Question

A uniform homogenous rope of mass ${}^{\prime }{m}^{\prime }$ and length ${}^{\prime }{l}^{\prime }$ is pulled by a force $F$ on a smooth inclined plane of angle $\theta$. Find the force on the rope at a distance $x$ from the end where force is applied.

• A. $F\left(\frac{x}{l}\right)$
• B. $F\left(\frac{l}{l-x}\right)$
• C. $F\left(\frac{l-x}{l}\right)$
• D. $2F\frac{(l-x)}{l}$

Answer 1

The correct option is C $F\left(\frac{l-x}{l}\right)$

Let $\lambda =$ mass per unit length of the rope.
$⟹\lambda =\frac{m}{l}$

Let $a$ be the acceleration of the rope on the inclined plane.
For the rope , using Newton's second law
$F-\lambda lg\sin \theta =\lambda la$
$a=\frac{F-\lambda lg\sin \theta }{\lambda l}$ (i)
Let $T$ be the tension in the rope at a distance $x$ from the end where force $F$ is applied.
FBD of $(l-x)$ part of the rope can be drawn as

For $(l-x)$ portion , using Newton's second law,
$T-\lambda (l-x)g\sin \theta =\lambda (l-x)a$ (ii)
Using (i) in (ii), we have
$T=\lambda (l-x)\left[\frac{F-\lambda lg\sin \theta }{\lambda l}\right]+\lambda (l-x)g\sin \theta$
$⟹T=F\left(\frac{l-x}{l}\right)$