In a hydraulic lift whose input line has a cross-sectional area of 1.00 cm2 and whose output line has a cross-sectional area of 22.0 cm2, what is the largest mass (kg) that can be lifted by an input force of 200. N?

1 answer

To solve this problem, we'll use Pascal's principle, which states that the pressure applied to a confined fluid is transmitted undiminished in every direction throughout the fluid.

First, we can calculate the pressure exerted by the input force. The pressure \( P \) is given by the formula:

\[ P = \frac{F}{A} \]

where \( F \) is the force applied and \( A \) is the area.

  1. Calculate the pressure at the input:

    Given,

    • Input Force \( F = 200 , \text{N} \)
    • Input Area \( A = 1.00 , \text{cm}^2 = 1.00 \times 10^{-4} , \text{m}^2 \) (conversion from cm² to m²)

    Now, we can calculate the pressure:

\[ P_{\text{input}} = \frac{F}{A} = \frac{200 , \text{N}}{1.00 \times 10^{-4} , \text{m}^2} = 2,000,000 , \text{Pa} , (or , 2.00 \times 10^6 , \text{Pa}) \]

  1. Calculate the output force using the same pressure:

    The pressure at the output will be the same due to Pascal's principle. Thus:

\[ P_{\text{output}} = P_{\text{input}} = 2.00 \times 10^6 , \text{Pa} \]

We can express the force exerted by the output area \( A_{output} \):

Given,

  • Output Area \( A_{output} = 22.0 , \text{cm}^2 = 22.0 \times 10^{-4} , \text{m}^2 \)

Now, we can calculate the output force \( F_{output} \):

\[ F_{output} = P_{\text{output}} \cdot A_{output} = (2.00 \times 10^6 , \text{Pa}) \cdot (22.0 \times 10^{-4} , \text{m}^2) \]

Calculating that gives us:

\[ F_{output} = 2.00 \times 10^6 , \text{Pa} \cdot 0.0022 , \text{m}^2 = 4400 , \text{N} \]

  1. Calculate the largest mass that can be lifted:

To find the mass \( m \), we use the relation between force and mass, given by:

\[ F = m \cdot g \]

where \( g \) (acceleration due to gravity) is approximately \( 9.81 , \text{m/s}^2 \).

Rearranging the formula gives us:

\[ m = \frac{F_{output}}{g} = \frac{4400 , \text{N}}{9.81 , \text{m/s}^2} \]

Calculating this gives:

\[ m \approx \frac{4400}{9.81} \approx 448.0 , \text{kg} \]

Therefore, the largest mass that can be lifted by an input force of 200 N in the hydraulic lift is approximately 448 kg.