For the hydraulic lift shown in Fig. 12.7, what must be the ratio of the diameter of the vessel at the car to the diameter of the vessel where the force F₁ is applied so that a 1520-kg car can be lifted with a force F₁ of just 125 N?

Given data can be listed below as,

• The force F₁ applied on the on the piston is F₁= 125N.
• Mass of the car is m=1520kg .

The Pascal’s law can be used to amplify the input force by adjusting the cross-sectional area of the piston of a hydraulic lift.

The expression for the area of vessel is given by,

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

Here, D is the diameter.

A force F₂ must be applied on the vessel of car which is equal to weight of the car, F₂ = mg

Here, m is the mass of the car and g is the acceleration due to gravity.

$F_2\text{\hspace{0.17em}=\hspace{0.17em}}\frac{A_2}{A_1}{F}_1$

Here, A₂ is the area of vessel at the car, A₁is the area of vessel at the piston, and F₁ is the force acting on the piston.

Substitute the expression of area in above equation,

$\begin{array}{c}{F}_2=\frac{A_2}{A_1}{F}_1\\ \text{=}\frac{\pi D_2^2/4}{\pi D_1^2/4}{F}_1\end{array}$

Here, D₁ is the diameter of vessel at the piston, and D₂ is the diameter of the car.

Further solving above equation,

$\begin{array}{c}{F}_2\text{=}{\left(\frac{D_2}{D_1}\right)}^2{F}_1\\ \frac{D_2}{D_1}=\sqrt{\frac{F_2}{F_1}}\\ \frac{D_2}{D_1}=\sqrt{\frac{mg}{F_1}}\end{array}$

Substitute 1520 kg for m, 9.8 m/s² for g, and 125 N for F₁ in above equation,

$\begin{array}{c}\frac{D_2}{D_1}\text{=}\sqrt{\frac{\left(1520\text{\hspace{0.17em}kg}\right)\left(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{\hspace{0.17em}N}}{1\text{\hspace{0.17em}kg}\cdot \text{m}/{\text{s}}^2}\right)}{125\text{\hspace{0.17em}N}}}\\ \frac{D_2}{D_1}\text{=}10.9\end{array}$

Thus, the ratio of two diameters is 10.9:1 .