Q. A rope of length $L$ is pulled by a constant force $F$. What is the tension in the rope at distance $x$ from the end when the force is applied?
A) $\frac{F(L - x)}{L}$
B) $\frac{FL}{L - x}$
C) $\frac{FL}{x}$
D) $\frac{Fx}{L - x}$
Explanation
(a) $F - F_{1} = (dm)a$
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Acceleration of rope $a = \frac{F}{m}$
$dm = \frac{m}{L}x$
$F - F_{1} = \frac{mx}{L} \times \frac{F}{m}$
$F - F_{1} = \frac{Fx}{L}$ or $F_{1} = \frac{F(L - x)}{L}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.