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Force Problem - How do i get the inclination?

  1. Sep 15, 2015 #1
    A 10.0 kg object is resting on a horizontal surface with coefficient of friction of 0.7. Determine the minimum force F that can move this object. At what inclination should force F be applied? Present your answer in nearest hundredths.

    I don't know how where to start?, How do i get the inclination ? thanks for the help
     
    Last edited: Sep 15, 2015
  2. jcsd
  3. Sep 15, 2015 #2
    I'm pretty sure there's an equation for this and you already have half of the variables. A quick google search would show you the answer. Try searching for coefficient of friction and in this case it's probably static friction.
     
  4. Sep 15, 2015 #3
    ok thanks!!!!!
     
  5. Sep 15, 2015 #4

    haruspex

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    Does that mean you found the answer?
    Really you should be figuring this out for yourself using free body diagrams and applying the force balance equations. You'll learn more that way than looking the answer up. Besides, it often happens that there is some subtle difference between the problem you are trying to solve and the one you find lying around on the net.
     
  6. Sep 15, 2015 #5
    oh no, i'm still figuring out the answer
     
  7. Sep 15, 2015 #6
    i'm still figuring out the answer
     
  8. Sep 15, 2015 #7

    haruspex

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    Ok. Have you drawn yourself a force diagram? What forces do you have, and in what directions do they act?
     
  9. Sep 15, 2015 #8
    i calculated that the normal force is 98N and the friction force is 68.6N . and i dont know what is next
     
  10. Sep 15, 2015 #9
    just take angle as some 'x' and force as F and write the equations .u wil get as
    Fcos x=0.7(w-F sinx)
    solving this u wil get the eq. as 10 cos x+7 sin x should me max for the force to be minimum.
    and solving x=34.99 degrees
     
  11. Sep 15, 2015 #10
    how did you get 10 cos x +7 sin x
     
  12. Sep 15, 2015 #11
    ohhh ok i got it now thanks
     
  13. Sep 15, 2015 #12
    kk fn.
     
  14. Sep 15, 2015 #13
    hahahahaha thanks a lot
     
  15. Sep 15, 2015 #14
    ok i'm a bit lost could you explain how did you get 10 cos x +7 sin x
     
  16. Sep 15, 2015 #15
    F cos x =0.7(w-F sin x)
    F cos x/0.7=w-F sin x
    F((cos x/0.7)+sin x)=w
    F(10 cos x+7 sin x)/7=w
    F=(7w)/(10 cos x+7 sinx)
     
  17. Sep 15, 2015 #16

    haruspex

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    You have not allowed for the applied force being at some angle to the horizontal. What will that do to the normal force?
     
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