SSM-Ch09 - Chap. 9 Sec. 9.1 Vector Differential Calculus....

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Chap . 9 Vector Differential Calculus . Grad , Div , Curl Sec . 9 . 1 Vectors in 2 - Space and 3 - Space Problem Set 9 . 1 . Page 370 1 . Components . Length . Unit vector . According to the definition of components you have to calculate the differences of the coordinates of the terminal point Q minus the corresponding coordinates of the initial point P of the vector. Thus, v 1 u 5 u 3 u 2, etc. Since the z -coordinates of P and Q are zero, the vector v is a vector in the xy -plane; it has no component in the z -direction. Sketch the vector, so that you see what it looks like as an arrow in the xyz -coordinate system in space. Calculate the length of the vector by (2) (with a 3 u 0) | v | u v 1 2 U v 2 2 u u 5 u 3 U 2 U u u 2 u 2 U 2 u 4 U 16 u 20 . To obtain the unit vector in the direction of v , multiply v by 1/| v |; this is a scalar multiplication. It gives u u u 1/| v | U v u u 1/ 20 2, u 4 ² u ± 1/ 5 , u 4/5 ² . In the answer on p. A24 we have carried along a component 0, to emphasize that the xy -plane is part of the xyz -space. 15 . Addition and scalar multiplication . It makes no difference whether you first multiply and then add, or whether you first add the given vectors and then multiply their sum by the scalar 5. This problem and Example 2 in the text illustrate formula (6a). 29 . Forces were foremost among the applications that have suggested the concept of a vector, and forming the resultant of forces has motivated vector addition to a large extent. Thus, each of Problems 24-28 amounts to the addition of three vectors. “Equilibrium” means that the resultant of the given forces is the zero vector. Hence in Prob. 29 you must determine v such that v U p U q U u u 0 . Hence v u u u p U q U u U u u ± 2 u 4 U 2, 2 u 4 U 2, 2 U 0 U 7 ² u ± 0, 0, u 9 ² . 37 . Force in a rope . This is typical of many problems in mechanics. Choose an xy -coordinate system with the x -axis pointing horizontally to the right and the y -axis pointing vertically downward. Then the given weight is a force w u ± w 1 , w 2 ² u ± 0, w ² pointing vertically downward. You have to determine the unknown force in the left rope, call it u u ± u 1 , u 2 ² , and the unknown force v u ± v 1 , v 2 ² in the right rope. The three forces are in equilibrium (they have the resultant 0 U because the system does not move. Thus, u U v U w u 0 . (A) You have two choices of giving u a direction, and similarly for v . The choice is up to you. Suppose you choose the vectors to point from the point where the weight is attached upward to the points where the ropes are fixed. Then (see the figure on p. 370) u u ± u 1 , u 2 ² u ± u | u | cos u , u | u | sin u ² v u ± v 1 , v 2 ² u ± | v | cos u , u | v | sin u ² w u ± w 1 , w 2 ² u ± 0, w ² , (B) where the components preceded by a minus sign are those that point in the negative direction of the corresponding coordinate axis (to the left or upward). From (A) and (B) you obtain for the horizontal
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86 Linear Algebra, Vector Calculus
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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.

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SSM-Ch09 - Chap. 9 Sec. 9.1 Vector Differential Calculus....

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