A uniform rod, [AB], of length 2 m and weight 120 N is smoothly hinged at end A to a vertical wall - Leaving Cert Applied Maths - Question 7 - 2014

For horizontal forces: $T \cos(60^{\circ}) = X$

For vertical forces: $T \sin(60^{\circ}) + Y = 120$

Taking moments about point A gives: $T \times 2 = 120 \times 1 \sin(60^{\circ})$

From the moment equation, $T \times 2 = 120 \times 1 \times \frac{\sqrt{3}}{2}$ Solving gives: $T = \frac{120 \times 1 \times \sqrt{3}}{2} = 30 \sqrt{3}$

Using the horizontal and vertical force equations:

For horizontal: $X = \sqrt{3}$

For vertical: $Y = 75$

The magnitude of the reaction at point A can be calculated as: $R = \sqrt{X^2 + Y^2} = \sqrt{(\sqrt{3})^2 + (75)^2} = 79.37 \, N$