Question

Determine the force (in N) exerted by only the fluid on the window of an instrument probe at this depth if the window is circular and has a diameter of 3.65 cm.We are given density: 1.00x10^3 kg/m^3air pressure: 101.3 kPafluid depth: 29mdiameter: 3.65cm

Answer 1

We are asked to determine the force exerted only by the fluid on a circular area that is submerged in a fluid. To do that we need to determine the pressure on the area, to do that we will use the following formula:

$P=\rho gh$

Where:

$\begin{gathered} \\ \rho=\text{ density} \\ g=\text{ acceleration of gravity} \\ h=\text{ depth} \end{gathered}$

We don't not add the air pressure to the formula because we need to know the force exerted only by the hydrostatic pressure of the fluid.

Substituting the values we get:

$P=(1\times10^3\frac{\operatorname{kg}}{m^3})(9.8\frac{m}{s^2})(29m)$

Solving the operations we get:

$P=284200Pa$

Now we use the following relationship for pressure:

$P=\frac{F}{A}$

Where:

$\begin{gathered} F=\text{ force} \\ A=\text{ area} \end{gathered}$

Now we solve for the force by multiplying both sides by the Area:

$PA=F$

Now, we determine the area using the following formula:

$A=\pi\frac{D^2}{4}$

Substituting the values:

$A=\frac{\pi(3.65\operatorname{cm})^2}{4}$

We convert the cm into meter by dividing by 100:

$A=\frac{\pi(0.0365m)^2}{4}$

Solving the operations we get:

$A=0.00105m^2$

Now we substitute in the formula for the force:

$(284200Pa\text{)}(0.00105m^2)=F$

Solving the operations:

$297.4N=F$

Therefore, the force is 297.4 Newtons.