Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 3, Problem 28P

Figure P3.28 illustrates typical proportions of male (m) and female (f) anatomies. The displacements d 1 m and d 1 f from the soles of the feet to the navel have magnitudes of 104 cm and 64.0 cm. respectively. The displacements d 2 m and d 2 f from the navel to outstretched fingertips have magnitudes of 100 cm and 86.0 cm, respectively. Find the vector sum of these displacements d 3 = d 1 + d 2 for both people.

Figure P3.28

Chapter 3, Problem 28P, Figure P3.28 illustrates typical proportions of male (m) and female (f) anatomies. The displacements

Expert Solution & Answer
Check Mark
To determine

The vector sum of the given displacements for both male and female.

Answer to Problem 28P

The vector sum of the displacements for male is 170.12cm in a direction of 57.24° from positive x axis and the vector sum of the displacements for female is 145.72cm in a direction of 58.59° from the positive x axis.

Explanation of Solution

Section 1:

To determine: The vector sum of displacements for male in unit vector notation.

Answer: The unit vector notation for vector sum of displacements for male is (92.05i^)cm+(143.07j^)cm .

Given information:

The magnitude of male displacements, d1m and d2m is 104cm and 100cm respectively and the magnitude of female displacements, d1f and d2f is 84.0cm and 86.0cm respectively.

The displacements of male in unit vector notation is given as,

d1m=(104j^)cm

d2m=(100cos23°i^)cm+(100sin23°j^)cm

Formula to calculate the vector sum for male is given as,

dsm=d1m+d2m (I)

Substitute (104j^)cm for d1m and (100cos23°i^)cm+(100sin23°j^)cm for d2m in equation (I).

dsm=(104j^)cm+(100cos23°i^)cm+(100sin23°j^)cm=(104j^)cm+(92.0i^)cm+(39.07j^)cm=(92.05i^)cm+(143.07j^)cm

Section 2:

To determine: The magnitude of the vector sum of displacements for male.

Answer: The magnitude of the vector sum of the displacements for male is 170.12cm .

Given information:

The magnitude of male displacements, d1m and d2m is 104cm and 100cm respectively and the magnitude of female displacements, d1f and d2f is 84.0cm and 86.0cm respectively.

Formula to calculate the magnitude of the vector sum of the displacements for male is,

|dsm|=(dsmx)2+(dsmy)2 (II)

Substitute 92.05cm for dsmx and 143.07cm for dsmy in equation (II).

|dsm|=(92.05cm)2+(143.07cm)2=170.12cm

Conclusion:

Therefore, the magnitude of the vector sum of the displacements for male is 170.12cm .

Section 3:

To determine: The direction for the vector sum of the displacements for male.

Answer: The direction for the vector sum of the displacements for male is 57.24° from the positive x axis.

Given information:

The magnitude of male displacements, d1m and d2m is 104cm and 100cm respectively and the magnitude of female displacements, d1f and d2f is 84.0cm and 86.0cm respectively.

Formula to calculate the direction for the vector sum of the displacements for male is,

θm=tan1(dsmydsmx)

Substitute 92.05cm for dsmx and 143.07cm for dsmy in above equation.

θm=tan1(143.07cm92.05cm)=57.24°

Conclusion:

Therefore, the direction for the vector sum of the displacements for male is 57.24° from the positive x axis.

Section 4:

To determine: The vector sum of displacements for female in unit vector notation.

Answer: The unit vector notation for vector sum of displacements for female is (75.93i^)cm+(124.37j^)cm .

Given information:

The magnitude of male displacements, d1m and d2m is 104cm and 100cm respectively and the magnitude of female displacements, d1f and d2f is 84.0cm and 86.0cm respectively.

The displacements of female in unit vector notation is given as,

d1f=(84.0j^)cm

d2f=(86.0cos28°i^)cm+(86.0sin28°j^)cm

Formula to calculate the vector sum for female is given as,

dsf=d1f+d2f (III)

Substitute (84.0j^)cm for d1f , (86.0cos28°i^)cm+(86.0sin28°j^)cm for d2f in equation (III).

dsf=(84.0j^)cm+(86.0cos28°i^)cm+(86.0sin28°j^)cm=(84.0j^)cm+(75.93i^)cm+(40.7j^)cm=(75.93i^)cm+(124.37j^)cm

Section 5:

To determine: The magnitude of the vector sum of displacements for female.

Answer: The magnitude of the vector sum of the displacements for female is 145.72cm .

Given information:

The magnitude of male displacements, d1m and d2m is 104cm and 100cm respectively and the magnitude of female displacements, d1f and d2f is 84.0cm and 86.0cm respectively.

Formula to calculate the magnitude of the vector sum of the displacements for female is,

|dsf|=(dsfx)2+(dsfy)2 (IV)

Substitute 75.93cm for dsfx and 124.37cm dsfy in equation (IV).

|dsf|=(75.93cm)2+(124.37cm)2=145.72cm

Conclusion:

Therefore, the magnitude of the vector sum of the displacements for female is 145.72cm .

Section 6:

To determine: The direction for the vector sum of the displacements for female.

Answer: The direction for the vector sum of the displacements for female is 58.59° from the positive x axis.

Given information:

The magnitude of male displacements, d1m and d2m is 104cm and 100cm respectively and the magnitude of female displacements, d1f and d2f is 84.0cm and 86.0cm respectively.

Formula to calculate the direction for the vector sum of the displacements for male is,

θf=tan1(dsfydsfx)

Substitute 75.93cm for dsfx and 124.37cm dsfy in above equation.

θm=tan1(124.37cm75.93cm)=58.59°

Conclusion:

Therefore, the direction for the vector sum of the displacements for female is 58.59° from the positive x axis.

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Chapter 3 Solutions

Physics for Scientists and Engineers with Modern Physics

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