Consider an inclined plane whose upper half is rough while lower half is smooth. Friction of coefficient of the rough surface is 0.8 . Angle of inclination is 37°. A heavy rope of mass m is placed on its inclined plane such that it remains in equilibrium. What minimum fraction of rope is required on tough surface of equilibrium ? Ans is 15/16....but please explain how?

Dear Student,

Please find below the solution to the asked query:

The situation is as shown in the figure.



Here, let "x" is the length of the rope on the rough surface, and "L" is the length of the rope.

Now, the force acting down the plane by gravity,

$F_{down}=mg \sin \theta = mg \sin {37}^0$

The friction force acts up the plane, the friction force is only due to the part on the rough surface. So,


$F_{fri}=\mu m\text{'}g \cos \theta = \mu \left(\frac{mx}{L}\right)g \cos {37}^0$
For equilibrium condition, these two forces should be equal. Therefore,

$$\begin{gathered} F_{down}=F_{fri} \Rightarrow \mu \left(\frac{mx}{L}\right)g \cos {37}^0=mg \sin {37}^0 \Rightarrow \mu \left(\frac{x}{L}\right) cos {37}^0= sin {37}^0 \\ \Rightarrow \frac{x}{L}=\frac{\tan {37}^0}{\mu } \Rightarrow \frac{x}{L}=\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{0.8} \Rightarrow \frac{x}{L}=\frac{3}{3.2} \Rightarrow \frac{x}{L}=\frac{30}{32} \\ \Rightarrow \frac{x}{L}=\frac{15}{16} \end{gathered}$$

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