Question
A horizontal uniform rope of length $$L,$$ resting on a frictionless horizontal surface, is pulled at one end by force $$F.$$ What is the tension in the rope at a distance $$l$$ from the end where the force is applied?
A.
$$F\left( {1 - \frac{l}{L}} \right)$$
B.
$$2F\left( {1 - \frac{l}{{2L}}} \right)$$
C.
$$\frac{F}{L}$$
D.
$$\frac{F}{l}\left( {1 - \frac{l}{L}} \right)$$
Answer :
$$F\left( {1 - \frac{l}{L}} \right)$$
Solution :
Let $$a$$ be the acceleration of the rope and $$M$$ be its total mass. Then
$$\eqalign{ & T = \frac{M}{L}\left( {L - \ell } \right)a\,......\left( {\text{i}} \right) \cr & {\text{and}}\,F - T = \frac{M}{L} \times \ell a\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) and (ii)
$$\eqalign{ & \frac{{F - T}}{T} = \frac{\ell }{{L - \ell }} \Rightarrow F\left( {L - \ell } \right) - T\left( {L - \ell } \right) = T\ell \cr & \Rightarrow F\left( {L - \ell } \right) = T\left( {L - \ell + \ell } \right) = T \times L \cr & \Rightarrow T = F\left( {1 - \frac{\ell }{L}} \right) \cr} $$
Let $$a$$ be the acceleration of the rope and $$M$$ be its total mass. Then
$$\eqalign{ & T = \frac{M}{L}\left( {L - \ell } \right)a\,......\left( {\text{i}} \right) \cr & {\text{and}}\,F - T = \frac{M}{L} \times \ell a\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) and (ii)
$$\eqalign{ & \frac{{F - T}}{T} = \frac{\ell }{{L - \ell }} \Rightarrow F\left( {L - \ell } \right) - T\left( {L - \ell } \right) = T\ell \cr & \Rightarrow F\left( {L - \ell } \right) = T\left( {L - \ell + \ell } \right) = T \times L \cr & \Rightarrow T = F\left( {1 - \frac{\ell }{L}} \right) \cr} $$
