question_answer

Consider given pulley and mass system consisting of ideal pulley, ideal springs and ideal identical strings.

Let accelerations of \[{{m}_{1}},\]\[{{m}_{2}},\]\[{{m}_{3}}\]and \[{{m}_{4}}\]are \[{{a}_{1}},{{a}_{2}},{{a}_{3}}\] and \[{{a}_{4}},\]respectively immediately after cutting thread A which keeps the system stable. Now, choose the correct option.

A) \[{{a}_{4}}=0\]                       

B) \[{{a}_{4}}=\left( \frac{{{m}_{3}}+{{m}_{4}}-{{m}_{1}}-{{m}_{2}}}{{{m}_{4}}} \right)\cdot g\]

C) \[{{a}_{4}}=\frac{({{m}_{1}}+{{m}_{2}})g}{{{m}_{4}}}\]   

D) \[{{a}_{4}}=\left( \frac{{{m}_{3}}-{{m}_{1}}-{{m}_{2}}}{{{m}_{4}}} \right)\cdot g\]

Correct Answer: B

Solution :

Here, \[{{m}_{1}}+{{m}_{2}}+>{{m}_{3}}+{{m}_{4}}\]otherwise equilibrium is not possible.

Force balance for mass \[{{m}_{3}}\] is

Also, force balance for mass \[{{m}_{1}}\]is

Here, \[{{T}_{1}}={{m}_{2}}g\]

So,   \[T=({{m}_{1}}+{{m}_{2}})g\]

\[{{m}_{3}}g+{{T}_{2}}-T=0\]

\[\Rightarrow \]\[{{T}_{2}}=T-{{m}_{3}}g\]or \[{{T}_{2}}=({{m}_{1}}+{{m}_{2}}-{{m}_{3}})g\]

When lower thread A is cut equations of motion for all masses can be written as

\[{{m}_{1}}{{a}_{1}}={{m}_{1}}g+{{T}_{1}}-T\]

\[{{m}_{2}}{{a}_{2}}={{m}_{2}}g-{{T}_{1}}\]

\[{{m}_{3}}{{a}_{3}}={{T}_{2}}+{{m}_{3}}g-T\]

\[{{m}_{4}}{{a}_{4}}={{m}_{4}}g-{{T}_{2}}\]

Solving these equations, we have

\[{{a}_{1}}={{a}_{2}}={{a}_{3}}=0\]

and       \[{{a}_{4}}=\left( \frac{{{m}_{3}}+{{m}_{4}}-{{m}_{1}}-{{m}_{2}}}{{{m}_{4}}} \right)g\]