Question
Solve the previous problem if the friction coefficient between the $2.0\ kg$ block and the plane below it is $0.5$ and the plane below the $4.0\ kg$ block is frictionless.
✓
Answer
$m_1 = 4\ kg, m_2 = 2\ kg$
Frictional co $-$ efficient between $2\ kg$ block and surface $= 0.5$
$R = 10\ cm = 0.1m$
$I = 0.5\ kg-m^2$
$\text{m}_1\text{g}\sin\theta-\text{T}_1=\text{m}_1\text{a}\ \dots(1)$
$\text{T}_2-(\text{m}_2\text{g}\sin\theta+\mu\text{m}_2\text{g}\cos\theta)=\text{m}_2\text{a}\ \dots(2)$
$(\text{T}_1-\text{T}_2)=\frac{\text{la}}{\text{r}^2}$
Adding equation $(1)$ and $(2)$ we will get
$\text{m}_1\text{g}\sin\theta-(\text{m}_2\text{g}\sin\theta+\mu\text{m}_2\text{g}\cos\theta)+(\text{T}_1+\text{T}_2)=\text{m}_1\text{a}+\text{m}_2\text{a}$
$\Rightarrow4\times9.8\times\Big(\frac{1}{\sqrt2}\Big)-\Big\{\Big(2\times9.8\times\Big(\frac{1}{\sqrt2}\Big)+0.5\times2\times9.8\times\Big(\frac{1}{\sqrt2}\Big)\Big\}$
$=\Big(4+2+\frac{0.5}{0.01}\Big)\text{a}$
$\Rightarrow27.80-(13.90+6.95)=65\text{a}$
$\Rightarrow\text{a}=0.125\text{ms}^{-2}$
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