A. In FIGURE $P14.38$, how much force does the fluid exert on the end of the cylinder at$A$?

B. How much force does the fluid exert on the end of the cylinder at $B$?

(A) Force exerted on the fluid at $A$comes from three pressure sources. They are atmospheric pressure $p_{atm}$, fluid pressure at level $A$below the piston $p_f$, hydraulic pressure caused by the piston's (gray block below) weight $p_w$

The given figure is as follows:

$d=4.0cm$

$=(4.0\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$=0.04\mathrm{m}$

The hydraulic pressure from the piston's weight can be given by

$p_{\mathrm{w}}=\frac{mg}{A_{piston}}$

Here $m$is the mass of the piston, $g$acceleration due to gravity; $A_{piston}$is the area of the cylinder at the piston end

Area can be written in the terms of diameter is as follows.

$A_{piston}=\pi {\left(\frac{d_{piston}}{2}\right)}^2$

Here $d_{piston}$the piston's diameter.

Substitute in the hydraulic pressure equation.

$p_{\mathrm{w}}=\frac{4}{\pi }\frac{mg}{{d_{piston}}^2}$

Substitute $10kg$for mass, $0.04m$for the diameter, $9.8m/s^2$for acceleration due to gravity.

$p_{\mathrm{w}}=\frac{4}{\pi }\frac{(10\mathrm{kg})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{(0.04\mathrm{m}{)}^2}$

$=77,986\mathrm{Pa}$

$p_{\mathrm{f}}=\rho gh$

Here ? density of oil, $h$the height of $A$below the piston,

Substitute $900kg/m^3$for density,$70cm$for the height these values with,$9.8m/s^2$or acceleration due to gravity.

$p_{\mathrm{f}}=\left(900\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(70\mathrm{cm})$

$=\left(900\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left((70\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)$

$=6,174\mathrm{Pa}$

In order to solve for remaining values we need to find the area $A$at level $A$is

$A=\pi {\left(\frac{d_{\mathrm{A}}}{2}\right)}^2$

Here $d_A$is the diameter at level $A$. Substitute $20cm$for $d_A$gives us

$A=\pi \left(\frac{20\mathrm{cm}}{2}\right)$

$=\pi {\left(\left(\frac{20\mathrm{cm}}{2}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2$

$=0.0314{\mathrm{m}}^2$

Now we have all the values to solve for force as below

$F=\left(p_{\mathrm{atm}}+p_{\mathrm{w}}+p_{\mathrm{f}}\right)A$

Substitute $101300Pa$for localid="1648189377672" $p_{atm}$,$0.0314m^2$for $A$, $6,174Pa$for $p_f$and $77,986Pa$for $p_w$

$F=(101,300\mathrm{Pa}+77,986\mathrm{Pa}+6,174\mathrm{Pa})\left(0.0314{\mathrm{m}}^2\right)$

$=5,800\mathrm{N}$

$=5,800\mathrm{N}\left(\frac{1\mathrm{kN}}{1000\mathrm{N}}\right)$

$=5.8\mathrm{kN}$

The force at level $A$ is$5.8kN$

(B)The force at B is also given by

$F=\left(p_{\mathrm{atm}}+p_{\mathrm{w}}+p_{\mathrm{f}}\right)A$

However, the fluid pressure given by $p_{f=}\rho gh$is different as height is lower at $130cm$. So we can expect fluid pressure to increase

Substitute values$900kg/m^3$for density and $130cm$for height and $9.8m/s^2$ for $g$above

$p_{\mathrm{f}}=\left(900\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)(130\mathrm{cm})$

$=\left(900\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left((130\mathrm{cm})\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right)$

$=11,466\mathrm{Pa}$

At $11,466Pa$this fluid pressure is higher than level $A$. Now we can solve for force at the new level. Note that the cross sectional area at $B$and $A$are the same

$F=\left(p_{\mathrm{atm}}+p_{\mathrm{w}}+p_{\mathrm{f}}\right)A$

Substitute $101300Pa$for$p_{atm}$, $0.0314m^2$for $A$, $11,466Pa$for $p_f$and $77,986Pa$for $p_w$and

$F=(101,300\mathrm{Pa}+77,986\mathrm{Pa}+11,466\mathrm{Pa})\left(0.0314{\mathrm{m}}^2\right)$

$=6000N$

$=6,000\mathrm{N}\left(\frac{1\mathrm{kN}}{1000\mathrm{N}}\right)$

$=6.0\mathrm{kN}$

$B$seems to have a $6.0kN$force.