Forces and Motion - Pressure

A person weighing $600 N$ stands on one foot. If the area of the sole of their shoe is $0.015 m^2$, calculate the pressure exerted on the ground.

Given: Force $F = 600 N$ Area $A = 0.015 m^2$ Using the formula: $P = \frac{F}{A}$ $P = \frac{600}{0.015}$ $P = 40000 Pa$

The pressure is calculated by dividing the weight (force) by the area of contact. The result is $40,000 Pa$ or $40 kPa$.

Calculate the pressure exerted by water at the bottom of a swimming pool that is $2.5 m$ deep. (Density of water $\rho = 1000 kg/m^3$ and $g = 9.8 m/s^2$)

Given: $h = 2.5 m$ $\rho = 1000 kg/m^3$ $g = 9.8 m/s^2$ Using the formula for liquid pressure: $P = \rho gh$ $P = 1000 \times 9.8 \times 2.5$ $P = 24500 Pa$

The pressure at a depth in a liquid depends only on the density of the liquid, the depth, and gravity. Here, the water exerts a pressure of $24.5 kPa$ at the bottom.

A hydraulic lift has a small piston with an area of $0.002 m^2$ and a large piston with an area of $0.1 m^2$. If a force of $50 N$ is applied to the small piston, what is the weight of the load that can be lifted by the large piston?

Given: $F_1 = 50 N$ $A_1 = 0.002 m^2$ $A_2 = 0.1 m^2$ Using Pascal's Principle: $\frac{F_1}{A_1} = \frac{F_2}{A_2}$ $\frac{50}{0.002} = \frac{F_2}{0.1}$ $25000 = \frac{F_2}{0.1}$ $F_2 = 25000 \times 0.1$ $F_2 = 2500 N$

Since the pressure is the same throughout the hydraulic fluid, the ratio of force to area must be constant. The larger area on the second piston allows it to lift a much larger force ($2500 N$) than was applied ($50 N$).