By Thomas R. Kane

ISBN-10: 1483231283

ISBN-13: 9781483231280

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And Problem: Fig. 3b shows three coplanar vectors, a, b and c. Evaluate c-(b + 2a). FIG. 3) and the relationships 111'ill = n i · 112 = = 113*113 = 1 112*113 = ri3*ni = 0 112*112 Problem: Determine the scalar product of the force F and the unit vector n shown in Fig. 4. FIG. 4 Solution: Set up unit vectors m, n2, n3 as shown in Fig. 4,. 8) + (-8)(0) + (6) ( - 0 . 6 ) = - 3 . 4) dibi + Gt2&2 + \ The arc cosine is a multi-valued function. 13) never exceeds 180 degrees. Only one of the values of this function is, therefore, appropriate in any given case.

6) and the relationships ■h X ni = n2 X Ü2 = n3 X n3 = 0 ni X Π2 = n3, n2 X n3 = ni, n3 X ni = n2 Problem: Referring to Fig. 4, resolve F X n into its ni, n2, n3 components. 6n3 Hence, F X n = [ ( - 8 ) ( - 0 . 8. 6 = ( - 8 ) ( - 0 . 1 The parentheses in the expression a · (b X c) are un necessary because (a · b) X c is meaningless, (a · b) being a scalar. Hence one writes a b X c. , [a, b, c] = a b X c = a X b e Proof: Resolve each of the vectors a, b, c into m, n2, n3 com ponents (in, n2, n3 being a right-handed set of mutually perpen dicular unit vectors), and carry out the indicated operations.

7) B(2) P6(8) FIG. 7a FIG. 7b CENTROIDS AND MASS CENTERS) SECTION 2 . 5 53 sum of the strengths of these centroids; but this is the vector which locates the centroid of the set of centroids. Problem: The points Pi, . . , Ps shown in Fig. 7a have the strengths indicated in parentheses. Locate the centroid of this set of points. Solution: From symmetry considerations, the centroid of the subset Pi, . . , PA is known to be at point A; that of subset P 6 , . . , Ps, at B. The points A and B, with strengths 16 - 13 + 16 - 13 = 6 and - 7 + 8 - 7 + 8 = 2, form a set of centroids (see Fig.

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