Mechanical #Problem : 1

Hello guys, this time I have a mechanical problem. See this picture

Given that :

• The purple block is not move.
• $M_{A} < M_{B}$
• Gravitational acceleration = $g$
• The kinetic frictional cooficient = $\mu _k$

Ask :

1. If the block move to down, determine the acceleration and the string tension.

Hold on guys !!!

Stop on this line !!!

Do it by yourself before you look the key answer and the solution

Key Answer:

Number 1

$a = \frac {(m_{B} - m_{A} . \mu_{k})}{(m_{A} + m_{B})}g$



$T = \frac {m_{A} . m_{B} . (\mu_{k} + 1)} {(m_{A} + m_{B})}g$

Explanation:

First of all, we can divided that system into two dependent systems. See picture below

System 1

From system 1, we got :

$\begin{eqnarray*} \Sigma F &=& m_{A} . a \\ T - F_{k} &=& m_{A} . a \\ T - W_{A} . \mu _{k} &=& m_{A} . a \\ T - m_{A} . g . \mu _{k} &=& m_{A} . a \cdots (1) \end{eqnarray*}$

System 2

From system 2, we got :

$\begin{eqnarray*} \Sigma F &=& m_{B} . a \\ W_{B} - T &=& m_{B} . a \\ m_{B} . g - T &=& m_{B} . a \cdots (2) \end{eqnarray*}$

Then, eliminate $(1)$ and $(2)$.
If we add them, we will get :

$\begin{eqnarray*} T - m_{A} . g . \mu _{s} &=& m_{A} . a \\ m_{B} . g - T &=& m_{B} . a \\ m_{B} . g - m_{A} . g. \mu_{k} &=& (m_{A} + m_{B}) . a \\ a &=& \frac {(m_{B} - m_{A} . \mu_{k})}{(m_{A} + m_{B})}g\end{eqnarray*} $

Then, we can determine the tension. See it : ...

$\begin{eqnarray*}T - m_{A} . g . \mu _{k} &=& m_{A} . a \\ T &=& m_{A} . g . \mu_{k} + m_{A} . \frac {(m_{B} - m_{A} . \mu_{k})}{(m_{A} + m_{B})}g \\ T &=& m_{A} . g . \frac {\mu_{k}.(m_{A} + m_{B}) + (m_{B} - m_{A} . \mu_{k})} {(m_{A} + m_{B})}\\ T &=& m_{A} . g . \frac {\mu_{k}.m_{A} + \mu_{k}.m_{B} + m_{B} - m_{A} . \mu_{k}} {(m_{A} + m_{B})}\\ T &=& m_{A} . g . \frac {\mu_{k}.m_{B} + m_{B}} {(m_{A} + m_{B})}\\ T &=& \frac {m_{A} . m_{B} . (\mu_{k} + 1)} {(m_{A} + m_{B})}g\\ \end{eqnarray*}$