PHYSICS OF EVERYDAY (LL) W/ACCESS CODE
PHYSICS OF EVERYDAY (LL) W/ACCESS CODE
18th Edition
ISBN: 9781307304015
Author: Griffith
Publisher: Mcgraw-Hill/Create
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Textbook Question
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Chapter 9, Problem 1SP

Suppose the input piston of a hydraulic jack has a diameter of 3 cm and the load piston a diameter of 24 cm. The jack is being used to lift a car with a mass of 1700 kg.

  1. a. What are the areas of the input and load pistons in square centimeters? (A = πr2)
  2. b. What is the ratio of the area of the load piston to the area of the input piston?
  3. c. What is the weight of the car in newtons? (W = mg)
  4. d. What force must be applied to the input piston to support the car?

(a)

Expert Solution
Check Mark
To determine

The area of input and load piston.

Answer to Problem 1SP

The area of input piston is 28.3cm2 and that of load piston is 1809.6cm2

Explanation of Solution

Given info: Diameter of input piston is 3cm, diameter of load piston is 24cm mass of car is 1700kg.

Write the equation to find the area of input piston.

Ai=πr2

Here,

Ai is the area of input piston

r is the radius

Substitute 3cm for r in equation to get Ai.

Ai=3.14×(3cm)2=28.2cm2

Write the equation to find the area of load piston.

Al=πr2

Here,

Al is the area of load piston

r is the radius

Substitute 3cm for r in equation to get Al.

Al=3.14×(24cm)2=1809.6cm2

Conclusion:

The area of input piston is 28.3cm2 and that of load piston is 1809.6cm2

(b)

Expert Solution
Check Mark
To determine

Ratio of area of load piston to area of input piston.

Answer to Problem 1SP

The ratio is 64:1

Explanation of Solution

Write the equation to find the ratio of area of load piston and input piston.

ratio=AlAi

Substitute 1809.6cm2 for Al and 28.3cm2 for Ai to get the ratio.

ratio=1809.6cm228.3cm2=64

Conclusion:

The ratio is 64:1

(c)

Expert Solution
Check Mark
To determine

The weight of car.

Answer to Problem 1SP

The weight of car is 1.411×104N.

Explanation of Solution

Write the equation to find the weight.

W=mg

Here,

W is the weight

m is the mass

g is the acceleration due to gravity

Substitute 1700kg for m and 9.8m/s2 for g in equation to get W .

W=1700kg×9.8m/s2=1.411×104N

Conclusion:

The weight of car is 1.411×104N.

(d)

Expert Solution
Check Mark
To determine

The force applied to piston to balance the car.

Answer to Problem 1SP

The force applied is 498.5kPa.

Explanation of Solution

Write the equation to find the pressure.

P=FA

Here,

P is the pressure

F is the force

A is the area

Substitute 1.411×104N for F and 28.3cm2 for A in equation to get P

P=1.411×104N28.3cm2=498.5kPa

Conclusion:

The force applied is 498.5kPa.

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Chapter 9 Solutions

PHYSICS OF EVERYDAY (LL) W/ACCESS CODE

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