Question

Figure (6-E6) shows two blocks in contact sliding down an inclined surface of inclination ${30}^{\circ }$ The friction coefficient between the block of mass 2.0 kg and the incline is ${\mu }_1$ and that between the block of mass 4.0 kg

Answer 1

From the free body diagram
$R=4g\cos {30}^{\circ }\Rightarrow R=4\times 10\times \frac{\sqrt{3}}{2}=20\sqrt{3}$

${\mu }_{2}R+4a-p-4g\sin {30}^{\circ }=0\Rightarrow 0.3\left(40\right)\cos {30}^{\circ }+4a-p-40\sin {20}^{\circ }=20$

$R_1=2g\cos {30}^{\circ }=10\sqrt{3}p+2a-{\mu }_1{R}_1-2g\sin {30}^{\circ }=0$
From equation, (ii)
$6\sqrt{3}+4a-p-20=0$

p = constant force
g = $10m{s}^2$
Fromequation (iv),
$p+2a+2\sqrt{3}-10=106\sqrt{3}+6a+30+2\sqrt{3}=0\Rightarrow 6a=30-8\sqrt{3}=30-13.85=16.15\Rightarrow a=\frac{16.15}{6}=2.69=2.7m/s^2$
(b) In this case, 4 kg block will travel with more acceleration because, coefficient of friction is less than that of 2 kg. So, they will move separately. Drawing the free body diagram of 2 kg mass only, it can be found that, a = $2.4,m/s^2$