A rope of length L and mass M is being pulled on a rough horizontal floor by a constant horizontal force F=Mg. The force is acting at one end of the rope in the same direction as the length of the rope. The coefficient of kinetic friction between rope and floor is 1/2 . Then, the tension at the midpoint of the rope is
(a) Mg/4
(b) 2Mg/5
(c) Mg/8
(d) Mg/2
Ans: (d)
Sol: For First half part
$ \displaystyle F- T -\frac{\mu M g}{2} = \frac{M}{2}a $ …(i)
For other half
$ \displaystyle T -\frac{\mu M g}{2} = \frac{M}{2}a $ …(ii)
subtracting (i) & (ii)
F – 2T = 0
T = F/2 = Mg/2