Force -displacement graph question

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SUMMARY

The discussion focuses on calculating the force-displacement graph for a crate with a mass of 45 kg being pulled upwards by two ropes, each with a mass of 0.75 kg/m, over a height of 12 m. The total mass of the ropes is determined to be 18 kg, leading to an initial normal force (Fn) of 617.4 N. After pulling the crate 12 m, the normal force reduces to 441 N. The work done in pulling the crate is calculated using the formula W = (A+B/2)(d), resulting in a total work of 6350.4 J.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of force calculations (Fn = mg)
  • Familiarity with work-energy principles
  • Ability to interpret force-displacement graphs
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  • Learn about graphical representation of force-displacement relationships
  • Explore the implications of mass distribution in pulley systems
  • Investigate the effects of friction on work calculations
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Homework Statement


A crate, having a mass of 45 kg is pulled steadily upwards by tw ropes through a height of 12 m. Each rope has mass of 0.75 kg/m. Draw the F-d graph and determine the work done in pulling the crate 12 m.

The Attempt at a Solution


Alright...

so we know that we are starting with
m1 = 45 kg
m2 = 0.75kg/m x 12 m
m2 = 9 kg x2
m2 = 18 kg

Now we need to find the force that it will have initially ...

Fn = mg
Fn = (18 kg + 45 kg)(9.8m/s^2)
Fn = 617.4 N

Now after the crate has been pulled 12 m ...
it'll only have m1 attached which is 45 kg

Fn=mg
Fn=(45kg)(9.8m/s^2)
Fn=441 N

Now i made this graph...

http://s1176.beta.photobucket.com/user/LolaGoesLala/media/ih.jpg.html

Then i used this formula..

W = (A+B/2)(d)
W = (617.4 N+ 441 N/2)(12 m)
W =6350.4 J
 
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