A particle of weight 8 N is attached at C to the ends of two light inextensible strings AC and BC - Edexcel - A-Level Maths Mechanics - Question 2 - 2013 - Paper 2

To find the tension in string AC (denote it as $T_A$), we first resolve the forces in vertical and horizontal directions.

Vertical Direction: The particle is in equilibrium, so: $T_A \sin 35^\circ + T_B \sin 25^\circ = 8 \text{ N}$

Horizontal Direction: Equating the horizontal forces gives: $T_A \cos 35^\circ = T_B \cos 25^\circ$

We can express $T_B$ in terms of $T_A$ using the horizontal equation: $T_B = T_A \frac{\cos 35^\circ}{\cos 25^\circ}$

Substituting this into the vertical equation: $T_A \sin 35^\circ + T_A \frac{\cos 35^\circ}{\cos 25^\circ} \sin 25^\circ = 8$

From here, isolate $T_A$ and solve for its value: $T_A(\sin 35^\circ + \frac{\sin 25^\circ \cos 35^\circ}{\cos 25^\circ}) = 8$

Compute this to find: $T_A = 8 / (\sin 35^\circ + \sin 25^\circ \frac{\cos 35^\circ}{\cos 25^\circ})$

Calculating the right side yields approximately: $T_A \approx 8.4 ext{ N}$