Q. A horizontal rope of length $y$ is pulled by a constant horizontal force $F$. What is the tension at a distance $x$ from the end where the force applied ?

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A

$\frac{F(y-x)}{y}$

B

$\frac{F \cdot y}{y-x}$

C

$\frac{F . y}{x}$

D

$\frac{F \cdot y}{y}$

Solution:

Let mass of rope is $M$
Acceleration of system, $a=\frac{F}{m}$
Tension at a distance $x$ from the end,
$T=\frac{(y-x) M a}{y}$
$T=\frac{(y-x)}{y} M \cdot \frac{F}{M}=\frac{(y-x) F}{y}$

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