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

2019
Views:

152

Pages:

1
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 85. Free-Body Diagram: (a) Using the law of cosines on triangle ABC: 2 2 R 2 = ( 2R ) + ( 2R ) − 2 ( 2R )( 2R ) cosθ 1 = 8 − 8cosθ 7 8 θ = 28.955° Also, cosθ = 2R cos (θ + α ) = R cos α 2R ( cosθ cos α − sin θ sin α ) = R cos α tan α = 2 cosθ − 1 2cos 28.955° − 1 = 2 sin θ 2 sin 28.955° or α = 37.8° α = 37.761° (b) From the free-body diagram: 2R R = sin φ sin θ sin φ = 2sin θ = 2 sin 28.955° φ = 75.522° Now using the law of sines on the force triangle: NA NB W = = sin [90° − (φ − α )] sin 90° − (θ + α )  sin (θ + α ) + (φ − α )  NA NB mg = = cos(φ − α ) cos (θ + α ) sin (θ + φ ) NA NB mg = = cos 37.762° cos 66.716° sin104.478° Solving for N A and N B : N A = 0.816 mg N B = 0.408 mg 66.7° 37.8°

COSMOS Chapter 4 Solution 85

of 1

Description

Free up your schedule!

Take 5 seconds to unlock