Block $A$in $FIGUREEX7.5$is sliding down the incline. The rope is massless, and the massless pulley turns on frictionless bearings, but the surface is not frictionless. The rope and the pulley are among the interacting objects, but you’ll have to decide if they’re part of the system.

mass of block $A$is $m_1$

mass of block $B$is $m_2$

the inclination of block $A$is $\theta$

the inclination of block $B$is $\varphi$

Tension in the rope is$T$

Resolving forces in block $A$, we get

role="math" localid="1650135815167" $\begin{array}{r}{F}_{\text{net}}=T-W_1=m_{1}a\\ T-m_{1}g\sin \theta =m_{1}a\left(1\right)\end{array}$

Resolving forces in block $B$, we get

role="math" localid="1650136106578" $\begin{array}{r}{F}_{\text{net}}=m_{2}g\sin \varphi -T=m_{2}a\\ T=m_{2}g\sin \varphi -m_{2}a\left(2\right)\end{array}$

Now substituting equation $\left(2\right)in\left(1\right),$we get

role="math" localid="1650136140070" $\begin{array}{c}{m}_{2}g\sin \varphi -m_{2}a-m_{1}g\sin \theta =m_{1}a\\ m_{2}g\sin \varphi -m_{1}g\sin \theta =m_{1}a+m_{2}a\\ \left(m_2\sin \varphi -m_1\sin \theta \right)g=\left(m_1+m_2\right)a\\ a=\left(\frac{m_2\sin \varphi -m_1\sin \theta }{m_1+m_2}\right)\end{array}$

Therefore the Tension of the rope is given by

role="math" localid="1650136181366" $\begin{array}{r}T=m_{2}g\sin \varphi -m_{2}a\\ T=m_{2}g\sin \varphi -m_2\left(\frac{m_2\sin \varphi -m_1\sin \theta }{m_1+m_2}\right)g\\ T=\left(\sin \varphi -\frac{m_2\sin \varphi -m_1\sin \theta }{m_1+m_2}\right)m_{2}g\end{array}$

The rope and the pulley are among the interacting objects, and they’re part of the system when$\theta =\varphi$