One end of a light inextensible string is attached to a block P of mass 5 kg - Edexcel - A-Level Maths Mechanics - Question 7 - 2009 - Paper 1

We substitute $a = 0.6g$ back into equation (1): $T - 3g = 5(0.6g)$ $T - 3g = 3g$ $T = 6g$

For block Q, the forces acting on it include its weight and the normal force exerted by block R. The weight of block Q is: $W_Q = 15g$

The normal force (N) acting on block Q can be calculated using: $N = 15g - F$ where $F$ is the force experienced by block Q due to block R. From the previous calculation, we have: $F = 5a = 5 \times 0.6g = 3g$ Therefore, substituting: $N = 15g - 3g = 12g$

To find the force exerted on the pulley by the string, we consider the tension in the string. The net force acting on the pulley due to the string is: $F = 2T \cos(90^{\circ} - \alpha)$

With $T = 6g$ and given that $\cos(90^{\circ} - \alpha) = \sin \alpha$: $F = 2 \times 6g \cdot \sin \alpha$ We need to evaluate $\sin \alpha$ which can be derived from $\sin \beta$ as $\alpha = 90^{\circ} - \beta$ and is equivalent to:\n$F = 12g \cdot \sin(\beta) = 12g \cdot \frac{4}{5} = 9.6g \, (approx. \, 105 \, N).$