Friction and Inclined Planes Questions and Answers

From $\tan\alpha=\dfrac{3}{4}$ the 3, 4, 5 triangle gives $\sin\alpha=\dfrac{3}{5}=0.6$ and $\cos\alpha=\dfrac{4}{5}=0.8$.

Perpendicular to the plane the block does not move, so $R=mg\cos\alpha$ and the friction, which opposes the sliding and so acts up the plane, is $\mu mg\cos\alpha$.

Along the plane, $mg\sin\alpha-\mu mg\cos\alpha=ma$. The mass cancels from every term:

$a=g(\sin\alpha-\mu\cos\alpha)=9.81\left(0.6-\tfrac{1}{4}(0.8)\right)=9.81(0.6-0.2)=9.81(0.4)=3.924$ ms$^{-2}$,

which is 3.9 ms$^{-2}$ correct to 2 significant figures. Notice the 3.5 kg never enters the answer.

The weight of the particle is $8g$ newtons, acting vertically downwards. Resolving it along the plane, the component down the slope is

$$8g\sin 30^{\circ}=8(9.81)\left(\frac{1}{2}\right)=4g=39.24 \text{ N},$$

which is 39 N correct to 2 significant figures.

The sine goes with the angle of inclination for the component along the plane; the cosine gives the component perpendicular to it, $8g\cos 30^{\circ}=67.96$ N, which the normal reaction balances.

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.