A hydraulic jack has an input piston of area 0.0600 m² and an output piston of area 0.690 m2. How much force (in N) on the input piston is required to lift a car weighing 1.40 x 104 N?
A hydraulic jack has an input piston of area 0.0600 m² and an output piston of area 0.690 m2. How much force (in N) on the input piston is required to lift a car weighing 1.40 x 104 N?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Hydraulic Jack Force Calculation
**Problem Statement:**
A hydraulic jack has an input piston of area \(0.0600 \, \text{m}^2\) and an output piston of area \(0.690 \, \text{m}^2\). How much force (in N) on the input piston is required to lift a car weighing \(1.40 \times 10^4\) N?
**Hint:**
You can use the principle of hydraulic systems which states that \( \frac{F_{\text{input}}}{A_{\text{input}}} = \frac{F_{\text{output}}}{A_{\text{output}}} \).
**Solution:**
To find the necessary force on the input piston, \( F_{\text{input}} \), we can rearrange the equation to solve for \( F_{\text{input}} \):
\[ F_{\text{input}} = F_{\text{output}} \times \frac{A_{\text{input}}}{A_{\text{output}}} \]
Input the given values:
\[ F_{\text{input}} = \left(1.40 \times 10^4 \, \text{N}\right) \times \frac{0.0600 \, \text{m}^2}{0.690 \, \text{m}^2} \]
Simplify the fraction:
\[ \frac{0.0600}{0.690} \approx 0.087 \]
So,
\[ F_{\text{input}} \approx \left(1.40 \times 10^4 \, \text{N}\right) \times 0.087 \approx 1218 \, \text{N} \]
Therefore, a force of approximately \(1218 \, \text{N}\) is required on the input piston to lift the car.
**Solution Box:**
\[ \boxed{1218 \, \text{N}} \]
This solution box indicates where students should input their answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F50bb9748-ad1e-4409-b934-aad7fabe70c9%2F9c9a39e7-2d36-4452-be2f-af6028c1a971%2F8eqklq.jpeg&w=3840&q=75)
Transcribed Image Text:### Hydraulic Jack Force Calculation
**Problem Statement:**
A hydraulic jack has an input piston of area \(0.0600 \, \text{m}^2\) and an output piston of area \(0.690 \, \text{m}^2\). How much force (in N) on the input piston is required to lift a car weighing \(1.40 \times 10^4\) N?
**Hint:**
You can use the principle of hydraulic systems which states that \( \frac{F_{\text{input}}}{A_{\text{input}}} = \frac{F_{\text{output}}}{A_{\text{output}}} \).
**Solution:**
To find the necessary force on the input piston, \( F_{\text{input}} \), we can rearrange the equation to solve for \( F_{\text{input}} \):
\[ F_{\text{input}} = F_{\text{output}} \times \frac{A_{\text{input}}}{A_{\text{output}}} \]
Input the given values:
\[ F_{\text{input}} = \left(1.40 \times 10^4 \, \text{N}\right) \times \frac{0.0600 \, \text{m}^2}{0.690 \, \text{m}^2} \]
Simplify the fraction:
\[ \frac{0.0600}{0.690} \approx 0.087 \]
So,
\[ F_{\text{input}} \approx \left(1.40 \times 10^4 \, \text{N}\right) \times 0.087 \approx 1218 \, \text{N} \]
Therefore, a force of approximately \(1218 \, \text{N}\) is required on the input piston to lift the car.
**Solution Box:**
\[ \boxed{1218 \, \text{N}} \]
This solution box indicates where students should input their answer.
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