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

2020
Views:

490

Pages:

2
Author:

hacedoras1kxy

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 43. Free-Body Diagram: Geometry: Equation of the slot: y = x2 4  2x   dy  = 1.20000   = slope of slot at C =  4  dx   ( x = 2.4 in.)  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 P = 1 lb: ΣM E = 0: P ( yE ) − ( Q sin β )( yE − yD ) − ( Q cos β )( 2.4 in.) = 0 (1 lb )( 3.44 in.) − ( Q sin12.6804° )(1.24 in.) − ( Q cos12.6804° )( 2.4 in.) = 0 Q = 1.31616 lb ΣFx = 0: P − N C cosθ − Q sin β = 0 1 lb − NC cos 39.806° − (1.31616 lb ) sin12.6804° = 0 NC = 0.92563 lb ΣFy = 0: N B + NC sin θ − Q cos β = 0 N B + ( 0.92563 lb ) sin 39.806° − (1.31616 lb ) cos12.6804° = 0 N B = 0.69148 lb (a) N = 0.691 lb , N = 0.926 lb B C 39.8° Q = 1.316 lb 77.3° (b)