University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2020
Views:

348

Pages:

2
Author:

agatoary6zt

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 86. Free-Body Diagram: (a) Note that the rod is a three-force body. Using the law of cosines on triangle ABC: 2 R 2 = ( 2R ) + L2 − 2 ( 2 R ) L cosθ cosθ = 3R 2 + L2 4RL (1) Also, L cos 45° = 2 R cos(θ + 45°) 2 L cos 45° = 2R(cosθ cos 45° − sin θ sin 45°) 2 L = cosθ − sin θ 4R Using (1) and that sin θ = 1 − cos 2 θ sin θ = ( 4 RL )2 − ( 3R 2 + L2 ) 4RL 2 , this gives continued COSMOS: Complete Online Solutions Manual Organization System L 3R 2 + L2 = − 4R 4RL ( ( 4 RL )2 − ( 3R 2 + L2 ) 2 4RL ) )2 L2 − 3R 2 + L2 = − 3 2 + L2 ) 2 Squaring both sides and simplifying: ( ) 9R 4 = 16R 2 L2 − 9R 4 + 6R 2 L2 + L4 , or L4 − 10R 2 L2 + 18R 4 = 0 Solving for L2 : L2 = ( ) ± 7 R 2 and taking the largest root ( ) L2 = 5 + 7 R 2 (b) or L = 2.77 R Using the value of L obtained in (a) and (1) cosθ = ( ) 3R 2 + 5 + 7 R 2 ( 4R 5 + 7 ) 1/2 R θ = 15.7380° Now using the law of sines on triangle ABC in the free body diagram: 2R R = sin φ sin θ sin φ = 2sin θ = 2 sin15.7380° φ = 32.852° Now using the law of sines on the force triangle: NA NB mg = = cos (135° − 32.852° ) sin ( 45° − 15.7380° ) sin (15.7380° + 32.852° ) Solving for N A and N B : N A = 1.303 mg N B = 0.652 mg 60.7° 12.15°