Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

need help with part c mechanics?

i did part a and b of these, they were pretty easy but i can't seem to get c. i'm not sure if im missing a cancellation somewhere or something but anyone have any ideas?

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  • NCS
    Lv 7
    1 month ago

    "constant speed" means that the motive force is equal to the resisting force:

    m₂*g*sinΘ = µ*m₂*g*cosΘ + m₁*g

    Divide through by m₂*g and set k = m₁/m₂ --

    sinΘ = µ*cosΘ + k

    Now I think I'd square both sides:

    sin²Θ = µ²cos²Θ + 2k*µ*cosΘ + k²

    and substitute for sin²Θ --

    1 - cos²Θ = µ²cos²Θ + 2k*µ*cosΘ + k²

    0 = (µ² + 1)cos²Θ + 2k*µ*cosΘ + k² - 1

    Now we've got a quadratic, the solution to which is

    cosΘ = [-2k*µ ± √(4k²µ² - 4(µ²+1)(k² - 1))] / 2(µ²+1)

    focus on the discriminant:

    √(4k²µ² - 4µ²k² + 4µ² - 4k² + 4 + 4µ²)

    = √(8µ² - 4k² + 4) = 2*√(2µ² - k² + 1)

    and so

    cosΘ = [-2kµ ± 2*√(2µ² - k² + 1)] / 2(µ²+1)

    cosΘ = [-kµ ± √(2µ² - k² + 1)] / (µ²+1)

    I'm pretty sure we can toss out the "-" in "±" since that will give us a negative solution (-k - 2 = -m₁/m₂ - 2 < -2). So

    cosΘ = [-kµ + √(2µ² - k² + 1)] / (µ²+1)

    cosΘ = [-(m₁/m₂)µ + √(2µ² - (m₁/m₂)² + 1)] / (µ²+1)

    and finally

    Θ = arccos( [-(m₁/m₂)µ + √(2µ² - (m₁/m₂)² + 1)] / (µ²+1) )

    It may be that you get a more elegant solution by substituting for cos² instead of sin² -- but I doubt it. And by all means, recheck my operations.

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