A bar of mass m is pulled by means of a thread up an inclined plane forming an angle `alpha` with the horizontal (figure). The coefficient of friction is equal to k. Find the angle `beta` which is thread must form with the inclined plane for the tension of the thread to be minimum. What is it equal to?

Explanation

Let us fix the `x-y` co-ordinate system to the wedge, taking the x-axis up, along the incline and the y-axis perpendicular to it (figure).
Now, we draw the free body diagram for the bar.
Let us apply Newton's second law in projection from along x and y axis for the bar:
`Tcos beta-mg sin alpha-f r=0` (1)
`Tsin beta+N-mg cos alpha=0`
or, `N=mg cos alpha-T sin beta` (2)
But `f r=kN` and using (2) in (1), we get
`T=mg sin alpha+k mg cos alpha//(cos beta+k sin beta)` (3)
For `T_(min)` the value of `(cos beta+k sin beta)` should be maximum
So, `(d(cos beta+k sin beta))/(d beta)=0` or `tan beta=k`
Putting this value of `beta` in Eq. (3) we get,
`T_(min)=(mg(sin alpha+k cos alpha))/(1//sqrt(1+k^2)+k^2//sqrt(1+k^2))=(mg(sin alpha+kcos alpha))/(sqrt(1+k^2))`