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17 Pages

### CH-1

Course: PHYSICS 101, Spring 2011
School: Qatar University
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Word Count: 1976

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Sabah Dr. AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 1.1 Coordinate Systems A coordinate system used to specify locations in space consists of: A fixed reference point O , called the origin. A set of specified axes with appropriate scales and labels on the axes. Instructions that tell us how to label a point in space relative to the origin and axes. A common coordinate system is called the...

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Sabah Dr. AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 1.1 Coordinate Systems A coordinate system used to specify locations in space consists of: A fixed reference point O , called the origin. A set of specified axes with appropriate scales and labels on the axes. Instructions that tell us how to label a point in space relative to the origin and axes. A common coordinate system is called the Cartesian Coordinate system, Sometimes called the Rectangular Coordinate system, This system labeled with the coordinate (X,Y). The Cartesian Coordinate can be expressed through equations relating the coordinates (X,Y) to the Plane polar coordinates system (r, ). As follows: r x2 y2 1 1.2 Some properties of vectors A scalar quantity has only magnitude and no direction. A vector quantity has both magnitude and direction. Addition & Subtraction: Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 A B B A A ( B C ) ( A B) C A B A ( B) A B B A Multiplication of a vector by a scalar: mA nB nB mA n( A B) nA nB k ( A B) kA kB Where n , m & k are scalar quantities. 1.3 Examples Example (1): Find the sum of the two vectors A and B lying in x - y plane and given by: A 3i 5 j & B 2i 3 j Solution: A B (3i 5 j ) (2i 3 j ) 3i 2i 5 j 3 j 5i 2 j 2 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Example (2): Find the sum of three vectors A , B and C lying in x - y plane and given by: A 2i 2 j , B 20i 6 j , C i j Solution: A B C ( 2i 2 j ) ( 20i 6 j ) (i j ) 2i 20i i 2 j 6 j j 23i 7 j Example (3): Find the following: a ) A 3B 5C , b) 2 A 0.5 B 0C where A 3i j k , B 2i 4 j 3k , C i j 2k Solution: a) A 3B 5C (3i j k ) 3(2i 4 j 3k ) 5(i j 2k ) 3i j k 6i 12 j 9k 5i 5 j 10k (3i 6i 5i ) j 12 j 5 j k 9k 10k 14i 8 j 18k b) 2 A 0.5B 0C 2(3i j k ) 0.5(2i 4 j 3k ) 0(i j 2k ) 6i 2 j 2k i 2 j 1.5k 0 6i i 2 j 2 j 2k 1.5k 7i 4 j 0.5k 3 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Example (4): Find the following: 5 R S 0.8 K where R i j , S 2i 3 j , K j k Example (5): Find the following: 1 3M 1 N O 2 3 M 3i k , N 2 j , O 3i 2k 4 1.4 Components of a vector and unit vectors Any vector A lying in the x-y Plane and making any arbitrary angle with the positive x-axis, we can represent this vector by the following expression: A Ax A y or A Ax i Ay j where A x Ax i A cos i and A y Ay j A sin j The magnitude of vector A is : A 2 2 Ax Ay and the direction of vector A is : Ay tan Ax tan 1 ( Ay Ax ) At this expression A Ax i Ay j we call i and j unit vectors , where i is a unit vector parallel to the x-axis and j is a unit vector parallel to the y-axis. 5 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 1.5 Vector Analysis A Ax A y where A x Ax i A cos i and A y Ay j A sin j A x A sin (i ) A y A cos ( j ) A x A sin (i ) A y A cos ( j ) A x A cos (i ) A y A sin ( j ) And So on 6 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 1.6 Analysis of Forces at Equilibrium Foreces at Equilibrium : x axis : F x 0 ....... .......... ... .......... . 0 y axis : F y 0 ......... .......... . .......... .... 0 7 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Example F3 F3 cos F2 sin F2 F1 F3 sin F2 cos F4 Solution: F3 cos x axis : F x F2 sin 0 F1 F2 cos F3 sin 0 y axis : F y 0 F1 F3 sin F2 cos F4 F2 sin F3 cos F4 0 8 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 1.7 Examples Example (1): Find the magnitude of force F1 and F4 .The system is at equilibrium. F2 150 N F3 180N 30 F3 sin 50 F2 cos 30 50 F1 F3 sin 0 F1 F2 sin 30 F3 cos 50 0 F1 F3 cos 50 F2 sin 30 180 cos 50 150 sin 30 115.7 75 40.7 N F y 0 F2 cos 30 F3 sin 50 F4 0 F4 F2 cos 30 F3 sin 50 150 cos 30 180 sin 50 129.9 137.9 267.8 N Therefore, F1 = 40.7 i F2 sin 30 F4 Solution: x F1 F3 cos 50 F4 F Graph: N F4 = 267.8 j N 9 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Example (2): Find the tension in each for cord this system. The system is at equilibrium. 40 Graph: 70 T2 T1 T2 T1 40 70 T1 sin 70 T2 sin 40 T1 cos 70 m 5kg T2 cos 40 mg Solution: F x 0 T1 cos 70 T2 cos 40 0 T1 cos 70 T2 cos 40 T1 T2 F y cos 40 cos 70 0 T1 sin 70 T2 sin 40 mg 0 cos 40 sin 70 T2 sin 40 mg cos 70 T2 ( 2.12) T2 (0.64) 5 X 9.8 49 T2 T2 ( 2.76) 49 T2 T1 17.8 49 17.8 N 2.76 cos 40 40.3 N cos 70 Therefore, Magnitude of T1 is T1= 40.3 N Magnitude of T2 is T2= 17.8 N 10 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Example (3): Find the magnitude of the force F1 and F2 where F1=F2 . The system is at equilibrium. F2 Graph: F1 20 F1 sin 20 20 F2 sin 20 200N F1 cos 20 F2 cos 20 200 Solution: F x 0 F1 cos 20 F2 cos 20 0 F cos 20 F cos 20 0 F y 0 F1 sin 20 F2 sin 20 200 0 F sin 20 F sin 20 200 0 F (sin 20)(1 1) 200 2 F sin 20 200 200 200 F 294.1 N 2 sin 20 2 X 0.34 F1 F2 F 294.1 N 11 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Example (4): Find magnitude of F2 and the weight (w) .The system is at equilibrium. F3 50N 35 Solution: F2 ? F3 sin 25 F2 cos 35 25 10 F1 cos10 F1 30 N w F2 sin 35 F3 cos 25 w F1 sin10 Solution: F x 0 F1 cos 10 F2 sin 35 F3 cos 25 0 F2 sin 35 F3 cos 25 F1 cos 10 F3 cos 25 F1 cos 10 45.3 29.5 27.72 N sin 35 0.57 F2 F y 0 F2 cos 35 F3 sin 25 F1 sin 10 w 0 w F2 cos 35 F3 sin 25 F1 sin 10 (27.72) cos 35 50 sin 25 30 sin 10 22.7 21.1 5.2 38.6 N 12 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Example (5): A man walking 25km southeast from his camp, as shown to you. Determine the components of the mans displacement. north west 30 Graph: east r cos 30 25km r sin 30 south Solution: x component : r cos 30i (25km) cos 30 (i ) 25 (0.87) i 21.75i Km east y component : r sin 30( j ) (25km) sin 30 ( j ) 25 (0.5) ( j ) 12.5 j Km south 13 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Example (6): On the first day, a girl walking 20m north from her house. On the second day, she walks 40m in a direction 60 northwest. Determine the components of her displacement for each day. north day 2 r2 40m day1 20m r1 60 east west south Solution: for day 1 : X component : 0 Y component : r1 ( j ) 20m j day1 20m r1 for day 2 : X component : r2 cos 60(i) 40 cos 60(i) 40(0.5)(i) 20 i m Y component : r2 sin 60( j ) 40 sin 60( j ) 40(0.87)( j ) 34.8 j m 14 r2 40m 60 r2 cos 60 r2 sin 60 Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Example (7): An Airplane starts from an airport takes the route as shown to you. It flies to city A located 175 km in a direction 30 north of east. Next, it files to city B located 150 km 20 west of north . Finally, it flies 190 km west to city C. Find the location of the plane regarding to the starting point O. north c Graph: b b 20 20 30 a a east 30 west c south Solution: x component Rx Rx a cos 30(i ) b sin 20(i ) c (i ) 175 cos 30 i 150 sin 20 i 190 i 151.6 i 51.3 i 190 i 89.7 i a sin 30 j b cos 20 j c y component R y b sin 20 i R y a sin 30( j ) b cos 20( j ) 175 sin 30( j ) 150 cos 20( j ) 87.5 ( j ) 140.95 ( j ) 228.5 j The location of the plane : R R x R y 89.7 i 228.5 j 15 a cos 30 i Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 Example (8): Find the resultant vector, its magnitude and its direction. C 5m Graph: B 40 B A 4m 4m 5m 3 sin 20 3m 40 c 20 3 cos 20 20 A B 5 sin 40 C 5 cos 40 3m Solution: Solution: The resultant vector : x component : R x 5 cos 40 i 3 cos 20 i 5(0.77) i 3(0.94) i 3.38 i 2.28 i 1.01 i y component : R y 4 j 5 sin 40 j 3 sin 20 j 4 j 5(0.64) j 3(0.34) j 3.21 j 1.03 j 8.24 j The magnitude : R 2 2 Rx R y (1.01) 2 (8.24) 2 The direction : R 8.24 tan 1 ( y ) tan 1 ( ) 83.01 Rx 1.01 16 68.92 8.3 m Whe .. F n Con necti ng 2 or mor e capa citor in serie s con necti on , We have the follo wing : 1 2 1 Z C3 T h e E q Dr. Sabah AL-Naimi Qatar University Physics Program FALL 2009 PHYS191 NOTE : The resultant vector of R is : R Rx i R y j The magnitude of R : R 2 2 Rx R y The Direction of R : Ry Ry tan tan 1 ( ) Rx Rx H.W. # 1 P. 71 76 3, 7, 10, 15, 18, 27, 29, 33, 35, 49, 59 17
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