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