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HW6-Solution

HW6-Solution - HW6 Solution Terry Cooke Cal Poly San Luis...

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Unformatted text preview: HW6 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 3, Solution 55 M AD AD (rB/A TBH ) Where AD rB/A 1 (4i 3k ) 5 (0.5 m)i (0.375) 2 1.125 m (0.75) 2 ( 0.75) 2 and d BH Then TBH 450 N (0.375i 0.75 j 0.75k ) 1.125 (150 N)i (300 N) j (300 N)k Finally MAD 4 0 1 0.5 0 5 150 300 3 0 300 1 [( 3)(0.5)(300)] 5 or M AD 90.0 N m Chapter 3, Solution 73 (a) M (35 lb)(7 in.) (25 lb)(9 in.) 245 lb in. 225 lb in. M 470 lb in. (b) With only one string, pegs A and D, or B and C should be used. We have tan 6 8 36.9 90 53.1 Direction of forces: With pegs A and D: With pegs B and C: (c) The distance between the centers of the two pegs is 82 62 10 in. 53.1 53.1 Therefore, the perpendicular distance d between the forces is d 10 in. 2 11 in. 1 in. 2 F 42.7 lb We must have M Fd 470 lb in. F (11 in.) Chapter 3, Solution 78 From the solution to Problem. 3.77 16 lb force: 40 lb force: P 20 lb M1 (480 lb in.)k M2 M3 8 5[(10 lb in.)i (30 lb in.) j (15 lb in.)k ] rC P (30 in.)i (20 lb)k (600 lb in.) j M M1 M2 M3 (480)k 8 5 (10i 30 j 15k ) 600 j (178.885 lb in)i (1136.66 lb in.) j (211.67 lb in.)k M (178.885) 2 1169.96 lb in axis (113.66) 2 (211.67) 2 M 1170 lb in. cos cos cos x y z M 0.152898i 0.97154 j 0.180921k M 0.152898 0.97154 0.180921 x 81.2 y 13.70 z 100.4 ...
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