Questions: What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resting on a second cylinder? The master cylinder has a 2.00-cm diameter and the second cylinder has a 24.0-cm diameter.

Solution Steps

We need to find the force exerted on the master cylinder of a hydraulic lift to support a car weighing 2000 kg. The master cylinder has a diameter of 2.00 cm, and the second cylinder (supporting the car) has a diameter of 24.0 cm.

The weight of the car can be calculated using the formula: \[ W = m \cdot g \] where \( m = 2000 \, \text{kg} \) is the mass of the car and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.

\[ W = 2000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19620 \, \text{N} \]

According to Pascal's principle, the pressure applied to a confined fluid is transmitted undiminished throughout the fluid. Therefore, the pressure in the master cylinder is equal to the pressure in the second cylinder.

The pressure \( P \) is given by: \[ P = \frac{F}{A} \] where \( F \) is the force and \( A \) is the area.

The area \( A \) of a circle is given by: \[ A = \pi \left(\frac{d}{2}\right)^2 \] where \( d \) is the diameter.

For the master cylinder: \[ A_1 = \pi \left(\frac{2.00 \, \text{cm}}{2}\right)^2 = \pi \left(1.00 \, \text{cm}\right)^2 = \pi \times 1.00^2 \, \text{cm}^2 = \pi \, \text{cm}^2 \]

For the second cylinder: \[ A_2 = \pi \left(\frac{24.0 \, \text{cm}}{2}\right)^2 = \pi \left(12.0 \, \text{cm}\right)^2 = \pi \times 144 \, \text{cm}^2 = 144\pi \, \text{cm}^2 \]

Using Pascal's principle, the force on the master cylinder \( F_1 \) is related to the force on the second cylinder \( F_2 \) by: \[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \]

Solving for \( F_1 \): \[ F_1 = F_2 \cdot \frac{A_1}{A_2} = 19620 \, \text{N} \cdot \frac{\pi \, \text{cm}^2}{144\pi \, \text{cm}^2} = 19620 \, \text{N} \cdot \frac{1}{144} \]

\[ F_1 = \frac{19620}{144} \, \text{N} = 136.25 \, \text{N} \]

Final Answer