COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 3, Problem 10P
To determine

The location of the sailboat using graphical method and component method of vector addition.

Expert Solution & Answer
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Answer to Problem 10P

The magnitude of the location of the sailboat is 29.3 NM and the angle is 342.8°.

Explanation of Solution

Starting from the origin, each vector is drawn beginning at the end of its preceding vector.

The graphical representation of the path of the sailboat is given below:

COLLEGE PHYSICS-CONNECT ACCESS, Chapter 3, Problem 10P

Write the expression for y-component of the position of the sailboat

    ry=r1sinθ1+r2sinθ2+r3sinθ3+r4sinθ4+r5sinθ5        (I)

Here, ry is the y-component of the resultant vector, r1,r2,r3,r4,r5 are the magnitudes of the vectors and θ1,θ2,θ3,θ4,θ5 are the angles of the vectors.

Write the expression for x-component of the position of the sailboat

    rx=r1cosθ1+r2cosθ2+r3cosθ3+r4cosθ4+r5cosθ5        (II)

Here, rx is the x-component of the resultant vector.

Write the expression for magnitude of the resultant vector

    r=(rx)2+(ry)2        (III)

Here, r is the magnitude of the resultant vector.

Write the expression for angle of the resultant vector

    r=tan1(ryrx)        (IV)

Here, r is the magnitude of the resultant vector.

Conclusion:

Substitute 45 NM for r1, 20 NM for r2, 30 NM for r3, 10 NM for r4 and 62 NM for r5, 0° for θ1, 60° for θ2, 0° for θ3, 60° for θ4 and 180° for θ5  in (I) to find ry

    ry={(45 NM)sin(0°)+(20 NM)sin(60°)+(30 NM)sin(0°)+(10 NM)sin(60°)+(62 NM)sin(180°)}=8.66 NM

Substitute 45 NM for r1, 20 NM for r2, 30 NM for r3, 10 NM for r4 and 62 NM for r5, 0° for θ1, 60° for θ2, 0° for θ3, 60° for θ4 and 180° for θ5  in (II) to find rx

    rx={(45 NM)cos(0°)+(20 NM)cos(60°)+(30 NM)cos(0°)+(10 NM)cos(60°)+(62 NM)cos(180°)}=28 NM

Substitute 28 NM for rx and 8.66 NM for ry in equation (III) to find r

    r=(28 NM)2+(8.66 NM)2=29.309 NM29.3 NM

Substitute 28 NM for rx and 8.66 NM for ry in equation (IV) to find θ

    θ=tan1(8.66 NM28 NM)=17.19°342.8°

Thus, the magnitude of the location of the sailboat is 29.3 NM and the angle is 342.8°.

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Chapter 3 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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