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Three concurrent forces P=150N, Q=250N and S=300N are acting at 120° with each other. Determine their resultant force magnitude and direction with respect to P. What is their equilibrant?
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written 3.0 years ago by
teamques10
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Answer: Given,
To find the resultant force
$\sum F_x= 150-250\cos{60}-300\sin{60} \\=-125N \\\text{and} \sum{F_y}=250\sin{60}-300\cos{30} = 43.30N $
$\therefore \text{ Resultant Force } R = \sqrt{ \left( \sum{F_x} \right)^2+\left( \sum{F_y} \right)^2} \\= \sqrt{ \left(-125\right)^2 +\left(43.3\right)^2} = 132.29N$
and Direction
$\theta = \tan^{-1}\left( \dfrac{\sum{F_y}}{\sum{F_x}} \right) \\= \tan^{-1}\left( \dfrac{-43.3}{-125} \right) = 198.98^0$
ie. resultant …
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