### Understanding Forces in a Hydraulic LiftA hydraulic lift has two pistons: a small one of cross-sectional area 5.00 cm² and a large one of cross-sectional area 300 cm². What minimum force must be applied to the small piston in order for the large piston to lift a truck of mass 4500 kg?#### Possible Answer Choices:1. \( 735 \, \text{N} \)2. \( 2.6 \times 10^6 \, \text{N} \)3. \( 75 \, \text{N} \)4. \( 2.7 \times 10^5 \, \text{N} \)**Explanation**:To determine the necessary force on the small piston, consider the following principle of a hydraulic system - Pascal’s Principle, which states that pressure applied to a confined fluid is transmitted undiminished throughout the fluid.First, calculate the force needed to lift the truck using the following formula:\[ F_{\text{large}} = \text{mass} \times \text{gravity} \]Given: \[ \text{mass} = 4500 \, \text{kg} \]\[ \text{gravity} = 9.81 \, \text{m/s}^2 \]\[ F_{\text{large}} = 4500 \times 9.81 = 44145 \, \text{N} \]Next, use the pressure equivalence in the fluid:\[ \frac{F_{\text{small}}}{A_{\text{small}}} = \frac{F_{\text{large}}}{A_{\text{large}}} \]Given areas:\[ A_{\text{small}} = 5.00 \, \text{cm}^2 = 5.00 \times 10^{-4} \, \text{m}^2 \]\[ A_{\text{large}} = 300 \, \text{cm}^2 = 300 \times 10^{-4} \, \text{m}^2 \]Use the relationship to find \( F_{\text{small}} \):\[ F_{\text{small}} = \left( \frac{A_{\text{small}}}{A_{\text{large}}} \right) \times F_{\text{large}} \]Calculate:\[ F_{\text

Given: area of small piston, a = 5 cm2 area of large piston, A = 300 cm2 mass on large piston, M =…

A hydraulic lift has two pistons: a small one of cross-sectional area 5.00 cm- and a large one of cross-sectional area 300 cm². What minimum force must be applied to the small piston in order for the large piston to lift a truck of mass 4500 kg? O 735 N O 2.6 x 106 N O 75 N O 2.7 x 105 N

A hydraulic lift has two pistons: a small one of cross-sectional area 5.00 cm- and a large one of cross-sectional area 300 cm². What minimum force must be applied to the small piston in order for the large piston to lift a truck of mass 4500 kg? O 735 N O 2.6 x 106 N O 75 N O 2.7 x 105 N

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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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Question

### Understanding Forces in a Hydraulic Lift

A hydraulic lift has two pistons: a small one of cross-sectional area 5.00 cm² and a large one of cross-sectional area 300 cm². What minimum force must be applied to the small piston in order for the large piston to lift a truck of mass 4500 kg?

#### Possible Answer Choices:
1. \( 735 \, \text{N} \)
2. \( 2.6 \times 10^6 \, \text{N} \)
3. \( 75 \, \text{N} \)
4. \( 2.7 \times 10^5 \, \text{N} \)

**Explanation**:

To determine the necessary force on the small piston, consider the following principle of a hydraulic system - Pascal’s Principle, which states that pressure applied to a confined fluid is transmitted undiminished throughout the fluid.

First, calculate the force needed to lift the truck using the following formula:
\[ F_{\text{large}} = \text{mass} \times \text{gravity} \]
Given:
\[ \text{mass} = 4500 \, \text{kg} \]
\[ \text{gravity} = 9.81 \, \text{m/s}^2 \]
\[ F_{\text{large}} = 4500 \times 9.81 = 44145 \, \text{N} \]

Next, use the pressure equivalence in the fluid:
\[ \frac{F_{\text{small}}}{A_{\text{small}}} = \frac{F_{\text{large}}}{A_{\text{large}}} \]

Given areas:
\[ A_{\text{small}} = 5.00 \, \text{cm}^2 = 5.00 \times 10^{-4} \, \text{m}^2 \]
\[ A_{\text{large}} = 300 \, \text{cm}^2 = 300 \times 10^{-4} \, \text{m}^2 \]

Use the relationship to find \( F_{\text{small}} \):
\[ F_{\text{small}} = \left( \frac{A_{\text{small}}}{A_{\text{large}}} \right) \times F_{\text{large}} \]

Calculate:
\[ F_{\text

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