The diagram shows a light inextensible string having one end fixed, passing under a smooth movable pulley C of mass km kg and then over a fixed smooth pulley - Leaving Cert Applied Maths - Question 4 - 2021

Using the relation obtained for acceleration: $T = mg + 2m \left(\frac{k-2}{k+4} \right)g.$ This leads to:

$T = \frac{(3k)}{(k+4)}mg.$

The reaction force R between D and the scale pan can be expressed considering the forces acting on D:

$R - mg = m \times 2a$

Substituting the value of $a$ obtained earlier:

$R = mg + 2m \left(\frac{(k-2)}{(k+4)} \right)g = \frac{(3k)}{(k+4)}mg.$

To analyze the forces:

• For the wedge (mass $4m$):

  • Weight: $4mg$ acting downwards.
  • Normal reaction from the ground, $R$ acting upwards.

• For the particle (mass $m$):

  • Weight: $mg$ acting downwards.
  • Normal force $N$ from the wedge.
  • An external horizontal force $F$ acting on the wedge.

For the particle on the wedge, applying the second law of motion in the normal direction:

$R = mg \cos(30)$

For horizontal direction, we have:

$F = R \sin(30) \Rightarrow F = mg \cos(30) \cdot\sin(30).$ Here: $F = \sqrt{3}/2 \cdot 4mg$ leading to: $F = 2\sqrt{3}mg.$

In the absence of force $F$, the system's equations change:

For the particle: $mg \cos(30) = ma_q \Rightarrow mg \cos(30) = m \cdot (p + q)$ For the wedge: $mg \sin(30) \Rightarrow R = m q \sin(30).$ From these equations, we find: $p = \frac{19}{17}, q = \frac{1}{17}.$