Question

A string of length $L$ and mass $M$ is lying on a horizontal table. A force $F$ is applied at one of its ends. Tension in the string at a distance $y$ from the end at which the force applied is

• A. zero
• B. $F$
• C. $\frac{F(L-y)}{L}$
• D. $\frac{F(L-y)}{M}$

Answer 1

Answer: (C)

Acceleration of the rope, $a=\frac{F}{M}\dots (i)$

Now, considering the motion of the part $AB$ of the rope
[which has mass $(M/L)y$ and acceleration given by eq. $(i)$]
assuming that tension at $B$ is $T$.
$F-T=\left(\frac{M}{L}y\right)\times a\text{ or }F-T=\frac{M}{L}y\times \frac{F}{M}=\frac{Fy}{L}$
$orT=F-F\frac{y}{L}=F\left(1-\frac{y}{L}\right)=\frac{F(L-y)}{L}$