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Angle of Inclination

  1. Feb 6, 2014 #1
    1. The problem statement, all variables and given/known data

    On how steep an incline(in degrees) can a car be parked without sliding down if the coefficient of static friction between the tires and the road is 0.8?


    2. Relevant equations

    F=ma

    3. The attempt at a solution

    My physics teacher showed how to find the acceleration for an object on an incline. He got the equation a=gsin(theta)-(mu)gcos(theta)

    (Sorry, I can't seem to find the sybmols for mu and theta on LaTex.


    I understand how he got the equation. It makes perfect sense. However, I'm having a hard time transforming the equation to get the angle of inclination. Adding mgcos(theta) to both sides yields (mu)gcos(theta)+a=gsin(theta). I would think that the acceleration is negligible since it's not moving, and I think the g's cancel out, correct? But this equation just gives 0.8cos(theta)=sin(theta). But where do I go from here? I don't know cos(theta), sin(theta), or theta. I've been looking at the theta as part of separate terms, but if they canceled out, then it would just be 0.8cos=sin. I tried the inverse of this on a hunch and got sin=36.9 degrees. This seems pretty reasonable, is this my answer? If so, why is it my answer?

    Thanks so much! :smile:
     
  2. jcsd
  3. Feb 6, 2014 #2
    You have the equation:

    ##0.8cosθ=sinθ##

    There a simple trigonometric identity relating sinθ and cosθ. You can use it to turn this into an equation involving only one term which has θ in it.
     
  4. Feb 6, 2014 #3
    Thanks! I thought so! But I've researched it, and looked back in my videos, and plugged cosines and sines of angles in to my calculator, and I can't find it anywhere.
     
  5. Feb 6, 2014 #4
    The identity you need to use is:

    ##tanθ=\frac{sinθ}{cosθ}##
     
  6. Feb 6, 2014 #5
    Okay, thank you! Looking at the diagram, I see how this equation makes sense. However, I still don't see how I have enough information. It looks like the hypotenuse length and the opposite length may be the same, does this mean that the tan and cos functions cancel out, leaving sin(0.8)? I don't think so, because that number seems far too small.
     
  7. Feb 6, 2014 #6
    You are trying to find the angle of inclination, which is θ.

    You have the equation ##0.8cosθ=sinθ##

    If you divide both sides of this equation by ##cosθ##, then you get:

    ##0.8=\frac{sinθ}{cosθ}##

    Now, we know that ##\frac{sinθ}{cosθ}=tanθ##

    So, we now have the following equation:

    ##tanθ=0.8##

    You are trying to find the angle of inclination, which is θ. You know that ##tanθ=0.8##. Can you see how you can now find what θ is?
     
  8. Feb 6, 2014 #7
    I understand how you got tan(theta)=0.8. I still can't seem to find a trigonometric relationship to get to the answer. If we multiply by the reciprocal we get 0.8(cos)/tan, but that doesn't make sense. I really do want to find the answer myself, so is there a way you can help show the relationship without giving away the answer?

    Edit: Is there a inverse relationship I can use? 0.8(inverse tangent)=38.66 degrees.

    Thanks so much!
     
  9. Feb 6, 2014 #8
    Yes, you know that ##tanθ=0.8##. Therefore:

    ##θ=tan^{-1}(0.8)##

    Which gives a value of ##θ=38.66°##.
     
  10. Feb 6, 2014 #9
    Ah, right. I forgot the "why" of the inverse function for a moment, and just went off a hunch. Thank you! I get it now.
     
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