# JEE Main & Advanced Physics NLM, Friction, Circular Motion Calculation of Necessary Force

Calculation of Necessary Force

Category : JEE Main & Advanced

Calculation of Necessary Force If W = weight of the body, $\theta = angle of friction$, $\mu =\tan \theta =$ coefficient of friction then we can calculate necessary force for different condition in the following manner: (1) Minimum pulling force P at an angle a 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 a 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 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 on body in downward direction along the surface of inclined plane to start its motion 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 a body down 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       $P\cos \alpha +F=W\sin \lambda$ \                     $F=W\sin \lambda -P\cos \alpha$ 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 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)                         $P\ge \frac{\mu \,mg}{\frac{1}{\sqrt{1+{{\mu }^{2}}}}+\frac{{{\mu }^{2}}}{\sqrt{1+{{\mu }^{2}}}}}\ge \frac{\mu \,mg}{\sqrt{1+{{\mu }^{2}}}}$             $\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{P}_{\min }}=\frac{\mu mg}{\sqrt{1+{{\mu }^{2}}}}$ Sample problems based on force against friction Problem 9. What is the maximum value of the force F such that the block shown in the arrangement, does not move ($\mu =1/2\sqrt{3}$) [IIT-JEE (Screening) 2003]                             (a) 20 N                                     (b) 10 N             (c) 12 N                                     (d) 15 N Solution: (a) Frictional force $f=\mu R$  $\Rightarrow \,\,F\cos 60=\mu (W+F\sin 60)$             $\Rightarrow \,\,\,F\cos 60=\frac{1}{2\sqrt{3}}\,\,\left( \sqrt{3}g+F\sin \,\,60 \right)$ $\Rightarrow \,\,\,F=20\,N$ Problem 10. A block of mass m rests on a rough horizontal surface as shown in the figure. Coefficient of friction between the block and the surface is m. A force F = mg acting at angle $\theta$ with the vertical side of the block pulls it. In which of the following cases the block can be pulled along the surface                         (a) $\tan \theta \ge \mu$                                    (b) $\cot \theta \ge \mu$             (c) $\tan \theta /2\ge \mu$                      (d) $\cot \theta /2\ge \mu$             Solution: (d) For pulling of block $P\ge f$                         $\Rightarrow \,\,\,mg\,\sin \theta \ge \mu R\Rightarrow mg\sin \theta \ge \,\mu \,(mg-mg\cos \theta )$             $\Rightarrow \,\,\sin \theta \ge \,\mu \,(1-\cos \theta )$             $\Rightarrow \,\,\,2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\ge \mu \,\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)\Rightarrow \cot \left( \frac{\theta }{2} \right)\ge \mu$

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