Answer
a) $5.83\;\rm kN$
b) $6.0\;\rm kN$
Work Step by Step
a) The given figure tells us that the liquid inside the system is oil, so we are going to use the typical oil density which is given in table 15.1.
As we see in the figure below, the pressure at point A is equal to the pressure at point C since the oil is incompressible and uniform.
$$P_A=P_C$$
$$P_A=\overbrace{P_a+P_m+P_L}^{P_C}$$
where $P_a$ is the atmospheric pressure, $P_m$ is the pressure due to the mass of the piston, and $P_L$ is the pressure due to the weight of the liquid column. Thus,
$$P_A=(1.013\times 10^5)+\dfrac{mg}{A_1}+h\rho_{oil} g$$
where $A_1$ is the cross-sectional area of the tube.
$$P_A=(1.013\times 10^5)+\dfrac{(10)(9.8)}{\pi (2\times 10^{-2})^2}+(0.7)(900)(9.8)$$
$$P_A=1.855\times 10^5 \rm N/m^2$$
Now we can find the force exerted on the cylinder at$A$,
$$F_{\rm at\;A}=P_AA_2$$
where $A_2$ is the cross-sectional area of the cylinder at point A.
$$F_{\rm at\;A}=(1.855\times 10^5)\pi(10\times 10^{-2})^2$$
$$F_{\rm at\;A}=\color{red}{\bf 5.83\times 10^3}\;\rm N$$
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b) By the same approach,
$$P_B=P_a+P_m+h_1\rho_{oil}g $$
$$P_B=(1.013\times 10^5)+\dfrac{(10)(9.8)}{\pi (2\times 10^{-2})^2}+(1.3)(900)(9.8)$$
$$P_B=1.91 \times 10^5 \rm N/m^2$$
Now we can find the force exerted on the cylinder at$A$,
$$F_{\rm at\;B}=P_BA_2$$
where $A_2$ is the cross-sectional area of the cylinder at point B.
$$F_{\rm at\;B}=(1.91 \times 10^5)\pi(10\times 10^{-2})^2$$
$$F_{\rm at\;B}=\color{red}{\bf 6.0\times 10^3}\;\rm N$$
