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?
Explanations & Calculations
"\\qquad\\qquad\n\\begin{aligned}\n\\small \\frac{F}{A}&=\\small \\frac{f}{a}\\\\\n\\small \\frac{1250kg\\times9.8ms^{-2}}{\\pi R^2}&=\\small \\frac{125\\,N}{\\pi r^2}\\\\\n\\small \\frac{R^2}{r^2}&=\\small \\frac{1250\\times9.8}{125}\\\\\n\\small \\frac{R}{r}&=\\small \\sqrt{98}\\\\\n\\small \\frac{\\frac{D}{2}}{\\frac{d}{2}}&=\\small 9.9\\\\\n\\small \\frac{D}{d}&\\approx\\small 10\n\\end{aligned}"
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