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Three Concurrent forces $P=150N, Q=250N$ and $S=300N$ are acting at $120^\circ$ with each other. Determine their resultant force magnitude and direction with respect to P. What is their equilibrant?

written 7.9 years ago by teamques10 ★ 68k

•   modified 4.5 years ago

engineering mechanics

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written 7.9 years ago by teamques10 ★ 68k

$∑f_x =150-250 \cos 60-300 \cos 60= -125N \\ ∑f_y=250 \sin 60-300 \sin 60= -43.3 N $

Resultant,

$R=\sqrt{(∑f_x )^2 + (∑f_y )^2} \\ R=\sqrt{(-125)^2+ (-43.3)^2} =132.28 N $

$θ= \tan^{-1} (\dfrac {-125}{-43.3}) = 70.89^\circ $

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