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Simple incline problem has me confused

  1. Dec 22, 2014 #1
    1. The problem statement, all variables and given/known data
    1. An inclined plane is used to raise a 100 kg box to a height of 5 m. How long must the inclined plane be if a force of 250 N is used to push the box up the inclined plane?

    2. Relevant equations
    i have search and searched all over google to try to figure this out.. I am not sure if I'm over thinking it or just confused myself even more. I found one equation that I think is correct, but as i was looking at other examples I tried to use this formula and it never worked out like I thought it should. Also, I am always finding examples that have an angle of inclination.. But the formula I think should work is.

    [tex]\frac{f}{w}=\frac{h}{l}[/tex]

    3. The attempt at a solution
    [tex]\frac{250}{100}=\frac{5}{l}[/tex]
    l=2???

    cheers
    keith
     
  2. jcsd
  3. Dec 22, 2014 #2

    NascentOxygen

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    Hi keithcuda. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    You won't be able to avoid involving either sinθ or cosθ in your calculations.
    Can you attach a pic showing the relevant diagram you have drawn for this problem?
     
    Last edited by a moderator: May 7, 2017
  4. Dec 22, 2014 #3
    You could draw a free body diagram to see what forces are opposing the motion of the box on an incline. For there to be an upward motion, that or those force(s) must be less than the upward force.
     
  5. Dec 22, 2014 #4
    i can try to draw pic and post it...

    That is where i am confused.. the question isn't giving me an angle of the incline.. so how do i include sinθ or cosθ?
     
  6. Dec 22, 2014 #5
    Maybe one of the opposing forces is related to the angle ##\theta## of the incline.
     
  7. Dec 22, 2014 #6

    NascentOxygen

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    θ
    So call the slope θ, or x or something.
     
  8. Dec 22, 2014 #7
    ?#3.jpg
     

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  9. Dec 22, 2014 #8

    NascentOxygen

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    OK. Next show a vector representing the force of gravity on that box. What magnitude would this force have? And its units?

    Have you given the slope a name, i.e., a letter?
     
  10. Dec 22, 2014 #9
    OK, so
    Ok, so I did some more googling...
    I found this, Fp=w(h/l)
    Fp=250N
    w=mg=100*9.8
    h=5
    l=slope
    250=980(5/l)
    l=19.6m ?
     
  11. Dec 22, 2014 #10
    let me know what I am doing wrong or missing and as soon as i leave work and get home I'll work on it some more

    cheers
     
  12. Dec 22, 2014 #11

    NascentOxygen

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    Once you have answered the questions in post #8 ....

    gravitational force on the box can be resolved into an equivalent pair of forces at right-angles to each other: one force acting parallel to the plane, and the second force acting perpendicular to the plane

    Forces are vectors, so to add them you must add them as vectors. Draw this "triangle of forces" to show these two perpendicular forces summing to equal the gravitational force on the box. Use trigonometry to calculate the magnitudes of the unknown sides of this triangle.
     
  13. Dec 22, 2014 #12
    Alright, I believe I know what you are talking about... But, I'm just getting home from my power plant after working a 17 hour shift. Once I get some sleep, I'll work on post #8 and #11.
     
  14. Dec 23, 2014 #13

    NascentOxygen

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    If you need help with the vector diagram, search on google images for "block on inclined plane". You'll get hundreds. :w

    http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg [Broken]
     
    Last edited by a moderator: May 7, 2017
  15. Dec 23, 2014 #14
    Thank you, I'm actually working on it right now :)
     
    Last edited by a moderator: May 7, 2017
  16. Dec 23, 2014 #15
    OK, so here is what I put together so far...
    incline_pdf.jpg
     
  17. Dec 23, 2014 #16

    NascentOxygen

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    So far so good ....
     
  18. Dec 23, 2014 #17
    I guess this is where I am really stuck. Without a given angel for theta, how to I proceed with my problem?
     
  19. Dec 23, 2014 #18
    g assumed at 9.81 m/s/s
    Choose any incline angle (and therefore incline length) but dont exceed 14.764 ° incline though, as you wont have any force left to accelerate the mass up the incline.
    The problem is, if the force is kept constant, at the top of the slope, the mass will still be moving (and accelerating).

    So, you accelerate the mass so far up the slope then remove the (250 N) force and allow it to decelerate under gravity, to come to a rest at the top of the slope.
    Ive ran a few examples through, and found that :
    Regardless of the incline length, the length you apply the force over turns out to be the same.
    The key is that you put the work in ( force * distance) which is equal to the final PE (m*g*h)
    So :
    250 * d = m * g * h
    You may do the rest.
     
  20. Dec 23, 2014 #19

    NascentOxygen

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    The problem includes a force 250N. Of the expressions you have derived, what can you equate to 250N?
     
  21. Dec 23, 2014 #20
    so, 250*d=100*9.8*5
    d=4900/250
    d=19.6
     
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