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 17P

(a)

To determine

The magnitude and the direction of D=A+B+C .

(a)

Expert Solution
Check Mark

Answer to Problem 17P

The magnitude of D=A+B+C is 2.83m in the direction of 315° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of D=A+B+C .

Answer: The magnitude of D=A+B+C is 2.83m .

Given information:

The three displacement vectors are A=(3i^3j^)m , B=(i^4j^)m and C=(2i^+5j^)m .

Formula to calculate the magnitude of D=A+B+C is,

D=(Ax+Bx+Cx)2+(Ay+By+Cy)2

Substitute 3m for Ax , 3m for Ay , 1m for Bx , 4m for By , 2m for Cx and 5m for Cy in above expression.

D=(3m+1m+(2m))2+(3m+(4m)+5m)2=2.83m

Conclusion:

Therefore, the magnitude of D=A+B+C is 2.83m .

Section 2:

To determine: The direction of D=A+B+C .

Answer: The direction of D=A+B+C is 315° counterclockwise from positive x -axis.

Given information:

The three displacement vectors are A=(3i^3j^)m , B=(i^4j^)m and C=(2i^+5j^)m .

Formula to calculate the direction of D=A+B+C is,

θ=tan1(Ay+By+CyAx+Bx+Cx)

Substitute 3m for Ax , 3m for Ay , 1m for Bx , 4m for By , 2m for Cx and 5m for Cy in above expression.

θ=tan1((3m+(4m)+5m)(3m+1m+(2m)))=tan1(2m2m)=45°

Thus, the angle measured from counterclockwise is 360°45°=315°

Conclusion:

Therefore, the direction of D=A+B+C is 315° counterclockwise from positive x -axis.

(b)

To determine

The magnitude and the direction of E=AB+C .

(b)

Expert Solution
Check Mark

Answer to Problem 17P

The magnitude of E=AB+C is 13.42m in the direction of 296.57° counterclockwise from the positive x -axis.

Explanation of Solution

Section 1:

To determine: The magnitude of E=AB+C .

Answer: The magnitude of E=AB+C is 13.42m .

Given information:

The three displacement vectors are A=(3i^3j^)m , B=(i^4j^)m and C=(2i^+5j^)m .

Formula to calculate the magnitude of E=AB+C is,

E=((Ax)+(Bx)+Cx)2+((Ay)+(By)+Cy)2

Substitute 3m for Ax , 3m for Ay , 1m for Bx , 4m for By , 2m for Cx and 5m for Cy in above expression.

E=((3m)+(1m)+(2m))2+(3m+4m+5m)2=180m=13.42m

Conclusion:

Therefore, the magnitude of E=AB+C is 13.42m .

Section 2:

To determine: The direction of E=AB+C .

Answer: The direction of E=AB+C is 296.57° counterclockwise from the positive x -axis

Given information:

The three displacement vectors are A=(3i^3j^)m , B=(i^4j^)m and C=(2i^+5j^)m .

Formula to determine the direction of E=AB+C is,

θ=tan1((Ay)+(By)+Cy(Ax)+(Bx)+Cx)

Substitute 3m for Ax , 3m for Ay , 1m for Bx , 4m for By , 2m for Cx and 5m for Cy in above expression.

θ=tan1((3m+(4m)+5m)(3m+(1m)+(2m)))=tan1(12m6m)=63.43°

Thus, the angle measured from counterclockwise is 360°63.43°=296.57°

Conclusion:

Therefore, the direction of E=AB+C is 296.57° counterclockwise from the positive x -axis.

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Chapter 3 Solutions

Physics for Scientists and Engineers with Modern Physics

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