Two bars with masses $m_{1}$ and $m_{2}$ are lying on a horizontal surface connected by a weightless non-stretchable thread. An external force F is applied horizontally. Determine the tension force of the thread if the coefficients of sliding friction between the surface and the bars are $\mu_{1}$ and $\mu_{2}$, respectively.
Decision:
If the magnitude $F$ of the force $\vec {F}$ does not exceed the maximum rest friction force of the first bar $F_{T1} = m_{1} \mu_{1} g$, then the bar rests even if the thread is not stretched. If $F_{T1} \leq F \leq F_{T1}+F_{T2}$, then the system is at rest and, therefore, the thread tension force $T = F - F_{T1}.$
If the condition $ F_{T1}+F_{T2}< F$ is fulfilled, the force $T$ will exceed the rest friction force of the second bar and the system will move with constant acceleration.
$a=\frac{F-g(m_{1} \mu_{1}+ m_{2} \mu_{2})}{m_{1}+m_{2}}$
and
$T=m_{2}a+F_{T1} = \frac{m_{2}}{ m_{1}+m_{2}[F+m_{a}g(\mu_{2}-\mu_{1})]}$.
Decision:
If the magnitude $F$ of the force $\vec {F}$ does not exceed the maximum rest friction force of the first bar $F_{T1} = m_{1} \mu_{1} g$, then the bar rests even if the thread is not stretched. If $F_{T1} \leq F \leq F_{T1}+F_{T2}$, then the system is at rest and, therefore, the thread tension force $T = F - F_{T1}.$
If the condition $ F_{T1}+F_{T2}< F$ is fulfilled, the force $T$ will exceed the rest friction force of the second bar and the system will move with constant acceleration.
$a=\frac{F-g(m_{1} \mu_{1}+ m_{2} \mu_{2})}{m_{1}+m_{2}}$
and
$T=m_{2}a+F_{T1} = \frac{m_{2}}{ m_{1}+m_{2}[F+m_{a}g(\mu_{2}-\mu_{1})]}$.