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Finding Applied Force at equilibrium

  1. Nov 7, 2012 #1

    pxw

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    1. A block with a mass of 6 kg is held in equilibrium on an inclined plane of angle θ=30 degrees by a horizontal force F. Find the magnitudes of normal force on the block and of F (ignore friction).

    So far, I've found Fg=6(-9.81)=-58.86 N

    I then found Fgy=-58.86(sin240)=50.97 and Fgx=-58.86(cos240)=29.43

    And I know that, since the block is in equilibrium, Fnet=0, so 0=Fn + Fgy + Fay

    I just seem to be having the problem of finding applied force so that I can find Fn and Fay.

    Any help would be greatly appreciated! Thanks in advance.
     
  2. jcsd
  3. Nov 7, 2012 #2

    PhanthomJay

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    You have assumed that the x axis is along the incline and the y axis is perpendicular to the incline, which is OK. Now you must break up the horizontal force F into its x and y components. Then use F_net in x direction = 0 , and F_net in y direction = 0, to solve for the 2 unknowns from the 2 equations.
    And welcome to PF!
     
  4. Nov 7, 2012 #3

    Simon Bridge

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    Did you draw a free body diagram for the block? Including all the forces?
    Did you define +y to be normal to the slope? Draw the axis on your fbd?
    Did you resolve all the forces to those axis?

    Why did you use angle 240 in the trig? Wouldn't 30 or 60 be more appropriate?

    The applied force is horizontal - so it has components in x and y directions that have to cancel out with the other forces in those directions.
    Note: sin(30)=cos(60)=1/2, sin(60)=cos(30)=√3/2
     
  5. Nov 7, 2012 #4

    pxw

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    Yes, I understand that I have to break up F into its x and y components, however, I'm not sure how to do that without a value... the equation I have is Fa(sin330)=Fay. Sticking that into the net force equation, I have 0=Fn - 50.97 + Fa(sin330). I can't seem to get any further than that...

    Thanks for your help and thanks for the welcome :D

    Yes, I did. I tilted my axes so that Fn was the y axis. I can't seem to resolve my Fa, since I don't have a value for it.

    My teacher taught us to use the larger angles because using them keeps the sine equation for the y resolution and the cosine equation for the x resolution. Taking the angle, then, I have my Fg to be at 240 degrees and my Fa to be at 330 degrees. Like I said, breaking down my Fa seems to be the big problem.
     
  6. Nov 7, 2012 #5

    PhanthomJay

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    That's one equation with 2 unknowns. You need another equation in the x direction.
    I agree with Simon that I would not use this convention for determining the angles and trig, but if you are comfortable with it, I guess its OK, although most texts won't do it that way.
     
  7. Nov 7, 2012 #6

    pxw

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    Oh, ok! I got it now! Thanks a lot! :D
     
  8. Nov 7, 2012 #7

    Simon Bridge

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    Well done :)

    fwiw: if ##\theta## is the angle of the angle to the horizontal, then ##F_{gx}=mg\cos(90-\theta)## would be correct to preserve the "cosine" rule. Whereas: ##mg\cos(270-\theta)=-F_{gx}## ; (which is what your teachers says to do) gives the opposite direction. You seem to have been unconsciously correcting for this as you go. It will trip you up later.

    It is usually easier to keep the signs right by just sketching the triangles on your diagram and using trig the way you learned in math class. People who prefer the "memorize equations" method of doing physics tend not to like this way. Probably your teacher is one of those.

    I urge you try the other way for a bit, in private, before settling on your final method. You'll need the skill anyway, to read online examples and text-books. Like PhantomJay says, it really boils down to what you find comfortable - that also works.
     
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