Problem 3.31

พิจารณา F.B.D. ของมวล m₁ พิจารณา F.B.D. ของมวล m₂

xT

yT

y

\(_{\theta _{1}}\)

\(_{\theta _{2}}\)

\(m_{1}gcos\theta _{1}n_{1}m_{2}gcos\theta _{2}m_{2}gsin\theta _{2}\)

\(m_{1}gsin\theta _{1}n_{2}\)

m₁g

m₂g

x

จะได้ \(ΣF_{x}= ma_{x}\) จะได้ \(ΣF_{x}= ma_{x}\)

\(T - m_{1}gsin\theta _{1} = m_{1}a \cdots \cdots (1) m_{2}gsin\theta _{2} - T = m_{2}a \cdots \cdots (2)\)

\((1) + (2) ; T - m_{1}gsin\theta _{1} + m_{2}gsin\theta _{2} - T = (m_{1} + m_{2})a\)

\(a =\frac{(m_{2}\sin \theta _{2} - m_{1}\sin \theta _{1})g}{m_{1} + m_{2}}\)

a แทนใน \((1)\) จะได้ \(T = \frac{m_{1}m_{2}g(\sin \theta _{1} + \sin \theta _{2})}{m_{1} + m_{2}}\;\blacksquare\)