JEE Main & Advanced Physics NLM, Friction, Circular Motion Minimum Mass Hung From String

Minimum Mass Hung From String (1) When a mass m1 placed on a rough horizontal plane: Another mass \[{{m}_{2}}\] hung from the string connected by pulley, the tension (T) produced in string will try to start the motion of mass \[{{m}_{1}}\]. At limiting condition \[T={{F}_{l}}\] \[\Rightarrow \,\,\,\,\,{{m}_{2}}g=\mu R\] \[\Rightarrow \,\,\,\,\,{{m}_{2}}g=\mu \,{{m}_{1}}g\] \[\therefore \,\,\,\,{{m}_{2}}=\mu {{m}_{1}}\] this is the minimum value of \[{{m}_{2}}\] to start the motion. Note: q In the above condition Coefficient of friction \[\mu =\frac{{{m}_{2}}}{{{m}_{1}}}\] (2) When a mass m1 placed on a rough inclined plane: Another mass \[{{m}_{2}}\] hung from the string connected by pulley, the tension (T) produced in string will try to start the motion of mass \[{{m}_{1}}\]. At limiting condition For \[{{m}_{2}}\,\,\,\,\,\,\,T={{m}_{2}}g\]                     ...... (i) For  \[{{m}_{1}}\,\,\,\,\,\,\,T={{m}_{1}}g\sin \theta +F\] Þ \[T={{m}_{1}}g\sin \theta +\mu R\] \[\Rightarrow \,\,\,\,\,T={{m}_{1}}g\sin \theta +\mu {{m}_{1}}g\cos \theta \]                     ...... (ii)  From equation (i) and (ii) \[{{m}_{2}}={{m}_{1}}[\sin \theta +\mu \cos \theta ]\] this is the minimum value of \[{{m}_{2}}\] to start the motion Note: q In the above condition Coefficient of friction \[\mu =\left[ \frac{{{m}_{2}}}{{{m}_{1}}\cos \theta }-\tan \theta  \right]\] Sample problems based on hung mass Problem 23. Two blocks of mass \[{{M}_{1}}\,\,and\,\,{{M}_{2}}\] are connected with a string passing over a pulley as shown in the figure. The block \[{{M}_{1}}\] lies on a horizontal surface. The coefficient of friction between the block \[{{M}_{1}}\] and horizontal surface is \[\mu \]. The system accelerates. What additional mass m should be placed on the block \[{{M}_{1}}\] so that the system does not accelerate.                           (a) \[\frac{{{M}_{2}}-{{M}_{1}}}{\mu }\]                                 (b) \[\frac{{{M}_{2}}}{\mu }-{{M}_{1}}\]             (c) \[{{M}_{2}}-\frac{{{M}_{1}}}{\mu }\]                                 (d) \[({{M}_{2}}-{{M}_{1}})\mu \] Solution: (b) By comparing the given condition with general expression             \[\mu =\frac{{{M}_{2}}}{m+{{M}_{1}}}\Rightarrow m+{{M}_{1}}=\frac{{{M}_{2}}}{\mu }\Rightarrow m=\frac{{{M}_{2}}}{\mu }-{{M}_{1}}\] Problem 24. The coefficient of kinetic friction is 0.03 in the diagram where mass \[{{m}_{2}}=20\,kg\] and \[{{m}_{1}}=4\,kg\]. The acceleration of the block shall be \[(g=10m{{s}^{-2}})\]             (a) \[1.8\,\,m{{s}^{-2}}\]          (b) \[0.8\,\,m{{s}^{-2}}\]          (c) \[1.4\,\,m{{s}^{-2}}\]           (d) \[0.4\,\,m{{s}^{-2}}\]                         Solution: (c) Let the acceleration of the system is a                         From the F.B.D. of \[{{m}_{2}}\]                                                 \[T-F={{m}_{2}}a\Rightarrow T-\mu {{m}_{2}}g={{m}_{2}}a\]             \[\Rightarrow T-0.03\times 20\times 10=20\,a\Rightarrow T-6=20a\] ..... (i)             From the FBD of \[{{m}_{1}}\]             \[{{m}_{1}}g-T={{m}_{1}}a\]                         \[\Rightarrow 4\times 10-T=4a\Rightarrow 40-T=4a\]          .... (ii)             Solving (i) and (ii) \[a=1.4m/{{s}^{2}}.\]