Question

A rope of length $L$ has mass per unit length $\lambda$ which varies according to the function $\lambda \left(x\right)=e^{x/L}.$ The rope is pulled by a constant force of $1\text{N}$ on a smooth horizontal surface. The tension in the rope at $x=\frac{L}{2}$ is:

• A. $0.5\text{N}$
• B. $0.38\text{N}$
• C. $0.62\text{N}$
• D. None

Answer 1

The correct option is B $0.38\text{N}$


Total mass of rope,
$M={\int }_0^L\lambda \left(x\right)dx$
$\Rightarrow M={\int }_0^L{e}^{x/L}dx$
$\Rightarrow M=\frac{1}{L}[e^{x/L}{]}_0^L=\frac{1}{L}(e-1)$

Acceleration of rope, $a=\frac{F}{M}=\frac{1}{\frac{(e-1)}{L}}=\frac{L}{e-1}$
which is the same for all points, since the rope moves as a system.

Mass of the rope from $0$ to $L/2$ is
$m={\int }_0^{L/2}{e}^{x/L}dx=\frac{1}{L}[e^{x/L}{]}_0^{L/2}$
$\Rightarrow m=\frac{e^{1/2}-1}{L}$

Using Newton's second law for the left half of the rope at $x=L/2$,
$T=ma=\frac{e^{1/2}-1}{L}\times \frac{L}{e-1}=\frac{\sqrt{2.7}-1}{2.7-1}=0.38\text{N}$