HW-1-ANS - HW 3 Solution Q1 ^ j ^ j ^ i i^ 2 ^ ^ ^ i(i j 2 ^ 2(i ^ ^ j j 2 2 2 ^ ^ ^ j ^ j F 3Ni 3Nj 3N i ^ 3N i ^ 2 2 ^ 3 2 Ni So the each

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Unformatted text preview: HW 3 Solution Q1. ^ j ^' j ^ i' i^ 2 ^ ^ ^ i (i ' j ') 2 ^ 2 (i ' ^ ') ^ j j 2 2 2 ^ ^ ^ j ^ j F 3Ni 3Nj 3N i ' ^ ' 3N i ' ^ ' 2 2 ^ 3 2 Ni ' So, the each representation should be: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) Correct Correct Correct Correct Incorrect Incorrect Incorrect Incorrect Correct Incorrect Q2. F F1 F2 F3 ^ ^ ^ ^ ^ ^ 2 Ni 3Nj 10 Nj 3Ni 1Nj 5 Nk ^ ^ ^ 5 Ni 6 Nj 5 Nk Q3. Fnet F1 F2 F2 Fnet F1 ^ ^ ^ 2 Ni (10 Ni 10 Nj ) ^ ^ 8 Ni 10 Nj Q4. 2 2 1. v (30) (40) 50 2. Unit Vector n 1 3^ 4 v i ^ j 5 5 v 3^ 4 j 3. v v n 50( i ^) 5 5 Q.5 1. A 8 B 4 90 2. rA (4)2 (8)2 8.944 rB (90)2 (6)2 90.200 3. ^ rA rB 90i 6 ^ 4i 8i j ^ ^ ^ 86i 2 ^ 86.023 j Q6. 1. ^ 1 (0.80)2 (0.60)2 1.0 ^ 2 (0.5)2 (0.866)2 1.0 2. ^ ^ ^ 1 2 1.30i 1.466 ^ j ^ ^ 1 2 (1.30)2 (1.466)2 1.959 ^ ^ So, sum of 1 and 2 is not a unit vector ^ u 1 ^ ^ 1 2 ^ ^ ^ (1 2 ) 0.644i 0.748 ^ j Q 7) The given equations are : Solving the above equations: =0 Substituting in (1) Solving (5) and (3) Substituting in (4) The equations can be solved. Q 8) Let be the vector perpendicular to = = and , then The magnitude of is , The unit vector perpendicular to and is: Q 9) Q 10) Q 11) y x ...
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This note was uploaded on 02/12/2012 for the course AMS 361 taught by Professor Staff during the Spring '08 term at SUNY Stony Brook.

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