7. (a) A uniform rod AB of length 1.5 m and mass 3 kg is smoothly hinged to a vertical wall at A - Leaving Cert Applied Maths - Question 7 - 2019

To find the coefficient of friction for the system at the point of slipping (G), we analyze the forces acting on rod GF:

1. Set Up the Equations:

   For rod FE:

   $Y \times 2l = W \times 1l \\ Y = \frac{1}{2}W$

2. Analyze Rod G:

   For GFE:

   $W \times \frac{1}{2}l + W \times 2l = X \times l \sqrt{3} + Y \times 3l$

   Set equations to relate forces Y, X, and the weight W.

3. Summarize Forces:

   • The horizontal force X is transmitted due to friction:

   $X = \mu R$

   With the normal force R determined from equilibrium conditions.

4. Combine and Solve:

   • Substitute for R in terms of W to find:

   $\mu = \frac{X}{R} = \frac{1}{2}W \cdot \frac{1}{\frac{2}{3}W} = \frac{3}{6} = 0.38$

   Thus, the coefficient of friction (μ) is approximately 0.38.

To find the horizontal and vertical forces acting at point E:

1. Derived Forces:

   From the previous parts, determine R and Y:

   • Vertical force:

   $Y = \frac{1}{2}W$

2. Horizontal Force Analysis:

   • The summation of forces at E gives:

   $R + \frac{1}{2}W = 2W \\ R = 2W - \frac{1}{2}W = \frac{3}{2}W$

   Thus, substituting in for X yields:

   $X = \mu R = \frac{3}{3\sqrt{3}}W\, \text {thus finding X in terms of W}$

3. Finalized Equations:

   • The final horizontal and vertical forces at E are expressed neatly, yielding:

   Horizontal force is $X$ and vertical force is $Y$.