A uniform beam AB of length 5 m and weight 1000 N is held in a horizontal position by two vertical forces, F1 and F2, positioned at A and C respectively - Leaving Cert Applied Maths - Question 7 - 2021

Using the equilibrium of forces, the sum of the vertical forces must equal zero:

$F_1 + F_2 = 1000$

Substituting F1: $375 + F_2 = 1000$

Thus, solving for F2 gives: $F_2 = 1000 - 375 = 625 \text{ N}$

The diagram should include:

• Weight of the rod (120 N) acting downwards at the center.
• Tension (T) in the string acting upwards and at an angle of 60°.
• Reaction force at the hinge, which has horizontal (X) and vertical (Y) components.

Resolving forces horizontally:

$X = T \cos 60°$ Resolving forces vertically:

$Y + T \sin 60° = 120$

The equation for moments about point P is:

$T \sin 60° \times 4 = 120 \times 2 \sin 30°$

From the moment equation:

$T \sin 60° \times 4 = 120 \times 2 \sin 30°$ Substituting values:

$T \sin 60° \times 4 = 120 \times 2 \times 0.5$

Simplifying yields:

$T \sin 60° = \frac{120}{2} = 60$

Thus,

$T \approx \frac{60}{\sin 60°} = 20\sqrt{3} \text{ N}$

The horizontal component is:

$X = T \cos 60° = 20\sqrt{3} \times 0.5 = 10\sqrt{3}$ The vertical component is:

$Y = 120 - T \sin 60° = 120 - 20\sqrt{3} \times \frac{\sqrt{3}}{2} = 120 - 90 = 30$ Thus, we have:

• Horizontal component: $10\sqrt{3}$
• Vertical component: $30$