JEE Main & Advanced Physics NLM, Friction, Circular Motion Motion of Two Bodies One Resting on the Other

Motion of Two Bodies One Resting on the Other When a body A of mass m is resting on a body B of mass M then two conditions are possible (1) A force F is applied to the upper body,      (2) A force F is applied to the lower body We will discuss above two cases one by one in the following manner: (1) A force F is applied to the upper body, then following four situations are possible (i) When there is no friction (a) The body A will move on body B with acceleration (F/m).                         \[{{a}_{A}}=F/m\] (b) The body B will remain at rest                         \[{{a}_{B}}=0\] (c) If L is the length of B as shown in figure A will fall from B after time t                               \[t=\sqrt{\frac{2L}{a}}=\sqrt{\frac{2mL}{F}}\]              \[\left[ \text{As}\,\,\text{s}=\frac{\text{1}}{\text{2}}a\,{{t}^{2}}\,\text{and}\,\,a=F/m\, \right]\] (ii) If friction is present between A and B only and applied force is less than limiting friction (F < Fl) (F = Applied force on the upper body, \[{{\operatorname{F}}_{1}}=\] limiting friction between A and B, \[{{F}_{k}}=\] Kinetic friction between A and B) (a) The body A will not slide on body B till \[F<{{F}_{l}}\] i.e. \[F<{{\mu }_{s}}mg\] (b) Combined system (m + M) will move together with common acceleration \[{{a}_{A}}={{a}_{B}}=\frac{F}{M+m}\] (iii) If friction is present between A and B only and applied force is greater than limiting friction \[\left( F > {{F}_{1}} \right)\] In this condition the two bodies will move in the same direction (i.e. of applied force) but with different acceleration. Here force of kinetic friction \[{{\mu }_{k}}mg\] will oppose the motion of A while will cause the motion of B.

\[F-{{F}_{k}}=m\,{{a}_{A}}\]            i.e.   \[{{a}_{A}}=\frac{F-{{F}_{k}}}{m}\]                    \[{{a}_{A}}=\frac{(F-{{\mu }_{k}}mg)}{m}\]

Free body diagram of A

\[{{F}_{k}}=M\,{{a}_{B}}\] i.e.        \[{{a}_{B}}=\frac{{{F}_{k}}}{M}\] \         \[{{a}_{B}}=\frac{{{\mu }_{k}}mg}{M}\]

Free body diagram of B

Note: q As both the bodies are moving in the same direction.             Acceleration of body A relative to B will be  \[a={{a}_{A}}-{{a}_{B}}=\frac{MF-{{\mu }_{k}}mg\,(m+M)}{mM}\]             So, A will fall from B after time \[t=\sqrt{\frac{2L}{a}}=\sqrt{\frac{2\,m\,ML}{MF-{{\mu }_{k}}mg\,(m+M)}}\] (iv) If there is friction between B and floor (where \[{{{F}'}_{l}}={\mu }'\,(M+m)\,g\]= limiting friction between B and floor, \[{{\operatorname{F}}_{k}}=\] kinetic friction between A and B) B will move only if \[{{F}_{k}}>{{{F}'}_{l}}\] and then \[{{F}_{k}}-{{{F}'}_{l}}=M\,{{a}_{B}}\] However if B does not move then static friction will work (not limiting friction) between body B and the floor i.e. friction force = applied force \[\left( = {{F}_{k}} \right)\] not \[{{{F}'}_{l}}\]. (2) A force F is applied to the lower body, then following four situations are possible (i) When there is no friction (a) B will move with acceleration (F/M) while A will remain at rest (relative to ground) as there is no pulling force on A.                         \[{{a}_{B}}=\left( \frac{F}{M} \right)\] and \[{{a}_{A}}=0\] (b) As relative to B, A will move backwards with acceleration (F/M) and so will fall from it in time t. \[\therefore \,\,\,\,\,\,t=\sqrt{\frac{2L}{a}}=\sqrt{\frac{2ML}{F}}\] (ii) If friction is present between A and B only and \[\operatorname{F}'< {{F}_{1}}\] (where F¢ = Pseudo force on body A and  Fl = limiting friction between body A and B) (a) Both the body will move together with common acceleration \[a=\frac{F}{M+m}\] (b) Pseudo force on the body A,   \[{F}'=ma=\frac{mF}{m+M}\] and \[{{F}_{l}}={{\mu }_{s}}mg\] (c) \[{F}'<{{F}_{l}}\]  Þ  \[\frac{mF}{m+M}<{{\mu }_{s}}mg\]  Þ \[F<{{\mu }_{s}}(m+M)\,g\] So both bodies will move together with acceleration  \[{{a}_{A}}={{a}_{B}}=\frac{F}{m+M}\] if \[F<{{\mu }_{s}}[m+M]\,g\] (iii) If friction is present between A and B only and F > Fl¢ (where Fl¢ = ms (m + M)g = limiting friction between body B and surface) Both the body will move with different acceleration. Here force of kinetic friction \[{{\mu }_{k}}mg\] will oppose the motion of B while will cause the motion of A.  

\[m{{a}_{A}}={{\mu }_{k}}mg\] i.e. \[{{a}_{A}}={{\mu }_{k}}g\]

Free body diagram of A

\[F-{{F}_{k}}=M{{a}_{B}}\] i.e.   \[{{a}_{B}}=\frac{[F-{{\mu }_{k}}mg]}{M}\]

Free body diagram of B

Note: q As both the bodies are moving in the same direction             Acceleration of body A relative to B will be             \[a={{a}_{A}}-{{a}_{B}}=-\left[ \frac{F-{{\mu }_{k}}g(m+M)}{M} \right]\]             Negative sign implies that relative to B, A will move backwards and will fall it after time             \[t=\sqrt{\frac{2L}{a}}=\sqrt{\frac{2ML}{F-{{\mu }_{k}}g(m+M)}}\] (iv) If there is friction between B and floor: The system will move only if \[F>F_{l}^{'}\] then replacing F by \[F-{{{F}'}_{l}}\]. The entire case (iii) will be valid. However if \[F<{{{F}'}_{l}}\] the system will not move and friction between B and floor will be F while between A and B is zero. Sample problems based on body resting on another Problem 19. A 4 kg block A is placed on the top of a 8 kg block B which rests on a smooth table. A just slips on B when a force of 12 N is applied on A. Then the maximum horizontal force on B to make both A and B move together, is (a) 12 N     (b) 24 N             (c) 36 N             (d) 48 N             Solution: (c) Maximum friction i.e. limiting friction between A and B, \[{{F}_{l}}\] = 12 N.             If F is the maximum value of force applied on lower body such that both body move together             It means Pseudo force on upper body is just equal to limiting friction             \[F'={{F}_{l}}\,\,\,\Rightarrow \] \[m\left( \frac{F}{m+M} \right)=\]\[\left( \frac{4}{4+8} \right)\,F=12\] \[\therefore \,F=36N.\] Problem 20. A body A of mass 1 kg rests on a smooth surface. Another body B of mass 0.2 kg is placed over A as shown. The coefficient of static friction between A and B is 0.15. B will begin to slide on A if A is pulled with a force greater than             (a) 1.764 N                                (b) 0.1764 N             (c) 0.3 N                                    (d) It will not slide for any F             Solution: (a) B will begin to slide on A if Pseudo force is more than limiting friction             \[F'>{{F}_{l}}\Rightarrow m\left( \frac{F}{m+M} \right)>{{\mu }_{s}}R\Rightarrow m\left( \frac{F}{m+M} \right)>0.15\,mg\therefore \,\,\,F>1.764\,N\] Problem 21. A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a horizontal floor. The coefficient of friction between A and B is 0.2, while that between B and floor is 0.5. When a horizontal force of 25 N is applied on the block B, the force of friction between A and B is [IIT - JEE 1993]             (a) Zero             (b) 3.9 N            (c) 5.0 N            (d) 49 N Solution: (a) Limiting friction between the block B and the surface             \[{{F}_{BS}}={{\mu }_{BS}}.R\]\[=0.5\left( m+M \right)\,g\] \[=0.5(2+8)10\]\[=50N\] but the applied force is 25 N so the lower block will not move i.e. there is no pseudo    force on upper block A. Hence there will be no force of friction between A and B.