- #1

member 428835

## Homework Statement

Consider a rod of length ##L## that is attached to a hinge making an angle ##\theta## with the horizontal. There is a force ##\vec{F}## acting vertically on the rod that creates constant angular velocity ##-\omega##. If there is fluid under the rod with constant density, what force is the rod exerting on the hinge?

## Homework Equations

Continuity. Navier-Stokes?

## The Attempt at a Solution

So I'm not really sure where to start. I let the rod have a distance coming out of the page, call this ##b##. Then total volume of fluid in the triangular region under the rod and above the horizontal is ##V(t) = b L^2 \cos \theta \sin \theta/2 = b L^2 \sin (2 \theta) /4 \implies V'(t) = -b L^2 \omega\cos (2 \theta) /2##, which is at least negative, implying fluid is leaving the region (good).

Now I know torque ##\vec{\tau} = \vec{r} \times \vec{F}##, so if we use polar coordinates we have ##\vec{\tau} = L \hat{r} \times F\hat{y}##. Now we know the position vector in polar coordinates is ##\vec{R} = r\cos\theta \hat{x}+r\sin\theta\hat{y} \implies \hat{r} = \cos\theta\hat{x}+\sin\theta\hat{y}##. Then ##\vec{\tau} = L \hat{r} \times F\hat{y} = L(\cos\theta\hat{x}+\sin\theta\hat{y}) \times -F\hat{y} = F\hat{y} \times L(\cos\theta\hat{x}+\sin\theta\hat{y}) = F\hat{y} \times L\cos\theta\hat{x} +F\hat{y} \times \sin\theta\hat{y} = -F L \cos\theta \hat{z}## where ##z## is out of the page. I'm not really sure how to consolidate this info. I'm thinking that the fluid would exert a sort of pressure force normal to the rod and in the positive ##\theta## direction. Then if we summed forces on the rod we would have the downward force ##-F \hat{y}##, the pressure force ##F_p \hat{\theta}## and the reacting force from the hinge to the rod ##F_h \hat{x}##. Since the hinge is not accelerating, these forces should balance, right?

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