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?

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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.
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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