Question

A bar of mass $m$ is pulled by means of a thread up an inclined plane forming an angle $\alpha$ with the horizontal as shown in figure above. The coefficient of friction is equal to $k$. Find the angle $\beta$ which the thread must form with the inclined plane for the tension of the thread to be minimum.

Answer 1

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 shown below).
Now, we draw the free body diagram for the bar.
Let us apply Newton's second law of projection form along x and y axis for the bar:
$T\cos \beta -mg\sin \alpha -fr=0$ (1)
$T\sin \beta +N-mg\cos \alpha =0$
or $N=mg\cos \alpha -T\sin \beta$ (2)
But $f_r=kN$ and using (2) in (1), we get
$T=mg\sin \alpha +\frac{kmg\cos \alpha }{(\cos \beta +k\sin \beta )}$ (3)
For $T_{min}$ the value of $(\cos \beta +k\sin \beta )$ should be maximum
So, $\frac{d(\cos \beta +k\sin \beta )}{d\beta }=0$ or $\tan \beta =k$
Putting this value of $\beta$ in equation (3) we get:
$T_{min}=\frac{mg(\sin \alpha +k\cos \alpha )}{\frac{1}{\sqrt{1+k^2}}+\frac{k^2}{\sqrt{1+k^2}}}=\frac{mg(\sin \alpha +k\cos \alpha )}{\sqrt{1+k^2}}$