Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 3, Problem 20P

(a)

To determine

The magnitude and direction A+B.

(a)

Expert Solution
Check Mark

Answer to Problem 20P

The magnitude of A+B is |A+B|=2.0units_, at 60° CW from x axis.

Explanation of Solution

Write the expression for the magnitude of A+B

|A+B|=[(A+B)x]2+[(A+B)y]2 (I)

Write the expression for direction of A+B

θ=tan1oppositeadjacent (II)

Conclusion:

Substitute 1.0 for (A+B)x, 3.0 for (A+B)y in equation (I)

|A+B|=[1.0]2+[3.0]2=2.0units

Substitute 3.0 for opposite side, and 1.0 for adjacent side in equation (II)

θ=tan13.01.0=60°

Therefore the magnitude of A+B is |A+B|=2.0units_, at 60° CW from x axis.

(b)

To determine

The magnitude and direction AB.

(b)

Expert Solution
Check Mark

Answer to Problem 20P

The magnitude of AB is |AB|=2.0units_, at 60° CW from +x axis.

Explanation of Solution

Write the expression for the magnitude of AB

|AB|=[(AB)x]2+[(AB)y]2 (I)

Write the expression for direction of A+B

θ=tan1oppositeadjacent (II)

Conclusion:

Substitute 1.0 for (AB)x, 3.0 for (AB)y in equation (I)

|A+B|=[1.0]2+[3.0]2=2.0units

Substitute 3.0 for opposite side, and 1.0 for adjacent side in equation (II)

θ=tan13.01.0=60°

Therefore, the magnitude of AB is |AB|=2.0units_, at 60° CW from +x axis.

(c)

To determine

The x and y component of BA.

(c)

Expert Solution
Check Mark

Answer to Problem 20P

The x component is Bx=1.0units_, and y component is Ay=3.0units_.

Explanation of Solution

Consider the figure given below representing individual vectors as well as resultant vectors.

Physics, Chapter 3, Problem 20P

The resultant vector BA lies in the third quadrant can be resolved into two components Bx and Ay(since, A in the positive y axis has no x component, and B in negative x axis has no y component). The magnitude of Bx=1.0units, and Ay is, Ay=3.0units.

Conclusion:

Therefore, the x component is Bx=1.0units_, and y component is Ay=3.0units_

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