01_InstSolManual_PDF_Part12

01_InstSolManual_PDF_Part12 - 5 9.6 m. A x 5 1 12.0 m 2 cos...

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1.43. Set Up: In each case, use a sketch (Figure 1.43) showing the components and the resultant in order to determine the quadrant in which the resultant vector lies. The component vectors add to give the resultant. Figure 1.43 Solve: (a) and (b) and (My calculator gives and (c) and (My calculator gives and (d) and (My calculator gives and 1.44. Set Up: For each vector, use the relations and Solve: For vector For vector For vector 1.45. Set Up: For parts (a) and (c), apply the appropriate signs to the relations and For (b) and (d), find the magnitude as and the direction as u5 tan 2 1 1 R y / R x 2 . R 5 " R x 2 1 R y 2 R y 5 A y 1 B y . R x 5 A x 1 B x C y 5 1 6.0 m 2 sin 1 240° 2 52 5.2 m. 1 6.0 m 2 cos 1 240° 2 52 3.0 m; C x 5 C S : B y 5 1 15.0 m 2 sin 1 320° 2 52 9.6 m. B x 5 1 15.0 m 2 cos 1 320° 2 5 11.5 m; B S : A y 5 1 12.0 m 2 sin 1 90° 2 37° 2
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Unformatted text preview: 5 9.6 m. A x 5 1 12.0 m 2 cos 1 90° 2 37° 2 5 7.2 m; A S : R y 5 R sin u . R x 5 R cos u u 5 f 1 360°.) f 5 2 37° u 5 323° tan u 5 2 54.7 N 71.3 N A 5 "1 71.3 N 2 2 1 1 2 54.7 N 2 2 5 89.9 N; u 5 f 1 180°.) f 5 2 53° u 5 127° tan u 5 2000 km 2 1500 km A 5 "1 2 1500 km 2 2 1 1 2000 km 2 2 5 2500 km; u 5 f 1 180°.) f 5 52° u 5 232° tan u 5 2 31 m / s 2 24 m / s A 5 "1 2 24 m / s 2 2 1 1 2 31 m / s 2 2 5 39 m / s; u 5 37°. tan u 5 A y A x 5 6.0 lb 8.0 lb A 5 " A x 2 1 A y 2 5 "1 8.0 lb 2 2 1 1 6.0 lb 2 2 5 10.0 lb; u u u u ( a ) ( c ) ( b ) ( d ) A y A x y x A O O O A y A x y x A A y A x y x A A y A x x A y O A S 1-12 Chapter 1...
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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