# Force on a chamber

## Homework Statement

A chamber has water entering horizontally with a velcoity of 80m/s with an inlet of an area of $0.1m^$ and leaving through an opening of area $0.15m^2$. The exiting flow makes an angle of 30 degrees with respect to the entering flow. What force is needed to hold the chamber in place

2. The attempt at a solution
Well the total force on the chamber is
$$\vec{F}=\dot{m_{2}}\vec{v_{2}}-\dot{m_{1}}\vec{v_{1}}$$
and $$\dot{m_{2}}=\rho A_{2}$$
and $$\dot{m_{1}}=\rho A_{1}$$

the component of the force is

$$F_{x}=\rho A_{2} v_{2} \cos\theta - \rho A_{1} v_{1}$$
$$F_{y}=\rho A_{2} v_{2}\sin\theta$$

Now heres the thing, we dont know v2...
But since the mass is conserved the volume flow rate in is equal to the volume flow rate out
$$\int_{S_{inlet}} \vec{v_{1}}\cdot\hat{n}dA=\int_{S_{outlet}} \vec{v_{2}}\cdot\hat{n}dA$$
$$v_{1}A_{1}=v_{2}A_{2}$$

From this we can find the X and Y components and hence the force and the direction. Is this all correct?

Thank you for your input and suggestions!

Q_Goest
Homework Helper
Gold Member
Hi stunner,
Now heres the thing, we dont know v2...
But since the mass is conserved the volume flow rate in is equal to the volume flow rate out
$$\int_{S_{inlet}} \vec{v_{1}}\cdot\hat{n}dA=\int_{S_{outlet}} \vec{v_{2}}\cdot\hat{n}dA$$
$$v_{1}A_{1}=v_{2}A_{2}$$

From this we can find the X and Y components and hence the force and the direction. Is this all correct?
Works for me. To be nit picky, the mass flow in = mass flow out (not volume flow) unless there's mass being stored inside the control volume, so we can equate VA(in) = VA(out) only if density(in) = density(out). Note that if this were a gas for example, and density changed, we'd have to determine velocity from the change in density and mass flow. But yea, you got it right.

thanks a lot

you can mark is solved