Given:
The mass of the block A, m_A=2 kg
The mass of the block B, m_B=4 kg
The angle of inclination, θ=30°
The coefficient of the friction between block A and the surface, μ=0.2
The acceleration due to gravity, g=9.8 m/s²
To find:
The acceleration of the objects and the tension in the string.
Explanation:
The forces acting on block A are the tension in the string directed up the incline, the friction, and the gravitational force acting down the incline.
Thus the net force acting on the block A is given by,
[tex]\begin{gathered} m_Aa=T-m_Ag-f \\ =T-m_Ag\sin\theta-\mu\times m_Ag\cos\theta \end{gathered}[/tex]Where a is the acceleration of the system and T is the tension in the string.
On rearranging the above equation,
[tex]T=m_Aa+m_Ag\sin\theta+(\mu\times m_Ag\cos\theta)\text{ }\to\text{ \lparen i\rparen}[/tex]The forces acting on block B are the downward gravitational force and the upward tension in the string.
Thus the net force acting on block B is given by,
[tex]m_Ba=m_Bg-T[/tex]On rearranging the above equation,
[tex]T=m_Bg-m_Ba\text{ }\to\text{ \lparen ii\rparen}[/tex]From equations (i) and (ii),
[tex]m_Aa+m_Ag\sin\theta+(\mu\times m_Ag\cos\theta)=m_Bg-m_Ba[/tex]On rearranging the above equation,
[tex]\begin{gathered} m_Aa+m_Ag\sin(\theta)+(\mu m_Ag\cos(\theta))=m_Bg-m_Ba \\ \implies m_Aa+m_Ba=m_Bg-m_Ag\sin(\theta)-(\mu m_Ag\cos(\theta)) \\ \implies(m_A+m_B)a=g[m_B-m_A\sin(\theta)-(\mu m_A\cos(\theta))] \\ \implies a=\frac{g[m_B-m_A\sin(\theta)-(\mu m_A\cos(\theta))]}{m_A+m_B} \end{gathered}[/tex]On substituting the known values in the above equation,
[tex]\begin{gathered} a=\frac{9.8[4-2\sin30\degree-(0.2\times2\times\cos30\degree)]}{4+2} \\ =4.33\text{ m/s}^2 \end{gathered}[/tex]On substituting the known values in the equation (ii),
[tex]\begin{gathered} T=4\times9.8-4\times2 \\ =31.2\text{ N} \end{gathered}[/tex]Final answer:
The acceleration of the system is 4.33 m/s²
The tension in the string is 31. 2 N
