# 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)

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

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

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

#### Other Topics

You need to login to perform this action.
You will be redirected in 3 sec