Question

A rope of length $L$ has its mass per unit length $\lambda$ vary 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 $x=L/2$ is: (Take $e=2.7$)

• A. $0.50\text{N}$
• B. $0.38\text{N}$
• C. $0.62\text{N}$
• D. None

Answer 1

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

Considering a small element of length $dx$ at distance $x$ from one end $\left(\text{O}\right)$.


Mass of small element $dm=\lambda dx=e^{x/L}dx$
$\therefore$ Mass of complete rope $M=\underset{x=0}{\overset{x=L}{\int }}dm$
$\Rightarrow M=\underset{0}{\overset{L}{\int }}{e}^{\frac{x}{L}}dx$
$\therefore M=\frac{1}{L}{\left[e^{\frac{x}{L}}\right]}_0^L=\frac{1}{L}(e-1)$

From FBD of rope:


$\therefore a=\frac{F}{M}=\frac{1}{\frac{1}{L}(e-1)}=\frac{L}{e-1}$

For finding tension at $x=\frac{L}{2}$, take the FBD of left half:


Mass of the part $M^{\prime }=\underset{x=0}{\overset{x=L/2}{\int }}dm=\underset{0}{\overset{L/2}{\int }}{e}^{\frac{x}{L}}dx$
$\therefore M^{\prime }=\frac{1}{L}(e^{\frac{1}{2}}-1)$

From Newton's 2nd law in direction of acceleration:
$T=M^{\prime }\times a^{\prime }=M^{\prime }a$
$\Rightarrow T=\frac{(e^{\frac{1}{2}}-1)}{L}\times \frac{L}{(e-1)}$
$\therefore T=\frac{e^{\frac{1}{2}}-1}{e-1}=\frac{\sqrt{2.7}-1}{2.7-1}=0.38\text{N}$