COSMOS Chapter 4 Solution 33 | Michigan State University - Edubirdie

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

2024
Views:

249

Pages:

2
Author:

Quintin Ware

COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 33. Free-Body Diagram: For both parts (a) and (b) ΣM D = 0: − RP − RC x = 0 Cx = −P ΣFx = 0: B cosθ − P = 0 B = ΣFy = 0: (1) P cosθ (2)  P  Cy −   sin θ + P = 0  cosθ  C y = P ( tanθ − 1) (a) (3) The magnitudes of the forces at B and C are equal: B = C x2 + C y2 2 2 2  P    = ( − P ) +  P ( tan θ − 1)   cosθ  or 1 = 1 + tan 2 θ − 2 tan θ + 1 cos 2 θ ( ) continued COSMOS: Complete Online Solutions Manual Organization System 1 , this gives cos 2 θ 1 1 =1+ − 2 tan θ , or 2 cos θ cos 2 θ 1 tan θ = , so 2 tan 2 θ + 1 = Noting that θ = 26.565° θ = 26.6° (b) Using (2) B= P , or 2/ 5 ∴ B= 5 P 2 26.6° ∴ C= 5 P 2 26.6° and using (1) and (3) C x = − P, C = Cy = − P 2 2 P 5 P  = 2  2 ( −P )2 +  −

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