A uniform beam, ab, is held in a horizontal position by two vertical inelastic strings attached at a and b respectively - Leaving Cert Applied Maths - Question 7 - 2007

A diagram depicting the ladder should include:

• Weight of the Ladder (W = 80 N) acting downward at its center of gravity (midpoint of the ladder).
• Normal Reaction Force (R) from the ground acting upward at the base of the ladder.
• Frictional Force ($R_f = \mu R$) acting horizontally at the base, opposing the motion.
• Normal Force from the Wall ($R_w$) acting horizontally towards the ladder from the wall.

The forces can be illustrated as follows:

------           (R_w)
|    |
|    |
|    | (W)
|    | 
------
|    
|----|  (R)

1. Vertical Forces: The normal reaction from the ground must equal the weight of the ladder.

   $R = W = 80 \text{ N}$

2. Horizontal Forces: From equilibrium, the horizontal forces satisfy:

   $R_f = R_w = \mu R$

   Thus:

   $R_w = 80\mu$

3. Take Moments About Point a: Taking moments about point a, we have:

   $R_w (5 \sin(60°)) - W(\frac{5}{2}\cos(60°) = 0$

   Substituting for $R_w$ and $W$:

   $80 \mu (5 \sin(60°)) = 80 (\frac{5}{2}\cos(60°)$

   Simplifying,

   $80 \mu (\frac{\sqrt{3}}{2}) = 80 (\frac{5}{4})$

   This simplifies to:

   $\mu = \frac{1}{2\sqrt{3}}$