Solutions-to-Exam-1-Review-Exercises

# Solutions-to-Exam-1-Review-Exercises - Review for Exam I In...

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Unformatted text preview: Review for Exam I In the numbering of these solutions, the Chapter is shifted down by one. For example 12.1.47 refers to Problem 47 in Section 1 on Chapter 13. Solutions 12.1.44 The desired vector is: 4 h cos 7 π 3 , sin 7 π 3 i = 4 h 1 2 , √ 3 2 i = h 2 , 2 √ 3 i . 12.1.46 We denote by A , B , C the points in the figure. x y C (7, 8) P ( x , y ) B (5, 4) A (2, 2) Let P = ( x ,y ). We compute the following vectors:--→ PC = h 7- x , 8- y i--→ AB = h 5- 2 , 4- 2 i = h 3 , 2 i The vectors--→ PC and--→ AB are equivalent, hence they have the same components. That is: 7- x = 3 8- y = 2 ⇒ x = 4 , y = 6 ⇒ P = (4 , 6) 12.1.47 We denote the points in the figure by A , B , C and D . x y C (2, 3) A (- 3, 2) D ( a , 1) B (- 1, b ) We compute the following vectors:--→ AB = h- 1- (- 3) ,b- 2 i = h 2 ,b- 2 i--→ DC = h 2- a, 3- 1 i = h 2- a, 2 i Since--→ AB =--→ PC , the two vectors have the same components. That is, 2 = 2- a b- 2 = 2 ⇒ a = b = 4 12.1.54 See the following three figures: y x A w v s w r v y x w v B s w r v y x w v s w r v C 12.1.59 We denote k F 1 k = f 1 and k F 2 k = f 2 . x 210 °- 45 ° 20 F 2 F 3 F 1 Since there is no net force on the object, we have F 1 + F 2 + F 3 = 0 (1) We find the forces: F 1 = f 1 h , 1 i = h ,f 1 i F 2 = f 2 h cos(- 45 ◦ ) , sin(- 45 ◦ ) i = f 2 h √ 2 2 ,- √ 2 2 i = h . 707 f 2 ,- . 707 f 2 i F 3 = 20 h cos 210 ◦ , sin 210 ◦ i = h- 17 . 32 ,- 10 i We substitute the forces in (1): h ,f 1 i + h . 707 f 2 ,- . 707 f 2 i + h- 17 . 32 ,- 10 i = h , i h . 707 f 2- 17 . 32 ,f 1- . 707 f 2- 10 i = h , i Equating corresponding components we obtain . 707 f 2- 17 . 32 = f 1- . 707 f 2- 10 = The first equation gives f 2 = 24 . 5. Substituting in the second equation and solving for f 1 gives f 1- . 707 · 24 . 5- 10 = 0 ⇒ f 1 = 27 . 32 The magnitude of the forces F 1 and F 2 are f 1 = 27 . 32 lb and f 2 = 24 . 5 lb respectively. 12.1.61 We denote by A the point in the figure. x y q 1 q 2 90 ° - q 2 q 2- 90 ° 90 ° - q 1 O P A By the parallelogram law we have r =-→ OA +-→ AP (2) We find the vectors-→ OA and-→ AP : • The vector-→ OA has length L 1 and it makes an angle of 90 ◦- θ 1 with the x-axis. • The vector-→ AP has length L 2 and it makes an angle of- (90 ◦- θ 2 ) = θ 2- 90 ◦ with the x-axis. Hence,-→ OA = L 1 h cos(90 ◦- θ 1 ) , sin(90 ◦- θ 1 ) i = L 1 h sin θ 1 , cos θ 1 i = h L 1 sin θ 1 ,L 1 cos θ 1 i-→ AP = L 2 h cos( θ 2- 90 ◦ ) , sin( θ 2- 90 ◦ ) i = L 2 h sin θ 2 ,- cos θ 2 i = h L 2 sin θ 2 ,- L 2 cos θ 2 i Substituting into (1) we obtain r = h L 1 sin θ 1 ,L 1 cos θ 1 i + h L 2 sin θ 2- L 2 cos θ 2 i r = h L 1 sin θ 1 + L 2 sin θ 2 ,L 1 cos θ 1- L 2 cos θ 2 i Thus, the x component of r is L 1 sin θ 1 + L 2 sin θ 2 and the y component is L 1 cos θ 1- L 2 cos θ 2 ....
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Solutions-to-Exam-1-Review-Exercises - Review for Exam I In...

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