Question

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 $\frac{1}{2}$. Then, the tension at the mid-point of the rope is

• A. Mg/4
• B. 2Mg/5
• C. Mg/8
• D. Mg/2

Answer 1

The correct option is D Mg/2

As the center of mass of the rope is at the mid point of it, the tension is equal to the $Appliedforce-Frictionforce$.

$\Rightarrow T=F-\mu N=Mg-\frac{1}{2}Mg=\frac{1}{2}Mg$

Ans:(D)