01_InstSolManual_PDF_Part15

01_InstSolManual_PDF_Part15 - y R x 2 5 R 5" R x 2 1 R...

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Figure 1.48 Solve: (a) (b) is the resultant pull. (c) and are shown in Figure 1.48b. and so (d) The vector addition diagram is given in Figure 1.48c. Careful measurement gives an value that agrees with our results using components. 1.49. Set Up: Use coordinates for which is east and is north. Each of the professor’s displacement vectors make an angle of or with one of these axes. The components of his total displacement can thus be calculated directly from and Solve: (a) (b) From the scaled sketch in Figure 1.49, the graphical sum agrees with the calculated values. Figure 1.49 Reflect: Themagnitudeofhisresultantdisplacementisverydifferentfromthedistancehetraveled,whichis159.50km. B S C S A S R S 5.0 km 70 ° f 5 180° 2 20.2° 5 69.8° west of north tan 2 1 31 1 1.75 2 / 1 2 4.75 24 5 2 20.2°; u 5 tan 2 1 1 R y
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Unformatted text preview: y / R x 2 5 R 5 " R x 2 1 R y 2 5 5.06 km; 3.25 km 1 1 1 2 1.50 km 2 5 1.75 km 5 1.75 km north; R y 5 A y 1 B y 1 C y 5 R x 5 A x 1 B x 1 C x 5 1 1 2 4.75 km 2 1 5 2 4.75 km 5 4.75 km west; R y 5 A y 1 B y 1 C y . R x 5 A x 1 B x 1 C x 180° 0° 1 y 1 x R S u 5 75° tan u 5 87 N 23 N R 5 " R x 2 1 R y 2 5 90 N R S R y R x , R y 5 A y 1 B y 1 C y 5 50 N 1 69 N 1 1 2 32 N 2 5 1 87 N R x 5 A x 1 B x 1 C x 5 87 N 1 1 2 40 N 2 1 1 2 24 N 2 5 1 23 N R S 5 A S 1 B S 1 C S C y 5 C sin 233° 5 2 32 N. C x 5 C cos 233° 5 2 24 N; B y 5 B sin 120° 5 69 N; B x 5 B cos 120° 5 2 40 N; A y 5 A sin 30° 5 50 N; A x 5 A cos 30° 5 87 N; ( c ) y x 30 ° 30 ° 53 ° R C B A O Models, Measurements, and Vectors 1-15...
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