Uniform rope of mass M and length L is placed on a smooth horizontal surface. A horizontal force F is acting at one end of rope.calculate the tension in the rope at a distance x

Let the size of Force applied by the segment of the rope to one side of position x be F(x)F, and F(x+dx) be the greatness of the drive applied by the rope to one side of position (x+dx). At that point, the net drive following up on component dxof the rope is F(x)−F(x+dx), which by Newton`s second law of movement is equivalent to M/LdxF/M, i.e., F(x)−F(x+dx)=MLdxFM or( F(x+dx)−F(x))/dx=−M/LF/M=−F/L So, we have dF/dx=−F/LdF/dx=−F/L, or F(x)=−F/Lx+F(x)=−F/Lx+constant. The consistent is assessed by utilizing the way that at x=0x=0, F(x)=F, the connected constrain. In this way, the drive following up on the rope at a separation x from the end where the constrain FF is connected is :F(x)=−F/Lx+F=F(1−x/L).