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Inclined plane problem

  • Thread starter livvy07
  • Start date
  • #1
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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 Im doing wrong would be great :)
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
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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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