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

2021
Views:

210

Pages:

1
Author:

Campbell Wagner

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 87. Free-Body Diagram: Note that the wheel is a two-force body and therefore the force at C is directed along CA and perpendicular to the incline. The wheelbarrow is a three-force body. Let D be the intersection of the lines of action of the three forces acting on the wheelbarrow. Then, using the triangle DEG DE = EG tan 72° = ( 8 in.) tan 72° = 24.6215 in. DF = DE − EF = 24.6215 in. − 9 in. = 15.6215 in. Using triangle DFB: φ = tan −1 FB  40  = tan −1   = 68.667° DF  15.6215  From the force triangle: α = φ − 18° = 68.667° − 18° = 50.667° β = 180° − 50.667° − 18° = 111.333° Using the law of sines: B C 120 lb = = sin18° sin 50.667° sin111.333° B = 39.809 lb, C = 9.644 lb (a) (b) Noting that the force on each handle is B/2: 1 B = 19.90 lb 2 39.3° C = 99.6 lb 72.0° Reaction at C: