COSMOS Chapter 3 Solution 113 | Michigan State University - Edubirdie

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Michigan State University
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ERG 160 | Success in Science, Technology, Engineering, and Mathematics
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2024
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2
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COSMOS: Complete Online Solutions Manual Organization System Chapter 3, Solution 113. ΣFx = 0 Based on P cos α − 15 N = 0 ∴ P cos α = 15 N (1) ΣFy = 0 and P sin α − 10 N = 0 ∴ P sin α = 10 N (2) Dividing Equation (2) by Equation (1), tan α = 10 15 ∴ α = 33.690° Substituting into Equation (1), P= 15 N = 18.0278 N cos33.690° or P = 18.03 N (a) Based on ΣM B = 0 33.7° − (18.0278 N ) cos 33.690° ( d + 0.150 m ) sin 50° − (18.0278 N ) sin 33.690° ( d + 0.150 m ) cos 50° + (15 N ) ( 0.150 m ) sin 50°  − (10 N )( 0.150 m ) + 6.25 N ⋅ m = 0 −17.9186d = −3.7858 ∴ d = 0.21128 m or d = 211 mm . COSMOS: Complete Online Solutions Manual Organization System (b) Based on ΣM D = 0 − (18.0278 N ) cos 33.690° ( d + 0.150 m ) sin 50° − (18.0278 N ) sin 33.690°  ( d + 0.150 m ) cos50° + 0.150 m  + (15 N ) ( 0.150 m ) sin 50°  + 6.25 N ⋅ m = 0 −17.9186d = −3.7858 ∴ d = 0.21128 m or d = 211 mm This result is expected, since R = 0 and M RB = 0 for d = 211 mm implies that R = 0 and M = 0 at any other point for the value of d found in part a. .

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