Question

A uniform rope of length L, resting on a frictionless horizontal surface is pulled at one end by a force F . Derive an expression for the tension in the rope as a function of position along its length

Answer 1

Let the magnitude of force exerted by the portion of the rope to the left of position xx be F(x)F(x), and F(x+dx)F(x+dx) be the magnitude of the force exerted by the rope to the right of position (x+dx)(x+dx). Then, the net force acting on element dxdx of the rope is F(x)−F(x+dx)F(x)−F(x+dx), which by Newton’s second law of motion is equal to MLdxFMMLdxFM, i.e., F(x)−F(x+dx)=MLdxFMF(x)−F(x+dx)=MLdxFM, or F(x+dx)−F(x)dx=−MLFM=−FLF(x+dx)−F(x)dx=−MLFM=−FL. So, we have dFdx=−FLdFdx=−FL, or F(x)=−FLx+F(x)=−FLx+constant. The constant is evaluated by using the fact that at x=0x=0, F(x)=FF(x)=F, the applied force. So, the force acting on the rope at a distance x from the end where the force FF is applied is F(x)=−FLx+F=F(1−xL)F(x)=−FLx+F=F(1−xL).

Answer 2

Answer:

Explanation:

Let M be the mass of rope

Acceleration of rope a=FM

Now we divide the rope into two parts by applying a cut at P

T=$\frac{m}{l}$(L−x)a

F−T=M$\frac{m}{l}$xa

=$\frac{m}{l}$(L−x)$\frac{f}{m}$ =$\frac{m}{l}$x$\frac{f}{m}$

=F(1−$\frac{x}{l}$)

T=F(1−$\frac{x}{l}$)