Initial upward acceleration in a pulley problem

In summary: So the acceleration is half of what I had calculated which makes sense. In summary, the problem involves a worker laying bricks on a roof and using a pulley system to lower extra bricks in a barrel. The worker's weight is 74.2 kg and the total mass of the bricks and barrel is 151.5 kg. The worker accidentally gets pulled up by the barrel and then falls down, hitting the barrel during the descent. Using the equations F=ma and x=vot+1/2at^2, the initial upward acceleration experienced by the worker is approximately 10.209 m/s2. The height of the building can be calculated using the time of 2.95 seconds and the acceleration, and the time it
  • #1
NinjaPenguin
2
0

Homework Statement



Given data:

Gravity= 9.8 m/s2
Mass of bricks= 136 kg
Mass of barrel= 15.5 kg
Mass of barrel and bricks= 151.5 kg
Mass of worker= 74.2 kg
Time= 2.95s

The problem

The acceleration of gravity is 9.8 m/s2 .
To emphasize the fact that lessons learned in 303K carry over into real life, this problem is based on an accident report from the Worker’s Compensation Board in Bryant, Texas.

Dear Sir/Madam:
I am writing in response to your request for a fuller explanation on the cause of my accident. I trust the following details will be sufficient. On the day of the accident, I was working alone laying bricks on the roof of a new building. On completing my work, I had some bricks left over which were found to have a mass of 136 kg. Rather than carry the bricks down by hand, I decided to lower them in a barrel with a mass of 15.5 kg by using a pulley which was attached to the side
of the building at the roof.

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 bricks. You will note that my mass is 74.2 kg. To my surprise, I was jerked off the ground suddenly and forgot to let go of the rope. Needless to say, I proceeded at a rapid rate up towards the roof of the building. Somewhere up the building, I met the barrel which was now proceeding downward at an equally impressive speed. This did not slow me down and I continued my rapid ascent, hitting the pulley on the top floor. I estimate that all this happened in the matter of 2.95 s.
At the same time, however, the barrel of bricks hit the ground and the bottom fell out of the
barrel. Devoid of the bricks, the barrel now had a mass of 15.5 kg. I refer you again to my mass. As you might imagine, I began a rapid descent down the side of the building and met the barrel coming up.

My luck finally began to change slightly. The barrel slowed me enough to lessen my injuries when I fell into the pile of bricks. 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 several stories above me, I again lost
my composure and let go of the rope.

1. What was the initial upward acceleration
experienced by the worker?
Answer in units of m/s2.

2. What is the height of the building?
Answer in units of m.

3. How many seconds did it take (after starting
his descent) for the worker to hit the ground?
Answer in units of s.

Homework Equations



F=ma

x = vot + 1/2at*2

The Attempt at a Solution



1.
I tried finding the sum of the force up and down.
Force up = 151.5 (9.8)= 1484.7 N
Force down = 74.2(9.8)= 727.16 N

Then I subtracted up force from the down force to get the net.

1484.7 - 727.16 = 757.54 N

I then used F =ma to find a.

757.54 = 74.2a
757.54/74.2 = 10.209 = a

For 2
I would just plug in time(2.95) and the acceleration to find the height

For 3
I would plug in height and (9.8 m/s2) to find time

My answer of 10.209 m/s2 is not right though so I can't do the other two. Any help would be appreciated
 
Last edited:
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  • #3
Thank you. For some reason I didn't even think about that and we just did a lab on it.
 

Related to Initial upward acceleration in a pulley problem

What is initial upward acceleration in a pulley problem?

Initial upward acceleration in a pulley problem refers to the initial rate of change in the upward velocity of an object attached to a pulley system. It is the acceleration that occurs right at the beginning of the motion.

How is initial upward acceleration calculated in a pulley problem?

To calculate the initial upward acceleration in a pulley problem, you can use the equation a = (m1 - m2)g / (m1 + m2), where m1 and m2 are the masses of the objects on either side of the pulley and g is the acceleration due to gravity. This equation takes into account the forces acting on the objects and the mass of the pulley itself.

What factors affect the initial upward acceleration in a pulley problem?

The initial upward acceleration in a pulley problem is affected by several factors, including the masses of the objects, the mass and friction of the pulley, and the force of gravity. In addition, the angle at which the pulley is positioned and the type of pulley system (e.g. fixed, movable, or compound) can also impact the initial upward acceleration.

Why is initial upward acceleration important in a pulley problem?

Initial upward acceleration is important in a pulley problem because it determines how quickly an object attached to the pulley system will accelerate upwards. It also affects the tension in the rope or string connecting the objects, which can impact the stability and overall motion of the system.

How does initial upward acceleration differ from average upward acceleration in a pulley problem?

Initial upward acceleration and average upward acceleration are different in a pulley problem because initial upward acceleration only refers to the acceleration at the beginning of the motion, while average upward acceleration takes into account the entire motion and is calculated by dividing the change in velocity by the change in time. Average upward acceleration may be affected by external forces and friction, whereas initial upward acceleration is primarily determined by the forces within the pulley system itself.

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