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.
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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:
\[ P_{\text{input}} = \frac{F}{A} = \frac{200 , \text{N}}{1.00 \times 10^{-4} , \text{m}^2} = 2,000,000 , \text{Pa} , (or , 2.00 \times 10^6 , \text{Pa}) \]
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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:
\[ F_{output} = 2.00 \times 10^6 , \text{Pa} \cdot 0.0022 , \text{m}^2 = 4400 , \text{N} \]
- 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.