Question

A small body $A$ starts sliding down from the top of a wedge as shown in figure above, whose base is equal to $l=2.10m$. The coefficient of friction between the body and the wedge surface is $k=0.140$. At what value of the angle $\alpha$ will the time of sliding be the least?

Answer 1

Let $a$ be the acceleration of block when sliding and distance traveled by the block A from top of wedge to bottom of wedge.
i.e $d=l\sec \alpha$
Friction force, $f=kN=kmg\cos \alpha$
Therefore, $mg\sin \alpha -kmg\cos \alpha =ma........\left(1\right)$
$a=g\sin \alpha -kg\cos \alpha$

Now, from kinematical equation: $(d=ut+\frac{1}{2}a{t}^2)$
$l\sec \alpha =0+\left(\frac{1}{2}\right)a{t}^2$

$t=\sqrt{\frac{2l\sec \alpha }{(\sin \alpha -k\cos \alpha )}g}$ (using equation (1))

$=\sqrt{\frac{2l}{(\frac{\sin 2\alpha }{2}-k{\cos }^2\alpha )}g}......\left(2\right)$

For $t_{min}$, $\frac{d(\frac{\sin 2\alpha }{2}-k{\cos }^2\alpha )}{d\alpha }=0$

i.e. $\frac{2\cos 2\alpha }{2}+2k\cos \alpha \sin \alpha =0$

$\cos 2\alpha +k\sin 2\alpha =0$

$\tan 2\alpha =-\frac{1}{k}⟹\alpha ={49}^{\circ }$

Putting the values of $\alpha$, $k$ and $l$ in equation (2) we get $t_{min}=1s$