Question

Two identical uniform heavy ropes are joint with light string and pulled on a horizontal smooth surface as shown in figure by applying force F at one end of a heavy rope:

• A. Tension in the light string is F/2
• B. Tension at mid-point of rope 1 is 3F/4
• C. Tension at mid-point of rope 2 is F/2
• D. Tension at mid-point of rope 2 is F/4

Answer 1

Answer: (A), (B), (D)

Tension in the light string is F/2, Tension at mid-point of rope 1 is 3F/4, Tension at mid-point of rope 2 is F/4

Answer 2

The problem involves two identical uniform heavy ropes connected by a light string, pulled by a force F on a smooth horizontal surface.

Let:

• Mass of each rope = M
• Length of each rope = L
• Applied force = F

Since the surface is smooth, there is no friction. The light string has negligible mass. The entire system moves with a common acceleration.

1. Calculate the acceleration of the system: The total mass of the system is the sum of the masses of the two ropes, as the string has negligible mass. Total mass ($M_{total}$) = Mass of Rope 1 + Mass of Rope 2 = M + M = 2M. According to Newton's second law, $F = M_{total} \times a$. Therefore, the acceleration of the system ($a$) is: $a = \frac{F}{M_{total}} = \frac{F}{2M}$

2. Tension in the light string ($T_{string}$): The light string connects Rope 1 and Rope 2. The tension in the string is responsible for pulling Rope 2. Consider Rope 2 as a free body. The only horizontal force acting on it is the tension from the string. $T_{string} = M_{rope2} \times a$ $T_{string} = M \times \left(\frac{F}{2M}\right)$ $T_{string} = \frac{F}{2}$ Thus, the first statement "Tension in the light string is F/2" is correct.

3. Tension at the mid-point of Rope 1 ($T_{mid1}$): Rope 1 has mass M and length L. Its midpoint is at L/2 from the point where force F is applied and also L/2 from the string. The tension at the midpoint of Rope 1 is responsible for accelerating the mass of the segment of Rope 1 from its midpoint to the string, plus the mass of the string, plus the mass of Rope 2. Mass of Rope 1 from midpoint to string = $\frac{M}{2}$ (since it's a uniform rope, half the length means half the mass). Mass being pulled by $T_{mid1}$ = (Mass of Rope 1 segment) + (Mass of string) + (Mass of Rope 2) Mass being pulled by $T_{mid1}$ = $\frac{M}{2} + 0 + M = \frac{3M}{2}$ $T_{mid1} = \left(\frac{3M}{2}\right) \times a$ $T_{mid1} = \left(\frac{3M}{2}\right) \times \left(\frac{F}{2M}\right)$ $T_{mid1} = \frac{3F}{4}$ Thus, the second statement "Tension at mid-point of rope 1 is 3F/4" is correct.

4. Tension at the mid-point of Rope 2 ($T_{mid2}$): Rope 2 has mass M and length L. Its midpoint is at L/2 from the string and also L/2 from its free end. The tension at the midpoint of Rope 2 is responsible for accelerating the mass of the segment of Rope 2 from its midpoint to its free end (the leftmost end). Mass of Rope 2 from midpoint to free end = $\frac{M}{2}$. Mass being pulled by $T_{mid2}$ = $\frac{M}{2}$ $T_{mid2} = \left(\frac{M}{2}\right) \times a$ $T_{mid2} = \left(\frac{M}{2}\right) \times \left(\frac{F}{2M}\right)$ $T_{mid2} = \frac{F}{4}$ Thus, the fourth statement "Tension at mid-point of rope 2 is F/4" is correct. The third statement "Tension at mid-point of rope 2 is F/2" is incorrect.

Conclusion: The correct statements are:

1. Tension in the light string is F/2
2. Tension at mid-point of rope 1 is 3F/4
3. Tension at mid-point of rope 2 is F/4