Question

Replace the loading system shown by an equivalent resultant force and couple moment acting at point A.   

Answer 1

Total force in \(y-\) direction

$$ F_{y}=50+50 \sin 30=75 l b $$

Total force acting in \(x-\) direction

$$ F_{x}=100+50 \cos 30=143.30 \mathrm{lb} $$

Net force, \(F\)

$$ F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{75^{2}+143.30^{2}}=161.74 l b $$

angle \(\theta\) made by the force with positive \(x-\) axis

$$ \theta=\tan ^{-1}\left(\frac{F_{y}}{F_{x}}\right)=62.37^{\circ} $$

The net moment of forces about point \(\mathrm{A}\)

$$ \sum M=(50 \times 3)+(50 \cos 30 \times 6)-(50 \sin 30 \times 3)-(100 \times 6)=-265.19 l b \text { in. } $$

where clockwise moment is taken as positive and counter clockwise moment as negative.

Equivalent resultant force and couple moment acting at point A is graphically shown below