Particles A and B of masses 5 kg and 8 kg respectively are connected by a light inextensible string passing over a smooth pulley. Particle A lies on a smooth plane inclined at an angle $\tan^{-1}\left(\dfrac{4}{3}\right)$ to the horizontal and particle B hangs freely. The system is released from rest. Take $g = 9.81$ ms$^{-2}$.

If $\theta = \tan^{-1}\left(\dfrac{4}{3}\right)$ then the 3, 4, 5 triangle gives $\sin\theta = \dfrac{4}{5}$.

The string is light and inextensible over a smooth pulley, so the tension T is the same throughout and both particles share one acceleration a. B is the heavier, so B descends and A is drawn up the plane.

For A along the plane: $T - 5g\sin\theta = 5a$, that is $T - 4g = 5a$.

For B vertically: $8g - T = 8a$.

Adding removes T: $8g - 4g = 13a$, so $4g = 13a$ and

$a = \dfrac{4g}{13} = \dfrac{4(9.81)}{13} = 3.018$ ms$^{-2}$,

which is 3.0 ms$^{-2}$ correct to 2 significant figures.