Tension and Friction with Acceleration

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SUMMARY

The discussion centers on calculating the final velocity of a person descending 12 meters using a bedsheet rope that can withstand a maximum force of 600 N. The occupant weighs 800 N, leading to a net downward force of 200 N when considering the friction provided by the rope. Two conflicting calculations arise: one yielding a final velocity of -29.7 mph and the other -17 mph. The correct approach involves recognizing that the rope's tension creates an upward force of 600 N, resulting in a net force of -200 N, which should be used to determine the acceleration and final velocity accurately.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of forces and acceleration
  • Familiarity with kinematic equations
  • Ability to convert units (e.g., m/s to mph)
NEXT STEPS
  • Review Newton's second law of motion and its applications
  • Explore kinematic equations in detail, particularly Vf^2 - Vo^2 = 2ay
  • Learn about tension and friction in ropes and their effects on motion
  • Practice unit conversion techniques, especially for speed
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts of force, acceleration, and friction in real-world scenarios.

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Homework Statement



A building is on fire and the occupant of a room ties together some bed sheets, attaches them to something immovable in the room and throws the bedsheet ‘rope’ out the window. The person weighs 800 N and the bed sheets can only hold a force of 600 N before they break. If the person descends (starting with zero vertical velocity) 12 m (about 4 stories) using this bedsheet rope, applying just the right amount of friction on the rope to provide maximum braking without breaking the rope, how fast is the person going after 12 m of descent? Give you answer in miles per hour so you have a feel for how fast this is. (26.8 m/s = 60 mph)



Homework Equations



I'm sorry, I do not know how to use the LaTex Math Typesetting, but the equations are pretty basic.

Fnet = ma

Fnet = F1 + F2 + F3 + ... + Fn

Vf^2 - Vo^2 = 2ay


The Attempt at a Solution



Okay, so what I did is say that the person must be using a maximum downward force of 600N on the rope so it doesn't break. This makes then friction equal to 200N, since it acts in the opposite direction of the 800N downward of the occupant.
So, (-600N)/(9.8kg*m/s^2) = -7.35 m/s^2. This is person's acceleration downwards.
Plugging this into the final equation yields Vf = √(2*(-7.35 m/s^2)*(-12m)) = - 13.3 m/s or -29.7 mph.

HOWEVER, my classmate says that the rope pulls upward with 600N and since the rope is applying the force of friction, the there're 600N of friction. So that would make the net force -200N. Redoing the calculation gets -17 mph.

So, which is correct? My classmate's makes sense, but it seems off to me...so any input is very, very appreciated! We could also both be wrong haha. Thank you for taking your time!
 
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If the rope exerts some amount of friction, then by Newton's third law it is itself subject to the same amount of force (acting in the opposite direction).
 
So, then the friction must be equal to the tension limit of 600N?
 
It is in the interest of the occupant to use the max friction afforded by the rope.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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