three-forces-having-equal-magnitude-s-of-10N-20N-and-30N-make-angles-of-30-120-and-210-respectively-with-the-positive-direction-of-the-x-axis-By-scale-drawing-find-the-magnitude-and-the-direction

Question Number 62587 by Jmasanja last updated on 23/Jun/19

t h r e e f o r c e s h a v i n g e q u a l m a g n i t u d e

s o f 10 N, 20 N a n d 30 N m a k e a n g l e s

o f 30 °, 120 ° a n d 210 ° r e s p e c t i v e l y w i t h

t h e p o s i t i v e d i r e c t i o n o f t h e x a x i s .

B y s c a l e d r a w i n g f i n d t h e m a g n i t u d e

a n d t h e d i r e c t i o n o f t h e r e s u l t a n t

f o r c e

Commented by MJS last updated on 23/Jun/19

if they have equal magnitudes how can these forces

be 10, 20, 30 ?

Answered by MJS last updated on 23/Jun/19

F_{1} = f_{1} (\begin{matrix} \cos 30 ° \\ \sin 30 ° \end{matrix}) = f_{1} (\begin{matrix} \frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{matrix})

F_{2} = f_{2} (\begin{matrix} \cos 120 ° \\ \sin 120 ° \end{matrix}) = f_{2} (\begin{matrix} - \frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{matrix})

F_{3} = f_{3} (\begin{matrix} \cos 210 ° \\ \sin 210 ° \end{matrix}) = f_{3} (\begin{matrix} - \frac{\sqrt{3}}{2} \\ - \frac{1}{2} \end{matrix})

F_{1} + F_{2} + F_{3} = \frac{1}{2} (\begin{matrix} \sqrt{3} f_{1} - f_{2} - \sqrt{3} f_{3} \\ f_{1} + \sqrt{3} f_{2} - f_{3} \end{matrix}) =

[if f_{1} = f_{2} = f_{3} = f]

= \frac{f}{2} (\begin{matrix} 1 \\ \sqrt{3} \end{matrix})

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