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

\[m_2 gsin\theta -  T_2  =  m_1 a..........(1)\]

\[ T_1  - \left( m_1 g\sin\theta + \mu m_1 g\cos  \theta \right) =  m_1 a ............(2)\]

\[ T_2  -  T_1  = \frac{Ia}{r^2} .............(3)\]

Adding equations (1) and (2), we get

\[m_2 g\sin\theta - \left( m_1 g\sin\theta - \mu m_1 g\cos  \theta \right) + \left( T_1 - T_2 \right) =  m_1 a +  m_2 a\]

\[m_2 g\sin\theta - \left( m_1 g\sin\theta + \mu m_1 g\cos  \theta \right) - \frac{Ia}{r^2} =  m_1 a +  m_2 a\]

\[ m_2 g\sin\theta - \left( m_1 g\sin\theta + \mu m_1 g\cos  \theta \right) =  m_1 a +  m_2 a + \frac{Ia}{r^2}\]

\[4 \times 9 . 8 \times \frac{1}{\sqrt{2}} - \left\{ 2 \times 9 . 8 \times \frac{1}{\sqrt{2}} + 0 . 5 \times 2 \times 9 . 8 \times \frac{1}{\sqrt{2}} \right\} = \left( 4 + 2 + \frac{0 . 5}{0 . 01} \right)a\]

\[ \Rightarrow 27 . 80 - 13 . 90 + 6 . 95 = 56a\]

\[ \Rightarrow 27 . 8 - 20 . 8 = 56a\]

\[ \Rightarrow a = \frac{7}{56} = 0 . 125  m/s^2\]