Explanation Procedure used: To find x and y components of a vector: 1. Place the vector A such that θ is in standard position. 2. Calculate A x and A y from A x = A cos θ and A y = A sin θ . Use the reference angle if the direction of the component is noted. 3. Check the components to see if each is in the correct direction and has a magnitude that is proper for the reference angle. Adding vectors by components: 1. Add the x -component of the given vectors to obtain R x . 2. Add the y -component of the given vectors to obtain R y . 3. Find the magnitude of the resultant R . Use the equation R = R x 2 + R y 2 . 4. Find the standard-position angle θ for the resultant R . Find the reference angle θ r e f for the resultant R by using the equation θ r e f = tan − 1 | R y | | R x | . Then, use the reference angle and the quadrant of R to determine the standard position angle θ . Calculation: Given that, the angle between the two equal momentum of 15 kg m/ is 72 ° . Sketch the figure of the corresponding vectors as shown below in Figure 1. From Figure 1, it is observed that the angle between the two equal momentum of 15 kg m/ is 72 ° . Let the resultant vector be represented as R . Then, the x component of the resultant vector R is, R x = A x + B x = A cos θ A + B cos θ B = 15 cos 0 ° + 15 cos 72 ° = 15 + 4

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 9.3, Problem 35E

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The magnitude of the resultant vector R if the angle between the two equal momentum of 15 kg m/s is 72°.

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Chapter 9.3, Problem 34E

Chapter 9.3, Problem 36E

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Chapter 9 Solutions

Basic Technical Mathematics

Ch. 9.1 - For the vectors in Example 2, show that R = B + C...Ch. 9.1 - Prob. 2PE Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E

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The magnitude of the resultant vector R if the angle between the two equal momentum of 15 kg m/s is 72 ° .

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The magnitude of the resultant vector R if the angle between the two equal momentum of 15 kg m/s is 72°.

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Chapter 9 Solutions