Given:
The coefficient of friction between the plane and the mass A is,
[tex]\mu_k=0.15[/tex]The masses are,
[tex]m_A=m_B=2.7\text{ kg}[/tex]The angle of inclination is,
[tex]\theta=30\degree[/tex]To find:
a) the acceleration of the masses
b) The smallest value of the coefficient of friction which will keep the system from accelerating
Explanation:
a)
The free-body diagram is:
For the mass B,
[tex]m_Ba=m_Bg-T[/tex]For the other mass, the normal reaction is,
[tex]N=m_Agcos\theta[/tex]Again the horizontal motion gives,
[tex]\begin{gathered} m_Ba=T-f-m_Bgsin\theta \\ m_Ba=T-\mu_km_Bgcos\theta-m_Bgs\imaginaryI n\theta \end{gathered}[/tex]Combining the equations we get,
[tex]\begin{gathered} a=\frac{m_Bg-m_Agsin\theta-\mu_km_Agcos\theta}{m_A+m_B} \\ =\frac{2.7\times9.8-2.7\times9.8\times sin30\degree-0.15\times2.7\times9.8\times cos30\degree}{2.7+2.7} \\ =1.8\text{ m/s}^2 \end{gathered}[/tex]Hence, the acceleration is,
[tex]1.8\text{ m/s}^2[/tex]b)
For, the zero acceleration,
[tex]\begin{gathered} a=\frac{m_{B}g-m_{A}gs\imaginaryI n\theta-\mu_{k}m_{A}gcos\theta}{m_{A}+m_{B}}=0 \\ m_B-m_As\imaginaryI n\theta-\mu_km_Acos\theta=0 \\ \mu_k=\frac{m_B-m_As\mathrm{i}n\theta}{m_Acos\theta} \\ \mu_k=\frac{2.7-2.7\times sin30\degree}{2.7cos30\degree} \\ \mu_k=0.58 \end{gathered}[/tex]Hence, the required kinetic friction is 0.58.
