Question
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?
Answers
GPT-4o mini
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:
\[
P_{\text{input}} = \frac{F}{A} = \frac{200 \, \text{N}}{1.00 \times 10^{-4} \, \text{m}^2} = 2,000,000 \, \text{Pa} \, (or \, 2.00 \times 10^6 \, \text{Pa})
\]
2. **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:
\[
F_{output} = 2.00 \times 10^6 \, \text{Pa} \cdot 0.0022 \, \text{m}^2 = 4400 \, \text{N}
\]
3. **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**.
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:
\[
P_{\text{input}} = \frac{F}{A} = \frac{200 \, \text{N}}{1.00 \times 10^{-4} \, \text{m}^2} = 2,000,000 \, \text{Pa} \, (or \, 2.00 \times 10^6 \, \text{Pa})
\]
2. **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:
\[
F_{output} = 2.00 \times 10^6 \, \text{Pa} \cdot 0.0022 \, \text{m}^2 = 4400 \, \text{N}
\]
3. **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**.