# JEE Main & Advanced Physics NLM, Friction, Circular Motion Motion of Blocks Connected by Mass Less String

Motion of Blocks Connected by Mass Less String

Category : JEE Main & Advanced

 Condition Free body diagram Equation Tension and acceleration $T={{m}_{1}}a$ $a=\frac{F}{{{m}_{1}}+{{m}_{2}}}$ $F-T={{m}_{2}}a$ $T=\frac{{{m}_{1}}F}{{{m}_{1}}+{{m}_{2}}}$ $F-T={{m}_{1}}a$ $a=\frac{F}{{{m}_{1}}+{{m}_{2}}}$ $T={{m}_{2}}a$ $T=\frac{{{m}_{2}}F}{{{m}_{1}}+{{m}_{2}}}$ ${{T}_{1}}={{m}_{1}}a$ $a=\frac{F}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}$ ${{T}_{2}}-{{T}_{1}}={{m}_{2}}a$ ${{T}_{1}}=\frac{{{m}_{1}}F}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}$ $F-{{T}_{2}}={{m}_{3}}a$ ${{T}_{2}}=\frac{({{m}_{1}}+{{m}_{2}})F}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}$ $F-{{T}_{1}}={{m}_{1}}a$ $a=\frac{F}{{{m}_{1}}+{{m}_{2}}={{m}_{3}}}$ ${{T}_{1}}-{{T}_{2}}={{m}_{2}}a$ ${{T}_{1}}=\frac{({{m}_{2}}+{{m}_{3}})F}{{{m}_{1}}+{{m}_{2}}={{m}_{3}}}$ ${{T}_{2}}={{m}_{3}}a$ ${{T}_{2}}=\frac{{{m}_{3}}F}{{{m}_{1}}+{{m}_{2}}={{m}_{3}}}$

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