question_answer

Two weights\[{{w}_{1}}\]and\[{{w}_{2}}\]are suspended to the two strings on a frictionless pulley. When the pulley is pulled up with an acceleration g, then the tension in the string is:

A)  \[\frac{4{{w}_{1}}{{w}_{2}}}{{{w}_{1}}+{{w}_{2}}}\]                       

B) \[\frac{{{w}_{1}}{{w}_{2}}}{{{w}_{1}}+{{w}_{2}}}\]

C)   \[\frac{2{{w}_{1}}{{w}_{2}}}{{{w}_{1}}+{{w}_{2}}}\]                      

D) \[\frac{{{w}_{1}}+{{w}_{2}}}{2}\]

Correct Answer: A

Solution :

                 The free body diagram of the given situation is shown. T is the tension in strings,\[{{w}_{1}}\]and\[{{w}_{2}}\]are the weights acting downwards. Taking the upward and downward motion respectively, we get:                 \[T-{{m}_{1}}g={{M}_{1}}(g+a)\] \[\Rightarrow \]               \[T-2{{m}_{1}}g={{m}_{1}}a\]                            ...(1) For second weight                 \[{{m}_{2}}g-T={{m}_{2}}(g-a)\] \[\Rightarrow \]        \[2{{m}_{2}}g-T={{m}_{2}}a\]                                     ...(2) On solving Eqs. (1) and (2), we get                 \[T=\frac{4{{m}_{1}}{{m}_{2}}g}{({{m}_{1}}-{{m}_{2}})}\] Given,\[{{w}_{1}}={{m}_{1}}g,{{w}_{2}}={{m}_{2}}g\]hence, we get                 \[T=\frac{4{{w}_{1}}+{{w}_{2}}}{({{w}_{1}}+{{w}_{2}})}\]