Question

Find the minimum force 'F' and the angle '$\alpha$' at which it should be applied to a block of mass 'm' kept on a horizontal surface such that the block starts moving. Given that the coefficient of friction between the block and surface is $\mu$.

• A. $F_{min}=\frac{\mu mg}{\sqrt{1+{\mu }^2}},\alpha =ta{n}^{-1}\frac{1}{\mu }$
• B. $F_{min}=\frac{\mu mg}{\sqrt{1+{\mu }^2}},\alpha =ta{n}^{-1}\frac{1}{\mu }$
• C. $F_{min}=\frac{\mu mg}{\sqrt{1+{\mu }^2}},\alpha =ta{n}^{-1}\mu$
• D. None of these

Answer 1

The correct option is C

$F_{min}=\frac{\mu mg}{\sqrt{1+{\mu }^2}},\alpha =ta{n}^{-1}\mu$

Minimum force for motion and its direction

Let the force P be applied at an angle α with the horizontal.

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

For Vertical equillibrium

R + Psin$\alpha$ = mg

$\therefore R=mg-Psin\alpha$ ..........(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) and (ii)

P cos$\alpha \ge \mu (mg-Psin\alpha )$

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.,

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

$\therefore tan\alpha =\mu or\alpha =tan-1\left(\mu \right)=$angle of friction

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

As $tan\alpha =\mu$ so from the figure $sin\alpha =\frac{\mu }{\sqrt{1+{\mu }^2}}andcos\alpha =\frac{1}{\sqrt{1+{\mu }^2}}$

By substituting these values in equation (iii)

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

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