lb) as the weight of the barrel.
This assumes that the reduced weight of the barrel when it is empty will not significantly affect the overall motion and acceleration of the system.
Remember, always double-check the calculations and make sure to input the correct units for the quantities involved.