Question

can u explain me in detail that how to solve the questions of a rope with mass and they ask to find the tension at some distance x

Answer 1

The tension T at a distance x from the end where the force F is applied is given by: $T = F \left( \frac{L - x}{L} \right)$

Answer 2

To solve problems involving a uniform rope with mass where you need to find the tension at a specific distance 'x' from one end, follow these steps:

General Principle

The tension at any point in a rope is the force required to accelerate the segment of the rope that lies beyond that point (in the direction of motion).

Step-by-Step Solution

Let's consider a uniform rope of total length L and total mass M, pulled by a force F on a smooth horizontal surface. We want to find the tension T at a distance x from the end where the force F is applied.

1. Calculate the acceleration of the entire rope: Since the force F is applied to the entire mass M of the rope, the acceleration a of the rope can be found using Newton's Second Law: $F = M a$ Therefore, the acceleration of the rope is: $a = \frac{F}{M}$ This acceleration is uniform throughout the entire length of the rope.

2. Determine the linear mass density of the rope: The linear mass density (mass per unit length) λ for a uniform rope is constant: $\lambda = \frac{M}{L}$

3. Identify the segment of the rope being pulled by the tension: Imagine a point P at a distance x from the end A where the force F is applied. The tension T at this point P is responsible for pulling the remaining part of the rope, which is the segment from P to the other end B. The length of this remaining segment is (L - x).

   Here, the blue segment is x and the red segment is (L-x).

4. Calculate the mass of this segment: The mass m_segment of the rope segment of length (L - x) is: $m_{segment} = \lambda \times (L - x)$ Substituting the value of λ: $m_{segment} = \frac{M}{L} (L - x)$

5. Apply Newton's Second Law to this segment: The tension T at point P is the force that accelerates this m_segment with acceleration a. $T = m_{segment} \times a$

6. Substitute and simplify to find the tension: Substitute the expressions for m_segment and a into the equation for T: $T = \left( \frac{M}{L} (L - x) \right) \times \left( \frac{F}{M} \right)$ The mass M cancels out: $T = \frac{F}{L} (L - x)$ This can also be written as: $T = F \left( 1 - \frac{x}{L} \right)$

Summary of the Formula:

If a uniform rope of length L is pulled by a force F from one end, the tension T at a distance x from the end where the force is applied is given by: $T = F \left( \frac{L - x}{L} \right)$

Important Considerations:

• Definition of 'x': Always be careful about how x is defined in the problem. In this derivation, x is measured from the end where the force F is applied. If x were measured from the other (free) end, the length of the segment being pulled by tension T would be x, and the formula would become T = F(x/L).
• Vertical Ropes: If the rope is hanging vertically or being pulled vertically, gravity must also be considered. The tension at a point would then also support the weight of the rope segment below it, plus any additional force needed for acceleration.
• Non-uniform Ropes: For non-uniform ropes, the linear mass density λ would be a function of x, and you would need to use integration to find the mass of the segment. However, for typical JEE/NEET problems, ropes are usually assumed to be uniform unless stated otherwise.

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Explanation of the solution:

1. Calculate the common acceleration a of the entire rope using F = Ma.
2. Determine the linear mass density λ = M/L.
3. Identify the segment of the rope (L-x) that the tension T needs to accelerate.
4. Calculate the mass of this segment: m_segment = λ(L-x).
5. Apply Newton's second law to this segment: T = m_segment * a.
6. Substitute a = F/M and m_segment = (M/L)(L-x) to get T = F(L-x)/L.