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Box sliding down a ramp

  • Thread starter makerfeldt
  • Start date
  • #1

Homework Statement



A 19.0 kg box sits on a frictionless ramp with a 14.9° slope. The movers pull up a rope attached to the box to move it up. If the rope makes an incline with the ramp that is 42.0° to the horizontal, what is the smallest force F the mover must use to move the box up the incline?

Homework Equations



Newton's 2nd Law

The Attempt at a Solution


I was already given the answer to this as 53.8 N, but I can't get there (though I get very close)

I broke it down into 3 forces; the gravitational force, the normal force, and the force the mover exerts on the box. Getting the value 186.2 N for the value of Fgy, I obtained 47.88 N for Fgx by utilizing Fgy divided by cos∅ and then multiplying that value by sin∅. The F exerted by the mover along the x should equal that value of 47.88. Using that value, 47.88 / cos 56.9° (The angle the rope makes with the rotated axis I used) I get a monstrous value of 87.68 N. Any tips or pointers would be much appreciated.
 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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hi makerfeldt! welcome to pf! :smile:
A 19.0 kg box sits on a frictionless ramp with a 14.9° slope. The movers pull up a rope attached to the box to move it up. If the rope makes an incline with the ramp that is 42.0° to the horizontal

Using that value, 47.88 / cos 56.9° (The angle the rope makes with the rotated axis I used)
why did you add (14.9° + 42° = 56.9°) ? :confused:
 
  • #3
The initial angle given for the rope (42) was the angle it makes with the horizontal, but that is not the horizontal I used. I rotated the axis 14.9 so that I could make the ramp's surface the new horizontal. As a result, I reasoned that the new angle the rope would make with the horizontal would be 56.9. Is there a flaw in that reasoning?

Edit: I have seen the flaw in that reasoning, I should have done the opposite.

Thanks for the welcome by the way, I am sure I will be making many posts in the near future :)
 

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