Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)
12th Edition
ISBN: 9781259587399
Author: Eugene Hecht
Publisher: McGraw-Hill Education
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Chapter 13, Problem 26SP

The sole of a man’s size-10 shoe is around 11.0 in. by 4.00 in. Determine the gauge pressure under the feet of a 200-lb man standing upright. Give your answer in both lb/in . 2 and Pa. [Hint: 1.00  lb/in 2 = 6895  Pa . Check your work using 1.00  in . 2 = 6.45 × 10 4  m 2 and 1.00  lb = 4.448  N .]

Expert Solution & Answer
Check Mark
To determine

The gauge pressure under the feet of a 200-lb man standing upright if the size of the man’s sole is around 11.0 in. by 4.00 in.

Answer to Problem 26SP

Solution:

2.27 lb/in2, 15.7 kPa

Explanation of Solution

Given data:

The weight of the person is 200 lb.

The length of each foot is 11 in

The breadth of each foot is 4 in

Formula used:

Write the expression for pressure exerted by the shoe:

P=FA

Here, P is the pressure exerted, F is the force applied, and A is the area on which the force is applied.

Write the expression of the area of one shoe:

A=lb

Here, A is the area, l is the length, and b is the breadth.

Explanation:

Recall the expression of the area of one shoe:

A=lb

Here, A is the area of one shoe, l is the length of one shoe, and b is the breadth of one shoe.

Substitute 11 in for l and 4 in for b

A=(11 in)(4 in) = 44 in2

Here, the force applied by both feet is equal to the weight of the person which is equally divided on both feet. Hence, the force on one foot will be F2.

Thus, the expression of gauge pressure under the feet of the man standing upright will be:

P=F2A

Here, P is the gauge pressure exerted by the standing man, F is the weight of the man, and A is the area of one foot.

Substitute 200 lb for F and 44 in2 for A

P=200 lb2(44 in2)= 2.27 lb/in2

Convert the pressure from lb/in2 to Pa

P=(2.27 lb/in2)(6895 Pa1 lb/in2)=(15,651.65 Pa)(1 kPa1000 Pa)=15.65 kPa15.7 kPa

Conclusion:

Therefore, the gauge pressure under the feet of a 200-lb man standing upright is 2.27 lb/in2 or 15.7 kPa.

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Chapter 13 Solutions

Schaum's Outline of College Physics, Twelfth Edition (Schaum's Outlines)

Ch. 13 - 13.36 [II] Repeat Problem 13.35, but now find the...Ch. 13 - 13.37 [II] Compute the pressure required for a...Ch. 13 - 38. A covered cubic tank 5.00 m by 5.00 m by 5.00...Ch. 13 - 39. A cubic covered tank 5.00 m by 5.00 m by 5.00...Ch. 13 - 40. For the press in Fig. 13-3, the ratio of the...Ch. 13 - 13.41 [I] The output area of the piston in the...Ch. 13 - 13.42 [I] For the hydraulic press in Fig. 13-3,...Ch. 13 - 13.43 [II] The area of a piston of a force pump is...Ch. 13 - 13.44 [II] The diameter of the large piston of a...Ch. 13 - 45. An iron cube 20.0 cm on each side is submerged...Ch. 13 - 13.46 [I] The cube in the previous problem is...Ch. 13 - 47. A metal cube, 2.00 cm on each side, has a...Ch. 13 - 48. A solid wooden cube, 30.0 cm on each edge, can...Ch. 13 - 49. A metal object “weighs” 26.0 g in air and...Ch. 13 - 50. A solid piece of aluminum (ρ = 2.70 g/cm3) has...Ch. 13 - 51. A beaker contains oil of density 0.80 g/cm3. A...Ch. 13 - 13.52 [II] A tank containing oil of sp rests on a...Ch. 13 - Prob. 53SPCh. 13 - 13.54 [III] Determine the unbalanced force acting...Ch. 13 - 57. A piece of metal has a measured mass of 5.00...Ch. 13 - 13.56 [II] A balloon and its gondola have a total...Ch. 13 - 55. A 2.0-cm cube of metal is suspended by a fine...Ch. 13 - Prob. 58SPCh. 13 - 13.59 [II] What fraction of the volume of a piece...Ch. 13 - 13.60 [II] A cube of wood floating in water...Ch. 13 - 13.61 [III] Suppose we have a spring scale that...Ch. 13 - 13.62 [II] A glass of water has a ice cube...Ch. 13 - 13.63 [II] A glass tube is bent into the form of a...Ch. 13 - 13.64 [II] On a day when the pressure of the...
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