Motion of Connected Block Over A Pulley
Category : JEE Main & Advanced
| Condition | Free body diagram | Equation | Tension and acceleration |
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\[{{m}_{1}}a={{T}_{1}}-{{m}_{1}}g\] | \[{{T}_{1}}=\frac{2{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}g\] |
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\[{{m}_{2}}a={{m}_{2}}g-{{T}_{1}}\] | \[{{T}_{2}}=\frac{4{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}g\] | |
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\[{{T}_{2}}=2{{T}_{1}}\] | \[a=\left[ \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right]g\] | |
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\[{{m}_{1}}a={{T}_{1}}-{{m}_{1}}g\] | \[{{T}_{1}}=\frac{2{{m}_{1}}[{{m}_{2}}+{{m}_{3}}]}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] |
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\[{{m}_{2}}a={{m}_{2}}g+{{T}_{2}}-{{T}_{1}}\] | \[{{T}_{2}}=\frac{2{{m}_{1}}{{m}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}g\] | |
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\[{{m}_{3}}a={{m}_{3}}g-{{T}_{2}}\] | \[{{T}_{3}}=\frac{4{{m}_{1}}[{{m}_{2}}+{{m}_{3}}]}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}g\] | |
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\[{{T}_{3}}=2{{T}_{1}}\] | \[a=\frac{[({{m}_{2}}+{{m}_{3}})-{{m}_{1}}]g}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] |
| Condition | Free body diagram | Equation | Tension and acceleration |
When pulley have a finite mass M and radius R then tension in two segments of string are different  |
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\[{{m}_{1}}a={{m}_{1}}g-{{T}_{1}}\] | \[a=\frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}+\frac{M}{2}}\] |
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\[{{m}_{2}}a={{T}_{2}}-{{m}_{2}}g\] | \[{{T}_{1}}=\frac{{{m}_{1}}\left[ 2{{m}_{2}}+\frac{M}{2} \right]}{{{m}_{1}}+{{m}_{2}}+\frac{M}{2}}g\] | |
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Torque = \[=({{T}_{1}}-{{T}_{2}})R=I\alpha \] \[({{T}_{1}}-{{T}_{2}})R=I\frac{a}{R}\] \[({{T}_{1}}-{{T}_{2}})R=\frac{1}{2}M{{R}^{2}}\frac{a}{R}\] \[{{T}_{1}}-{{T}_{2}}=\frac{Ma}{2}\] | \[{{T}_{2}}=\frac{{{m}_{2}}\left[ 2{{m}_{1}}+\frac{M}{2} \right]}{{{m}_{1}}+{{m}_{2}}+\frac{M}{2}}g\] | |
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\[T={{m}_{1}}a\] | \[a=\frac{{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}g\] |
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\[{{m}_{2}}a={{m}_{2}}g-T\] | \[T=\frac{{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}g\] | |
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\[{{m}_{1}}a=T-{{m}_{1}}g\sin \theta \] | \[a=\left[ \frac{{{m}_{2}}-{{m}_{1}}\sin \theta }{{{m}_{1}}+{{m}_{2}}} \right]g\] |
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\[{{m}_{2}}a={{m}_{2}}g-T\] | \[T=\frac{{{m}_{1}}{{m}_{2}}(1+\sin \theta )}{{{m}_{1}}+{{m}_{2}}}g\] | |
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\[T-{{m}_{1}}g\sin \alpha ={{m}_{1}}a\] | \[a=\frac{({{m}_{2}}\sin \beta -{{m}_{1}}\sin \alpha )}{{{m}_{1}}+{{m}_{2}}}g\] |
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\[{{m}_{2}}a={{m}_{2}}g\sin \beta -T\] | \[T=\frac{{{m}_{1}}{{m}_{2}}(\sin \alpha +\sin \beta )}{{{m}_{1}}+{{m}_{2}}}g\] |
| Condition | Free body diagram | Equation | Tension and acceleration |
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\[{{m}_{1}}g\sin \theta -T={{m}_{1}}a\] | \[a=\frac{{{m}_{1}}g\sin \theta }{{{m}_{1}}+{{m}_{2}}}\] |
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\[T={{m}_{2}}a\] | \[T=\frac{2{{m}_{1}}{{m}_{2}}}{4{{m}_{1}}+{{m}_{2}}}g\] | |
 As \[\frac{{{d}^{2}}({{x}_{2}})}{d{{t}^{2}}}\] \[=\frac{1}{2}\frac{{{d}^{2}}({{x}_{1}})}{d{{t}^{2}}}\] \[\therefore \,\,{{a}_{2}}=\frac{{{a}_{1}}}{2}\] \[{{a}_{1}}=\] acceleration of block A \[{{a}_{2}}=\] acceleration of block B |
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\[T={{m}_{1}}a\] | \[{{a}_{1}}=a=\frac{2{{m}_{2}}g}{4{{m}_{1}}+{{m}_{2}}}\] |
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\[{{m}_{2}}\frac{a}{2}={{m}_{2}}g-2T\] | \[{{a}_{2}}=\frac{{{m}_{2}}g}{4{{m}_{1}}+{{m}_{2}}}\] \[T=\frac{2{{m}_{1}}{{m}_{2}}g}{4{{m}_{1}}+{{m}_{2}}}\] | |
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\[{{m}_{1}}a={{m}_{1}}g-{{T}_{1}}\] | \[a=\frac{({{m}_{1}}-{{m}_{2}})}{[{{m}_{1}}+{{m}_{2}}+M]}g\] |
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\[{{m}_{2}}a={{T}_{2}}-{{m}_{2}}g\] | \[{{T}_{1}}=\frac{{{m}_{1}}(2{{m}_{2}}+M)}{[{{m}_{1}}+{{m}_{2}}+M]}g\] | |
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\[{{T}_{1}}-{{T}_{2}}=Ma\] | \[{{T}_{2}}=\frac{{{m}_{2}}(2{{m}_{2}}+M)}{[{{m}_{1}}+{{m}_{2}}+M]}g\] |
Motion of massive string
| Condition | Free body diagram | Equation | Tension and acceleration |
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 force applied by the string on the block |
\[F=(M+m)a\] \[{{T}_{1}}=Ma\] | \[a=\frac{F}{M+m}\] \[{{T}_{1}}=M\frac{F}{(M+m)}\] |
 Tension at mid point of the rope |
\[{{T}_{2}}=\left( M+\frac{m}{2} \right)\,a\] | \[{{T}_{2}}=\frac{(2M+m)}{2(M+m)}F\] | |
 m = Mass of string T = Tension in string at a distance x from the end where the force is applied |
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\[F=ma\] | \[a=F/m\] |
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\[T=m\,\left( \frac{L-x}{L} \right)\,a\] | \[T=\left( \frac{L-x}{L} \right)\,F\] | |
 M = Mass of uniform string L = Length of string |
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\[{{F}_{1}}-T=\frac{Mxa}{L}\] | \[a=\frac{{{F}_{1}}-{{F}_{2}}}{M}\] |
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\[{{F}_{1}}-{{F}_{2}}=Ma\] | \[T={{F}_{1}}\left( 1-\frac{x}{L} \right)+{{F}_{2}}\left( \frac{x}{L} \right)\] | |
 Mass of segment BC \[=\left( \frac{M}{L} \right)x\] |
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\[T'=\frac{M}{L}(L-x)g+T\] \[T=F+\frac{M}{L}xg\] | \[T'=F+Mg\] \[T=F+\frac{M}{L}xg\] |
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