A small body A starts sliding down from the top of a wedge (figure) 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? What will it be equal to?

Explanation

Let us designate the x-axis (figure) and apply `F_x=mw_x` for body A:
`mg sin alpha-k mg cos alpha=mw`
or, `w=gsin alpha-k g cos alpha`
Now, from kinematical equation:
`l sec alpha=0+(1//2)wt^2`
or, `t=sqrt(2lsecalpha//(sin alpha-k cos alpha)g)`
`=sqrt(2l//(sin 2alpha//2-k cos^2alpha)g)`
(using Eq. (1)).
for `t_(min')` `(d((sin 2alpha)/(2)-kcos^2alpha))/(dalpha)=0`
i.e. `(2cos 2alpha)/(2)+2k cos alpha sin alpha=0`
or, `tan 2alpha=-1/kimpliesalpha=49^@`
and putting the values of `alpha`, `k` and `l` in Eq. (2) we get `t_(min)=1s`.