Question

A 55 kg cheerleader uses an oil-filled hydraulic lift to hold four 110 kg football players at a height of 1.0 m. If her piston is 16 cm in diameter, what is the diameter of the football players’ piston?

Answer 1

Answer:

$D = 55.2 cm$

Explanation:

As we know that the total mass of the all four players is given as

$M = 4\times 110$

$M = 440 kg$

diameter of the piston of cheer leader is given as

$d_1 = 16 cm$

are of cross-section is given as

$A_1 = \pi r^2$

$A_1 = \pi(0.08)^2 = 0.02 m^2$

mass of the cheer leader is given as

m = 55 kg

so the pressure due to cheer leader is given as

$P_{in} = \frac{mg}{A_1}$

$P_{in} = \frac{55 \times 9.81}{0.02}$

$P_{in} = 26835 Pa$

Now on the other side pressure must be same

so we have

$\frac{Mg}{A} + \rho gH = P_{in}$

$\frac{440 \times 9.8}{A} + (900)(9.8)(1) = 26835$

$A = 0.24 m^2$

$\pi r^2 = 0.24$

$r = 0.276 m$

so diameter on the other side is given as

$D = 2 r$

$D = 55.2 cm$