A uniform beam AB of length 3l and weight W is free to turn in a vertical plane about a hinge at A - Leaving Cert Applied Maths - Question 7 - 2015

(i) Normal Reactions

From the equilibrium conditions for joint Z, we have:

• The upward forces are the normal reactions R1 at X and R2 at Y.
• The downward forces are the weights; for ZY, this is 2W and for YZ, this is W:

Using equilibrium equations:

1. For X:
   $R_1 = W + 2W$
   Therefore,
   $R_1 = 3W$

2. For Y:
   R_2 = rac{5W}{4}

3. Completing calculations gives: R_2 = R_1 + rac{7W}{4}
   Thus, $R_2 = 3W + R_2$

(ii) Slipping Occurs at Y Before X

As θ increases, the sideways force caused by the weight at Y exceeds that at X:

• Thus, as θ increases, the frictional force at Y is surpassed by the weight, leading to slipping at Y before X.

(iii) Coefficient of Friction

For the point of slipping condition at θ = 90°:

• The frictional force at Y is given by: $R_2 imes ext{sin}(45) = F_{x} imes ext{cos}(45) + W imes ext{cos}(45)$
• Rearranging gives: R_2 = rac{5W}{4} + rac{W^2}{2}
  Thus, the coefficient of friction is: ext{μ} = rac{3}{5}