91影视

mass of block $A$is $m_1$

mass of block $B$is $m_2$

the inclination of block $A$is $胃$

the inclination of block $B$is $蠒$

Tension in the rope is$T$

Resolving forces in block $A$, we get

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

Resolving forces in block $B$, we get

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

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

role="math" localid="1650136140070" $\begin{array}{c}{m}_{2}g\sin 蠒鈭�m_{2}a鈭�m_{1}g\sin 鈦�胃=m_{1}a\\ m_{2}g\sin 蠒鈭�m_{1}g\sin 鈦�胃=m_{1}a+m_{2}a\\ \left(m_2\sin 蠒鈭�m_1\sin 鈦�胃\right)g=\left(m_1+m_2\right)a\\ a=\left(\frac{m_2\sin 蠒鈭�m_1\sin 鈦�胃}{m_1+m_2}\right)\end{array}$

Therefore the Tension of the rope is given by

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

The rope and the pulley are among the interacting objects, and they鈥檙e part of the system when$胃=蠒$