Question

If point A is in equilibrium under the action of the applied forces, the values of tensions $T_{AB}$ and $T_{AC}$ are respectively.

• A. 520 N and 300 N
• B. 300 N and 520 N
• C. 450 N and 150 N
• D. 150 N and 450 N

Answer 1

The correct option is A 520 N and 300 N

Method I:

By Lami's theorem

$\frac{T_{AB}}{\sin {120}^{\circ }}=\frac{T_{AC}}{\sin {150}^{\circ }}=\frac{600}{\sin {90}^{\circ }}$

$\therefore T_{AB}=600\sin {120}^{\circ }=519.61=520N$
and $T_{AB}=600\sin {120}^{\circ }=519.61=520N$

Method II:


In equilibrium,
Horizontal forces,
$\Sigma F_x=0,$
$T_{AC}\cos {30}^{\circ }-T_{AB}\cos {60}^{\circ }=0$

$T_{AC}=T_{AB}\frac{\cos {60}^{\circ }}{\cos {30}^{\circ }}=\frac{T_{AB}}{\sqrt{3}}..\left(i\right)$

Vertical forces,
$\Sigma F_y=0$
$T_{AC}\sin {30}^{\circ }+T_{AB}\sin {60}^{\circ }-600=0$

$T_{AC}\sin {30}^{\circ }+T_{AB}\sin {60}^{\circ }-600$

$T_{AC}+\sqrt{3}{T}_{AB}=\mathrm{1200...}\left(ii\right)$

From equation (i) and (ii)
$\frac{T_{AB}}{\sqrt{3}}+\sqrt{3}{T}_{AB}=1200$
$4T_{AB}=120\sqrt{3}$

$T_{AB}=300\sqrt{3}=519.61N=520N$

and $T_{AC}=\frac{T_{AB}}{\sqrt{3}}=\frac{300\sqrt{3}}{\sqrt{3}}=300N$