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Replace the force system by a single force w.r.to point C
1 Answer
written 3.5 years ago by |
$R_x=\sum F_x=20\ N\ (\rightarrow)$
$R_y=\sum F_y=50-60=-10\ N=10\ N(\downarrow)$
$R=\sqrt{R_x^2+R_y^2}=22.36\ N$
$\theta =tan^{-1}\Bigg(\dfrac{R_y}{R_x}\Bigg)=26.56^o$
Using Varignon's Theorem
$\sum M_c=M_c^R..............(i)$
$\sum M_c=-20(1)-50(9)+60(2)=-350\ N.m$
Considering point C as origin,
$x-intercept=\dfrac{\sum M_c}{R_y}=35\ m$
$y-intercept=\dfrac{\sum M_c}{R_x}=17.5\ m$