Solve Pulley on a Slope Homework: Mass of Counterweight in kg

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The discussion revolves around calculating the mass of a counterweight needed to lower a 60 kg actress playing Peter Pan down a slope in a stage production. The actress must descend 3.2 meters in 2.2 seconds, leading to an acceleration of 1.322 m/s². The force exerted by the actress is calculated to be 79.338 N, while the acceleration of the counterweight on the slope is determined to be 8.408 m/s². After some calculations and corrections, the correct mass of the counterweight is found to be approximately 51.91 kg. The participants confirm the understanding of the problem and the solution process, ensuring clarity on the physics involved.
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Homework Statement


In a stage production of Peter Pan, the 60 kg actress playing Peter has to fly in vertically, and to be in time with the music, she must be lowered, starting from rest, a distance of 3.2 m in 2.2 s. Backstage, a smooth surface sloped at 59º supports a counterweight of mass m.

What is the mass of the counterweight that must be used? The pulley is of negligible mass and is frictionless. The acceleration of gravity is 9.81 m/s². Answer in units of kg.


Homework Equations


d = vit + ½at²
acceleration = g*sin(Θ)
Mass = Force/Acceleration


The Attempt at a Solution



d = vit + ½at²
3.2 = ½a(2.2)²
a = 1.322
Using the distance equation and solving for acceleration, the acceleration of Peter would be 1.322 m/s²

F = mass x acceleration
F = 60 kg x 1.322
F = 79.338 N
This is the force of Peter as he falls.

acceleration of weight m = 9.81*sin(59)
= 8.408 m/s²

Mass of m = Force/Acceleration
= 79.338 N / 8.408 m/s²
= 9.435 kg

My final answer was very much incorrect. Is there another formula I was supposed to use?
 
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From what I understand the questions wants a counter mass m that is required to lower peter under the condtions.

EDIT: If i understood it correctly, I worked it out and got an answer of 51.91kg.

If this is the correct answer then I will walk you through my solution.
thanks.
 
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pooface said:
From what I understand the questions wants a counter mass m that is required to lower peter under the condtions.

Yes. I do believe that is what the problem in asking. I found the mass of the counterweight to be about 9kg. That's too small a weight to work, however.

EDIT: I checked the answers and 51.91 kg is the correct answer. =)
 
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How did you arrive at your solution?
 
Ok. So peter is going vertically down so let us convert his mass to weight by gravity which is equal to 9.81*60kg=588.6 N

You figured out the acceleration of the entire network to be 1.322 m/s^2.

From this acceleration that we need for the network to work, we need peters force downwards which you calculated to be 79.333 N.

This is the force that will be pulling peter down and pulling the mass m up the slope.

If peter's force down is 79.33 N with the desired acceleration, this value will be the NET force required for there to be an acceleration of 1.322 m/s^2.

So weight of peter due to gravity(588.6N) - weight of mass due to gravity = 79.33. (subtract because they are pulling opposite ways)
You know peters weight from step 1 and voila you have the weight of the mass m. divide by 9.81 and you have the mass m.

Do you understand?
 
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Yes, I understand it now. Thank you very much!
 
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