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 1

From the figure, we have:
$$\begin{gathered} m_{2}gs\mathrm{in}\theta -T_2=m_{1}a...\left(i\right) \\ {T}_1-\left(m_{1}g\sin \theta +\mu m_{1}g\cos \theta \right)=m_{1}a...\left(ii\right) \\ {T}_2-T_1=\frac{Ia}{r^2}...\left(iii\right) \end{gathered}$$
Adding equations (i) and (ii), we get:
$m_{2}g\sin \theta -\left(m_{1}g\sin \theta -\mu m_{1}g\cos \theta \right)+\left(T_1-T_2\right)=m_{1}a+m_{2}a$
$$\begin{gathered} m_{2}g\sin \theta -\left(m_{1}g\sin \theta +\mu m_{1}g\cos \theta \right)-\frac{Ia}{r^2}=m_{1}a+m_{2}a \\ {m}_{2}g\sin \theta -\left(m_{1}g\sin \theta +\mu m_{1}g\cos \theta \right)=m_{1}a+m_{2}a+\frac{Ia}{r^2} \end{gathered}$$
$$\begin{gathered} 4\times 9.8\times \frac{1}{\sqrt{2}}-\left\{2\times 9.8\times \frac{1}{\sqrt{2}}+0.5\times 2\times 9.8\times \frac{1}{\sqrt{2}}\right\}=\left(4+2+\frac{0.5}{0.01}\right)a \\ \Rightarrow 27.80-13.90+6.95=56a \\ \Rightarrow 27.8-20.8=56a \\ \Rightarrow a=\frac{7}{56}=0.125\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$