How Do You Solve an Inclined Plane Problem Involving Friction and Force?

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SUMMARY

The discussion centers on solving an inclined plane problem involving friction and force, specifically a box being pulled up a 285 m incline with a height of 16.7 m and a pulling force of 18,300 N. The weight of the box is calculated to be 18,300 N, leading to a mass of approximately 1,865.4 kg. When accounting for friction, which exerts a force of 17,400 N, the net force pulling the box is determined to be 900 N. The Ideal Mechanical Advantage (IMA) is calculated using the formula IMA = De/Dr, with De being the distance along the incline and Dr being the vertical rise.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of inclined plane mechanics
  • Familiarity with the concepts of force, weight, and mass
  • Basic understanding of Ideal Mechanical Advantage (IMA)
NEXT STEPS
  • Study the calculations for inclined planes with friction
  • Learn about the relationship between force, mass, and acceleration
  • Explore the concept of mechanical advantage in simple machines
  • Investigate the effects of varying friction coefficients on inclined planes
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and inclined planes, as well as educators looking for examples of real-world applications of these concepts.

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Homework Statement


A box is pulled up an incline which is 285 m long and in the process the box rises 16.7 m. The force pulling the truck is 18300 N.
A) If there was no friction, what is the weight and mass of the box? B)If friction provides a force of 17,400 N, what is the force pulling the box?
c) What is the IMA?

The Attempt at a Solution


A)The weight would just be 18300 N I believe, since it is the effort force. and w=mg, so 18300=m(9.81), so m=1865.4 g?
B) 18300-17400=900 N force pulling box
C) IMA=De/Dr To find the effort distance,would you square 16.7+square 285?
=81504/16.7
=4880m
Im really confused on inclined planes, so any tips on what I am doing wrong would be great :)
 
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There is no statement that it was constant velocity, but I think over this distance it can be assumed. That means that the force is not lifting the weight the whole way, but rather is pulling it up the incline at an angle to gravity given in the problem by the rise of 16.7 m divided by the 285 m.

If you add friction, why is it you aren't adding that to the 18,300?

And can you explain what you are calling IMA?
 

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