Find the tension in diagram below.

• For the $20 \, \text{N}$ force: $F_x = 0$ $F_y = -20 \, \text{N}$ (downwards)

• For Tension $T_1$: $T_{1x} = T_1 \cos(45^\circ)$ (to the right) $T_{1y} = T_1 \sin(45^\circ)$ (upwards)

• For Tension $T_2$: $T_{2x} = -T_2 \cos(30^\circ)$ (to the left) $T_{2y} = T_2 \sin(30^\circ)$ (upwards)

Sum of forces in the x-direction ($\sum F_x = 0$):

Sum of forces in the y-direction ($\sum F_y = 0$):

Multiply by 2 to clear denominators:

From equation (1), substitute $T_1 \sqrt{2} = T_2 \sqrt{3}$ into equation (2):

Factor out $T_2$:

To rationalize the denominator, multiply the numerator and denominator by $(\sqrt{3} - 1)$:

Using $\sqrt{3} \approx 1.732$:

Now, substitute the value of $T_2$ back into equation (1) to find $T_1$:

To rationalize, multiply by $\frac{\sqrt{2}}{\sqrt{2}}$:

Using $\sqrt{2} \approx 1.414$ and $\sqrt{6} \approx 2.449$:

The tensions are: