Question

The bars AB and AC are joined by a pin at A and a horizontal cable. The vertical cable carrying the 200-kg mass is attached to the pin at A. Determine the tension in the horizontal cable. Neglect the weights of the bars.

Answer 1

F_ACSin 60=F_ABSin 30

$F_{AB}=\sqrt3F_{AC}$

$F_{AC}Cos 60+F_{AB}Cos 30+Mg=0$

$\frac{F_{AC}}{2}+\sqrt3F_{AC}\frac{\sqrt3}{2}+200*9.81=0$

$F_{AC}=-981N$

$F_{AC}Cos30+T=0$

$T=-F_{AC}Cos30$

$=-(-981)Cos30$

$=849.57N$