Replace the force system by a single force w. r. to point C

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written 7.9 years ago by

teamques10 ★ 68k

• modified 4.5 years ago

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engineering mechanics

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2 Answers

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written 7.9 years ago by

teamques10 ★ 68k

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$∑f_x=20 N \\ ∑f_y=50-60= -10 N \\ \text { Resultant }, R=\sqrt{20^2+(-10)^2 } = 22.36 N \\ θ= \tan^{-1} (\dfrac {10}{20})=26.56^0 \\ M_c=60×2-50×9-20×1= -350 N.m $

Let Resultant act at perpendicular distance of x m from

$∴x = \dfrac {M_c}R = \dfrac {-350}{22.36}=15.65 m $

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109views

written 7.9 years ago by

teamques10 ★ 68k

enter image description here

$∑f_x=20 N \\ ∑f_y=50-60= -10 N \\ \text { Resultant }, R=\sqrt{20^2+(-10)^2 } = 22.36 N \\ θ= \tan^{-1} (\dfrac {10}{20})=26.56^0 \\ M_c=60×2-50×9-20×1= -350 N.m $

Let Resultant act at perpendicular distance of x m from

$∴x = \dfrac {M_c}R = \dfrac {-350}{22.36}=15.65 m $

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