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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 ?

Laws of Motion

Solution:

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