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 0.5. Then, tension at the midway of the rope is:

• A. $\frac{Mg}{4}$
• B. $\frac{3Mg}{4}$
• C. $\frac{Mg}{8}$
• D. $\frac{Mg}{2}$

Answer 1

Answer: (D)

$\frac{Mg}{2}$

a = $\frac{F-\mu N}{M}$ = $\frac{Mg-0.5Mg}{M}$ = $\frac{g}{2}$

T - $\mu$$\frac{Mg}{2}$ = $\frac{Ma}{2}$

T - $\frac{Mg}{4}$ = $\frac{Mg}{4}$

Hence T = $\frac{Mg}{2}$