Ch. 3 Solutions

Ch. 3 Solutions - CHAPTER 3 PROBLEM SOLUTIONS Problem 7....

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CHAPTER 3 PROBLEM SOLUTIONS Problem 7. Vectors A and B in Fig. 3-22 have the same magnitude, A . Find the magnitude and direction of (a) A ! B and (b) A + B . Solution The vectors A and B in Fig. 3-22 form two sides of a parallelogram, in which A ! B and A + B are the diagonals, as shown. Since the magnitudes of A and B are equal, the parallelogram is a rhombus, and the diagonals are perpendicular (the converse of this is also true; see Problem 60). Then A + B is along the perpendicular bisector of the base A ! B of an isosceles triangle, and vice versa. Using the given angles, we find the magnitudes (a) A ! B = 2 A sin20 ° = 0.684 A , and (b) A + B = 2 A cos20 ° = 1.88 A . In Fig. 3-22, (a) A ! B is up, and (b) A + B is to the right, but the directions could be specified relative to A , B , or some other coordinate system. (This problem can also be readily solved with components and unit vectors. Figure 3-22 suggests a coordinate system with x -axis to the right and y -axis up, as shown. Then A = A ( î cos 20 ° + ˆ j sin 20 ° ) and B = A ( î cos 20 ° ! ˆ j sin 20 ° ) , from which A ± B are easily obtained.) Problem 8. Vector A has magnitude 1.0 m and points at 35º clock-wise from the x -axis. Vector B has magnitude 1.8 m. What angle should B make with the x -axis in order that A + B be purely vertical? Solution The vectors A , B , and A + B form a triangle, but given only that B = 1.8 A and A + B are perpendicular to the x -axis, two are possible. In one, the angle opposite side B is ! = 125 ° , in the other = 55 ° , as sketched. For either, the law of sines gives = sin " 1 A sin # B $ % & ( ) = sin " 1 sin125 ° 1.8 $ % & ( ) = sin " 1 sin55 ° 1.8 $ % & ( ) = 27.1 ° . Therefore, B makes an angle of ± 62.9 ° with the negative x -axis or ± 117 ° with the x -axis (look at suitable right triangles in the sketch). (This result can also be obtained from components: since A + B is vertical, A x + B x = 0, or B x = B cos x = " A x = " A cos35 ° , so x = cos " 1 ( " A cos 35 ° = B ) = cos " 1 ( " cos 35 ° = 1 . 8 ) = ± 117 ° , where B y could be positive or negative.)
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