Question

A uniform rope of length $L$, resting on frictionless horizontal surface is pulled at one end by a force $F$. Then the tension $T$ in the rope at a distance $l$ from the end where the force $F$ is applied is

• A. $T=F$
• B. $T=\frac{FL}{L}$
• C. $T=\frac{FI}{L}$
• D. $T=F(1-\frac{1}{L})$

Answer 1

The correct option is D $T=F(1-\frac{1}{L})$

Let tension at $P$ be $T$. Imagine rope to be divided into two parts $A$ and $B$
Mass of rope$=M$
Mass of part $A=M_A=\frac{M}{L}I$
Mass of part $B=M_B=\frac{M}{L}(L-I)$
Let the rope move with acceleration $a$, then for part $A$
$F-T=M_{A}a$
or $F-T=\frac{MIa}{L}...\left(i\right)$
For part $B$ $T-O=M_{B}a$
or $T=\frac{M}{L}(L-I)a...\left(ii\right)$
Adding Eqs (i) and (ii) we get
$F=\frac{M}{L}a(1+L-I)=\frac{M}{L}a.L=Ma\Rightarrow a=\frac{F}{M}....\left(iii\right)$
From Eqs. (i), (ii), (iii)
$T=\frac{M}{L}(L-1).\frac{F}{M}=\frac{M}{L}(L-1)=F(1-\frac{1}{L})$