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Thread: Physics again! really lost

  1. #1
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    Physics again! really lost

    Ok here we go.

    A 1900kg car experiences a combinded force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at 27m/s. Going up a hill, the car's engine needs to produce 47 hp (1horsepower=745.7 Watts) more power to sustain the constant velocity than it does going down the same hill. At what angle is the hill inclined above the horizontal?

    Any starting tips.

  2. #2
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    have a look at my diagram first

    47hp = 35 047.9W

    at a speed of 27m/s this is equivalent to 1298.0703N (X in diagram)

    to remain at a constant velocity, all vectors have to be equal (taking g=10m/s^2)

    uphill:
    N.cos@+(E+1298).sin@ = F.sin@+19000
    N.sin@+F.cos@ = (E+1298).cos@

    downhill:
    N.cos@+F.sin@ = E.sin@+19000
    N.sin@+E.cos@ = F.cos@

    from these 4 equations you can solve for the 4 variables, N E F @, but i don't feel like doing it for you just yet
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  3. #3
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    i'm sure there's gotta be an easier way to do this other than solving for @ - it's proving to be quite a pain

  4. #4
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    Well as far as understanding your diagram, I'm on the mark. But as for your equations, I'm lost lol. But thanks for the help, I'm sure once I get my sleep/early morning coffee I'll be able to figure something out.

  5. #5
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    as there are parallel and perpendicular lines everywhere (add in the horizontal and vertical at the point on the hill) the @ can be moved around with alternate and vertically opposite angles. from there the vectors are resolved into their horizontal and vertical components

    to give you an understanding of how this works:

    the ascii diagram didnt work so imagine a right triangle with hypotenuse v, base x and vertical side y with angle between hypotenuse and base = @

    if the vector is v with angle @ then x and y are the horizontal and vertical components:

    cos@ = x/v
    x = v.cos@

    sin@ = y/v
    y = v.sin@

    that is what i have done in the earlier explanation and all horizontal components have to cancel in either direction as to vertical components in order to have no acceleration

  6. #6
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    ok so far so good. Ok so, I don't have F, E and @ variables but I have mg which gives me N?

    ok so F=E+X
    F=mgsin@

    so, mgsin@=E+1298. I can't solve for E, I don't have @.


    Ok, so N=mgcos@ right. That doesn't make sense, if mg and N=mg, then wouldn't the values cancel out. err.

    Ok, I've drawn something on your diagram, to see if I'm looking for the right angle. I used paint so it's a little "butchered" lol. I can see why this question was double starred for difficulty
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  7. #7
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    why did you include 19000 in there might I ask? W acts in the vertical, how does that influence the horizontal forces?

  8. #8
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    Ok is this way possible?
    Ok so let me see:

    P=FV

    P(uphill)=mgsin@ +F(friction)
    P(downhill)=mgsin@-F(friction)

    ok so I subtract them:

    47hp=mgsin@+F - (mgsin@-F(friction)
    47hp = 35047 Watts or 1298 N.

    1298N=mgsin@-mgsin@+F+F
    1298N=2F(friction)
    F= 649 N. Funny Friction doesn't cancel out!

    hmm,

    Cos@=Wx/W
    =649N/mg
    =649N/18639N
    =cos inverse(0.0348)
    =88 degrees!

    sounds kinda funny, that would be too steep.

  9. #9
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    you can't just say that N=mgcos@ because there are other vectors working in the vertical as well (parts of E & F)

    i included 19000 (equivalent of mg) just to get rid of more variables so it's easier to solve for the others - it is only in the vertical components that it is included

  10. #10
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    Slosh, help me lol, I'm out of ideas

  11. #11
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    you need to take those 4 equations and solve them simultaneously to get the variables:

    uphill:
    N.cos@+(E+1298).sin@ = F.sin@+19000 -- (1)
    N.sin@+F.cos@ = (E+1298).cos@ -- (2)

    downhill:
    N.cos@+F.sin@ = E.sin@+19000 -- (3)
    N.sin@+E.cos@ = F.cos@ -- (4)

    (4)/cos@:
    N.tan@+E=F since cos@ doesn't equal 0
    N.tan@=F-E

    (2)/cos@:
    N.tan@+F=E+1298
    N.tan@=E+1298-F

    therefore:
    F-E=E+1298-F (you didnt really need the equations to work that out)

    2E+1298=2F
    F=E+649

    put this into the earlier equation
    N.tan@+E=F
    N.tan@+E=E+649
    N.tan@=649

    you repeat this with equation 3:

    (3)/sin@:
    N.cot@+F=E+19000/sin@
    N.cot@=E+19000/sin@-F

    F=E+649 (from earlier) so:
    N.cot@=19000/sin@-649
    N.cos@=19000-649sin@

    now you have your two equations for N and @:

    N.tan@=649
    N.cos@=19000-649sin@
    Last edited by a_slosh; 10-17-2004 at 12:38 AM.

  12. #12
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    i just realised what an idiot i am i went all the way around with the maths to arrive back at the simplest answer

    from the last 2 equations:
    N.tan@=649
    N.cos@=19000-649sin@

    they cancel out to 649 = 19000sin@

    this is the exact same as working out the component of gravity that acts along the plane of the hill (mg = 19000)

    649 is half of the force difference between uphill and downhill because it acts with the engine and then against it, hence doubling

    it comes out to about 0.035 radians

    i could kick myself how easy that is after doing that work

  13. #13
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    wow, that looks complicated lol.

    Ok, does this way work?

    uphill:-mgsin@-F+(E+1298N)
    downhill: -mgsin-E+F

    solve for E in bottom and sub into uphill equation.

    E=-mgsin@+F

    sub into uphill-------> -mgsin@-F+(-mgsin@+F+1298)=0

    2mgsin@=1298
    mgsin@=649
    sin@=649/mg

    answer: 1.96 degrees = 2.0 degrees?

  14. #14
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    BTW, Thanks for the Help!

  15. #15
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    yup you've got the answer!

    i am so annoyed that i never saw that in the first place, you actually don't need to consider any of the other vectors besides mg as mg.sin@ is always the same, down the hill. so you can skip straight to 2mg.sin@=1298N

    well at least you now have a mathematical proof of it

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