Engineering Mechanics CBSGS(Old) MU May 2017

Solution :

Diagram

Here the magnitude of resultant is algebraic sum of all the parallel forces,

$\therefore R_{x}=-80-50-100+150$ $\therefore R_{×}=-80N$

$\therefore R_{x}=80 N\lt-$

$\therefore R_{y}=0$

$\therefore R=\sqrt{R x^{2}+R y^{2}}=\sqrt{80^{2}}=80 N$

Now position of resultant is finding out by, Applying Varignon's theorem at point 0

$$ M_{0}^{R}=R \times d $$

(Treating anticlockwise moment as positive) $$ \therefore R \times d =r \times 100 \sin 60^{\circ}-r \times 50 \sin45^{\circ} +r \times 80 \sin30^{\circ}+r \times 150 \sin30^{\circ} \\ $$

$$ \therefore 80 \times d =r \times 166.25 $$

$$ \therefore 80 \times d =0.5 \times 166.25 $$

$$ \therefore d =1.0391 m$$

d=1.0391 m from the point O