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Homework Help: Magnitude of Force within incline?

  1. Sep 27, 2007 #1
    Magnitude of Force within incline?????

    A block with mass of 5.5kg is held in equilibrium on a frcitionless incline of 34 degrees by the horizontal force F. The acceleration of gravity is 9.81m/s^2

    What is the magnitude of F? answer in units of N

    What is the magnitude of the normal force? answer in units of N


    I converted 5.5 kg to 53.9N

    then i used sin of 56 degrees * 53.9 to get the horizontal force of 44.68N

    then i used cos of 56 degrees * 53.9 to get the normal force of 30.14N

    however both answers were incorrect
     
  2. jcsd
  3. Sep 27, 2007 #2
    because you need to use sin of 34 not 56 degrees. you have the basic concept of weight the the triangle it forms. but that triangle is proportional the incline, so they have the same angles.
     
  4. Sep 27, 2007 #3
    ok but sin of 34 * 53.9 = 30.14N

    and cos of 34 * 53.9 = 44.68N

    so i get the same answers ..... and they are wrong... thats why i am stuck
     
  5. Sep 27, 2007 #4
    Yeah, use 34 degrees, not 56. Fh= Fg*(sin x)
    Your angle of incline will be the angle you use in the computation.

    so Fh= (54.0)*(sin 34) = 30.2 N
    and Fn= (54.0)*(cos 34) = 44.8 N

    Double check my work.. but I think thats how you do it..
     
  6. Sep 27, 2007 #5
    Is there additional information in the problem you didn't include? Because I believe that's correct...
     
  7. Sep 27, 2007 #6
    those answers are incorrect... thats why i am so lost
     
  8. Sep 27, 2007 #7
    the drawing shows the incline... the slope of the incline is 34 degrees... the block on it is 5.5 kg which is 53.9N

    it slopes up from left to right... there is a line horizontal that is labeled as F.. and above it is a arrow pointing down sayin 34 degrees as well....


    where this horizontal line F touches the block is in the top left corner of the block... then an angled line down off of the top left corner of the block slopes down the incline as well
     
  9. Sep 27, 2007 #8
    those look correct to me too
     
  10. Sep 27, 2007 #9
    i wish there was a way to draw out what they give u .... i tried to describe it the best i could
     
  11. Sep 27, 2007 #10
    this is the best that i can describe the drawing shows the incline... the slope of the incline is 34 degrees... the block on it is 5.5 kg which is 53.9N

    it slopes up from left to right... there is a line horizontal that is labeled as F.. and above it is a arrow pointing down sayin 34 degrees as well....


    where this horizontal line F touches the block is in the top left corner of the block... then an angled line down off of the top left corner of the block slopes down the incline as well

    its an online class so when i enter the answer it tells me its wrong
     
  12. Sep 27, 2007 #11
    did you round before the answer? maybe you need to round only after finding the answer. that can change it by .1 or .01 or whatever
     
  13. Sep 27, 2007 #12
    and the horizontal comes from the left?
     
  14. Sep 27, 2007 #13
    i tried answers ranging from 30..05 to 30.2 and the same for 44.65 to 44.70

    and they were wrong
     
  15. Sep 27, 2007 #14
    Yeah, since its an online course where you enter the answer that might throw it off. Try my answers, I was careful with significant figures...
     
  16. Sep 27, 2007 #15
    i am uploading the picture of the problem online.. i will have the link for u to see it in one minute
     
  17. Sep 27, 2007 #16
    But I believe since the horizontal force touches the top left corner, and not the center of the block, it does not equal F parallel, so the formula does not apply... Let me give it another shot on paper...
     
  18. Sep 27, 2007 #17
  19. Sep 27, 2007 #18
    Awesome, that would help a lot.
     
  20. Sep 27, 2007 #19
    check that link for the drawing of the problem
     
  21. Sep 27, 2007 #20
    Ok, give me a minute I'll figure it out if someone doesn't before me.
     
  22. Sep 27, 2007 #21
    thanks alot i really appreciate it.. i thought for sure the sin and cos of the angle would solve this for me... but it didnt and left me puzzled
     
  23. Sep 27, 2007 #22
    http://img234.imageshack.us/img234/1231/26425932hu7.jpg [Broken]
    the box is applying 53.9sin34deg along the surface to the lower left
    now we have that new triangle
    cos 34=53.9sin34/F, solve for F
    thats problem number 1 i think
     
    Last edited by a moderator: May 3, 2017
  24. Sep 27, 2007 #23
    Yeah definitely. Well I know the formula won't give you the answer becuase it finds what that dotted line equals, and we want the solid line labeled F. So I"m tryin to figure out how to do that...
     
  25. Sep 27, 2007 #24
    Yeah there you go, but its 54.0(sin 34), if you get picky. No?
     
    Last edited: Sep 27, 2007
  26. Sep 27, 2007 #25
    36.4 N for first answer?
     
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