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AnnaJa

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## Homework Statement

An open rectangular tank contains water up to about half of its depth. This tank accelerates at a=2.20 m/s[itex]^{2}[/itex] up a slope α (alpha), which causes the free water surface to form an angle θ (theta) with the original horizontal plane.

What is the α angle of the slope?

## Homework Equations

p=ρh(g[itex]\stackrel{+}{-}[/itex]a)

tanθ=[itex]\stackrel{a}{g}[/itex]

## The Attempt at a Solution

θ=tan[itex]^{-1}([/itex][itex]\stackrel{199}{1242}[/itex])=9.1°

angle between water surface and slope = γ

γ= α+θ

tanγ=([itex]\stackrel{a}{g}[/itex])

tanγ=([itex]\stackrel{2.2}{9.81}[/itex])

γ=12.64°

α=γ-θ=12.64-9.1=3.54°

I am not sure if this is the right approach?

Any help would be greatly appreciated.

Thank you, Anna

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