A beam AB has mass 12 kg and length 5 m - Edexcel - A-Level Maths Mechanics - Question 5 - 2008 - Paper 1

Using the equilibrium condition for vertical forces, we have: $T_A + T_C = W_{beam}$

Substituting the values: $T_A + 73.575 = 117.72$ $T_A = 117.72 - 73.575 = 44.145 ext{ N}$

Thus, the tension in the rope at A is approximately 44.1 N.

Considering the additional load of mass 16 kg at a distance y from A, we can express the new tension in the rope at C. The total weight (W_load) of the additional load is: $W_{load} = 16 imes 9.81 = 156.96 ext{ N}$

Taking moments about point A again: $T_C imes 4 = W_{beam} imes 2.5 + W_{load} imes y$

Substituting: $T_C imes 4 = 117.72 imes 2.5 + 156.96 imes y$

Solving for T_C gives: $T_C = \frac{117.72 imes 2.5 + 156.96y}{4}$

To ensure the rope at C does not break, we need: $T_C \leq 98 ext{ N}$

Substituting our expression for T_C: $\frac{117.72 imes 2.5 + 156.96y}{4} \leq 98$

Multiplying through by 4 results in: $117.72 imes 2.5 + 156.96y \leq 392$

Calculating the left side: $294.3 + 156.96y \leq 392$

Rearranging gives:

y \leq \frac{97.7}{156.96}\ y \leq 0.6226 ext{ m}$$ Thus, the load must be no more than 0.6226 m from A.