# 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?

### 1 Answer

- NCSLv 71 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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