Question

Figure (6−E6) shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is μ₁, and that between the block of mass 4.0 kg and incline is μ₂. Calculate the acceleration of the 2.0 kg block if (a) μ₁ = 0.20 and μ₂ = 0.30, (b) μ₁ = 0.30 and μ₂ = 0.20. Take g = 10 m/s².

Figure

Answer 1

(a) From the free body diagram



R = 4g cos 30°
$$\begin{gathered} \Rightarrow \mathrm{R}=4\times 10\times \frac{\sqrt{3}}{2} \\ =20\sqrt{3}\mathrm{N}\left(1\right) \end{gathered}$$

μ₂R + m₁a − p − m₁g sin θ = 0
μ₂R + 4a − p − 4g sin 30° = 0
⇒ 0.3 $\times$ (40) cos 30° + 4a − p − 40 sin 30° = 20 (2)



R₁ = 2g cos 30° $=10\sqrt{3}$ (3)
p + 2a − μ₁R₁ − 2g sin 30° = 0 (4)

From Equation (2),
$6\sqrt{3}+4a-p-20=0$

From Equation (4),
$$\begin{gathered} p+2a+2\sqrt{3}-10=10 \\ 6\sqrt{3}+6a+30+2\sqrt{3}=0 \\ \Rightarrow 6a=30-8\sqrt{3} \\ =30-13.85=16.15 \\ \Rightarrow a=\frac{16.15}{6} \\ =2.69=2.7\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

(b) In this case, the 4 kg block will move at a higher acceleration because the coefficient of friction is less than that of the 2 kg block. Therefore, the two blocks will move separately. By drawing the free body diagram of 2 kg mass, it can be shown that a = 2.4 m/s².