JEE Main & Advanced Physics NLM, Friction, Circular Motion Calculation of Required Force in Different Situation

Calculation of Required Force in Different Situation

Category : JEE Main & Advanced

If W = weight of the body, \[\theta =\] angle of friction, \[\mu =\tan \theta =\]coefficient of friction

Then we can calculate required force for different situation in the following manner : 

(1) Minimum pulling force P at an angle \[\alpha \] from the horizontal

By resolving P in horizontal and vertical direction (as shown in figure)

For the condition of equilibrium

\[F=P\cos \alpha \] and \[R=W-P\sin \alpha \]

By substituting these value in

\[F=\mu R\] \[P\cos \alpha =\mu \,(W-P\sin \alpha )\]

\[\Rightarrow \] \[P\cos \alpha =\frac{\sin \theta }{\cos \theta }\,\,(W-P\sin \alpha )\]              [As \[\mu =\tan \theta \]]

\[\Rightarrow \] \[P=\frac{W\sin \theta }{\cos \,(\alpha -\theta )}\]

(2) Minimum pushing force P at an angle \[\alpha \] from the horizontal

By Resolving P in horizontal and vertical direction (as shown in the figure)

For the condition of equilibrium

\[F=P\cos \alpha \] and  \[R=W+P\sin \alpha \]

By substituting these value in \[F=\mu R\]

\[\Rightarrow \] \[P\cos \alpha =\mu \,(W+P\sin \alpha )\]

\[\Rightarrow \] \[P\cos \alpha =\frac{\sin \theta }{\cos \theta }\,(W+P\sin \alpha )\]  [As \[\mu =\tan \theta \]]

\[\Rightarrow \] \[P=\frac{W\sin \theta }{\cos \,(\alpha +\theta )}\]

(3) Minimum pulling force P to move the body up on an inclined plane

By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure)

 

For the condition of equilibrium

\[R+P\sin \alpha =W\cos \lambda \]  

\[\therefore \] \[R=W\cos \lambda -P\,\sin \alpha \] and \[F+W\sin \lambda =P\cos \alpha \]

\[\therefore \] \[F=P\cos \alpha -W\sin \lambda \]

By substituting these values in\[F=\mu R\] and solving we get

\[P=\frac{W\sin \,(\theta +\lambda )}{\cos \,(\alpha -\theta )}\]

(4) Minimum force to move a body in downward direction along the surface of inclined plane

By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure)

For the condition of equilibrium

\[R+P\sin \alpha =W\cos \lambda \]

\[\therefore \] \[R=W\cos \lambda -P\sin \alpha \] and      \[F=P\cos \alpha +W\sin \lambda \]

By substituting these values in \[F=\mu R\] and solving we get 

\[P=\frac{W\sin (\theta -\lambda )}{\cos \,(\alpha -\theta )}\] 

(5) Minimum force to avoid sliding of a body down on an inclined plane

By Resolving P in the direction of the plane and perpendicular to the plane (as shown in the figure)

For the condition of equilibrium

\[R+P\sin \alpha =W\cos \lambda \]

\[\therefore \] \[R=W\cos \lambda -P\sin \alpha \] and             \[P\cos \alpha +F=W\sin \lambda \]

\[\therefore \] \[F=W\sin \lambda -P\cos \alpha \]

By substituting these values in \[F=\mu R\] and solving we get

\[P=W\,\left[ \frac{\sin \,(\lambda -\theta )}{\cos \,(\theta +\alpha )} \right]\] 

(6) Minimum force for motion along horizontal surface and its direction  

         

Let the force P be applied at an angle\[\alpha \]with the horizontal.

By resolving P in horizontal and vertical direction (as shown in figure)              

For vertical equilibrium     

\[R+P\sin \alpha =mg\]

\[\therefore \] \[R=mg-P\sin \alpha \]                                                          ...(i)

and for horizontal motion       

\[P\cos \alpha \ge F\]

i.e. \[P\cos \alpha \ge \mu R\]                                                    ...(ii)

Substituting value of R from (i) in (ii)     

\[P\cos \alpha \ge \mu \,(mg-P\sin \alpha )\]     

\[P\ge \frac{\mu \,mg}{\cos \alpha +\mu \sin \alpha }\]                                   ...(iii)

For the force P to be minimum \[(\cos \alpha +\mu \sin \alpha )\] must be maximum i.e.

\[\frac{d}{d\alpha }[\cos \alpha +\mu \sin \alpha ]=0\]  

\[\Rightarrow \] \[-\sin \alpha +\mu \cos \alpha =0\] 

\[\therefore \]  \[\tan \alpha =\mu \]    

or     \[\alpha ={{\tan }^{-1}}(\mu )=\text{angle of friction }\]

i.e. For minimum value of P its angle from the horizontal should be equal to angle of friction

As \[\tan \alpha =\mu \] so from the figure, \[\sin \alpha =\frac{\mu }{\sqrt{1+{{\mu }^{2}}}}\]

and  \[\cos \alpha =\frac{1}{\sqrt{1+{{\mu }^{2}}}}\]

By substituting these value in equation (iii)

\[P\ge \frac{\mu \,mg}{\frac{1}{\sqrt{1+{{\mu }^{2}}}}+\frac{{{\mu }^{2}}}{\sqrt{1+{{\mu }^{2}}}}}\]

\[\ge \frac{\mu \,mg}{\sqrt{1+{{\mu }^{2}}}}\]

\[\therefore \]  \[{{P}_{\min }}=\frac{\mu mg}{\sqrt{1+{{\mu }^{2}}}}\]  

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