Question
A uniform rope of length $L$, resting on a frictionless horizontal surface is pulled at one end by a force $F$. What is the tension in the rope at a distance 1 from the end where the force is applied?
✓
Answer
Let $M$ be the mass of uniform rope of length $L$.
Then Mass per unit length of rope $=\frac{M}{L}$
Acceleration in the rope $=\frac{F}{M}$
Let T be the tension in the rope at a distance l from the end where the force F is applied.
Mass of length $( L - l )$ of the rope is
$
M^{\prime}=\frac{M}{L}(L-l)
$
As tension T is the only force on the length ( $L - l$ ) of the rope, so
$
T=M^{\prime} \times \frac{F}{M}=\frac{M}{L}(L-l) \times \frac{F}{M}=\left(1-\frac{l}{L}\right) F
$
Then Mass per unit length of rope $=\frac{M}{L}$
Acceleration in the rope $=\frac{F}{M}$
Let T be the tension in the rope at a distance l from the end where the force F is applied.
Mass of length $( L - l )$ of the rope is
$
M^{\prime}=\frac{M}{L}(L-l)
$
As tension T is the only force on the length ( $L - l$ ) of the rope, so
$
T=M^{\prime} \times \frac{F}{M}=\frac{M}{L}(L-l) \times \frac{F}{M}=\left(1-\frac{l}{L}\right) F
$
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