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

2024
Views:

84

Pages:

2
Author:

lankstoetp8

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 44. Free-Body Diagram: Geometry: Equation of the slot: y = x2 4  2x   dy  = 1.20000   = slope of slot at C =  4    ( x = 2.4 in.)  dx C It follows for the angles that: α = tan −1 (1.2 ) = 50.194° θ = 90° − α = 90° − 50.194° = 39.806°  4.8 − 2.64   = 12.6804° 9.6   β = tan −1  Coordinates for C, D, and E: ( 2.4 )2 xC = 2.4 in., yC = xD = 2.4 in., yD = 1.84 in. + (1.6 in.) tan β 4 = 2.44 in. = 1.84 in. + (1.6 in.) tan12.6804° = 2.20000 in. xE = 0, yE = yC + ( 2.4 in.) tan θ = 1.44 in. + ( 2.4 in.) tan 39.806° = 3.4400 in. continued COSMOS: Complete Online Solutions Manual Organization System With Q = 2 lb: (a) ΣM E = 0: P ( yE ) − ( Q sin β )( yE − yD ) − ( Q cos β )( 2.4 in.) = 0 P ( 3.44 in.) − ( 2 lb ) sin12.6804° (1.24 in.) − ( 2 lb ) cos12.6804° ( 2.4 in.) = 0 P = 1.51957 lb (b) ΣFx = 0: or P = 1.520 lb P − NC cosθ − Q sin β = 0 1.51957 lb − NC cos 39.806° − ( 2 lb ) sin12.6804° = 0 NC = 1.40656 lb ΣFy = 0: or NC = 1.407 lb 39.8° N B + NC sin θ − Q cos β = 0 N B + (1.40656 lb ) sin 39.806° − ( 2 lb ) cos12.6804° = 0 N B = 1.05075 lb or N B = 1.051 lb