Answer :
The depth of the mercury in the tank is approximately [tex]\( 0.093 \)[/tex] inches.
To find the depth of the mercury in the tank, we first need to calculate the pressure exerted by the mercury on the bottom of the tank using the given force.
The pressure exerted by a fluid at a certain depth is given by the formula:
[tex]\[ P = \rho \cdot g \cdot h \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure (in pounds per square inch, psi)
- [tex]\( \rho \)[/tex] is the density of the fluid (in pounds per cubic inch, lb/in³)
- [tex]\( g \)[/tex] is the acceleration due to gravity (in inches per second squared, in/s²)
- [tex]\( h \)[/tex] is the depth of the fluid (in inches, in)
We're given:
- Total force on the bottom of the tank [tex]\( F = 165 \)[/tex] lb
- Density of mercury [tex]\( \rho = 0.490 \)[/tex] lb/in³
We need to find [tex]\( h \),[/tex] the depth of the mercury.
First, let's convert the given force into pressure using the formula:
[tex]\[ P = \frac{F}{A} \][/tex]
where [tex]\( A \)[/tex] is the area of the bottom of the tank.
Given that the tank is rectangular and has dimensions [tex]\( 5.00 \)[/tex] in by [tex]\( 9.00 \)[/tex] in, the area [tex]\( A \)[/tex] is:
[tex]\[ A = \text{length} \times \text{width} = (5.00 \times 9.00) \, \text{in}^2 \][/tex]
Now, we find the pressure [tex]\( P \):[/tex]
[tex]\[ P = \frac{165}{5.00 \times 9.00} \, \text{psi} \][/tex]
Next, we equate this pressure to the pressure exerted by the mercury:
[tex]\[ \frac{165}{5.00 \times 9.00} = 0.490 \times g \times h \][/tex]
To find [tex]\( h \),\\[/tex] we rearrange the equation:
[tex]\[ h = \frac{\frac{165}{5.00 \times 9.00}}{0.490 \times g} \][/tex]
Now, we need the value of the acceleration due to gravity [tex]\( g \)[/tex]. This is usually taken as [tex]\( 32.174 \)[/tex] ft/s², but since we're working in inches, we convert it to [tex]\( 386.174 \)[/tex] in/s².
Now, let's plug in the values and solve for [tex]\( h \):[/tex]
[tex]\[ h = \frac{\frac{165}{5.00 \times 9.00}}{0.490 \times 386.174} \, \text{in} \]\[ h \approx \frac{165}{5.00 \times 9.00 \times 0.490 \times 386.174} \, \text{in} \]\[ h \approx \frac{165}{1775.11} \, \text{in} \]\[ h \approx 0.093 \, \text{in} \][/tex]
So, the depth of the mercury in the tank is approximately [tex]\( 0.093 \)[/tex] inches.
