HW2-Solution

# HW2-Solution - HW2 Solution Terry Cooke Cal Poly San Luis...

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Unformatted text preview: HW2 - Solution Terry Cooke: Cal Poly San Luis Obispo Chapter 2, Solution 9 Using the triangle rule and the law of sines: (a) (b) 25 1600 N sin 25° P sin 75 P 3660 N 75 180 180 80 R sin 80 25 75 1600 N sin 25° R 3730 N Chapter 2, Solution 19 Using the force triangle and the laws of cosines and sines We have R2 180 120 (40 20 ) Then (10 kN)2 (15 kN)2 2(10 kN)(15 kN) cos120 475 kN 2 21.794 kN 21.794 kN sin120 15 kN sin120 21.794 kN 0.59605 36.588 50 86.588 R 21.8 kN R and 15 kN sin sin Hence: 86.6° Chapter 2, Solution 30 (a) P Py cos 55 350 lb cos 55 610.2 lb P 610 lb (b) Px P sin 55 (610.2 lb)sin 55 499.8 lb Px 500 lb Chapter 2, Solution 42 Select the x axis to be along a a Then Rx Fx (60 lb) (80 lb) cos (120 lb)sin (1) and Ry Fy (80 lb)sin (120 lb) cos (2) (a) Set Ry 0 in Eq. (2). (80 lb)sin (120 lb) cos 0 Dividing each term by cos gives: (80 lb) tan tan 120 lb 120 lb 80 lb 56.310 56.3 (b) Substituting for Rx in Eq. (1) gives: 60 lb (80 lb) cos 56.31 (120 lb)sin 56.31 204.22 lb Rx 204 lb ...
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