Two blocks of masses as 2.9 kg and 1.9 kg are suspended from a rigid support S by two inextensible w

Q: Wait, why does the tension change along the length of the lower wire?

Great question! In a massless string, tension is uniform. But when a wire has mass, the top of the wire has to pull up not just the block below it, but also the weight of the wire itself! As you move down the wire, there is less wire mass hanging below that specific point, so the tension decreases.

Q: Why did we use $g = 9.8$ instead of $10$?

Always look at the numbers given in the problem! The problem explicitly states "The acceleration due to gravity is $9.8 \text{ m/s}^2$". Furthermore, notice how $9.8 + 0.2$ perfectly adds up to $10.0$, making the calculations incredibly smooth. This is a classic exam setter's trick!

Q: For part (b), why did we include the entire mass of the lower wire, but only half the mass of the upper wire?

Because we "cut" the system at the mid-point of the upper wire. Everything physically located below that cut is part of the system. This includes the bottom half of the upper wire, the top block, the *entire* lower wire, and the bottom block.

Q: Why can't we just find the tension at the top and bottom of the wire and take the average?

Actually, for a uniform wire accelerating linearly, the tension *does* vary linearly from top to bottom. So, the tension at the midpoint is indeed the exact average of the tension at the top and the bottom! However, defining a system at the midpoint and using $F=ma$ is a much faster and less error-prone method.

Q: What if the whole system was accelerating downwards at $0.2 \text{ m/s}^2$?

If the system was accelerating downwards, the net force would be downwards. The equation would change from $T - mg = ma$ to $mg - T = ma$, which means $T = m(g - a)$. The tension would be less than the static weight because the support is effectively "yielding" to gravity.

Question

Two blocks of masses as $2.9	ext{ kg}$ and $1.9	ext{ kg}$ are suspended from a rigid support $S$ by two inextensible wires each of length $1	ext{ m}$ (see figure). The upper wire has negligible mass and the lower wire has a uniform mass of $0.2	ext{ kg/m}$. The whole system of blocks, wires and support have an upward acceleration of $0.2	ext{ m/s}^2$. The acceleration due to gravity is $9.8	ext{ m/s}^2$. (1989)

(a) Find the tension at the mid-point of the lower wire.
(b) Find the tension at the mid-point of the upper wire.

Sigma Percentile · Accepted Answer

The Art of Choosing Your System: Mastering Tension in Accelerating Strings

Have you ever looked at a complex physics problem with multiple blocks, heavy ropes, and accelerating frames, and felt a wave of panic? You are not alone. The key to dismantling these seemingly terrifying setups lies in one of the most powerful conceptual tools in classical mechanics: **The System Choice**.

In this thrilling problem, we are presented with a vertical arrangement. Two blocks of masses $2.9 	ext{ kg}$ and $1.9 	ext{ kg}$ are suspended from a rigid support by two wires. The entire assembly is accelerating upwards at $a = 0.2 	ext{ m/s}^2$. The twist? The upper wire is an idealized "massless" string, but the lower wire has a uniform linear mass density of $0.2 	ext{ kg/m}$.

Our mission is to find the tension at the exact mid-points of both the lower and upper wires. Let's embark on this journey and see how a simple shift in perspective makes the math beautifully elegant.

### The Trap of the Heavy Wire

When a wire has mass, the tension is no longer uniform throughout its length. Why? Because each segment of the wire must support the weight of all the segments below it, plus any attached blocks, *and* provide the force necessary to accelerate them.

The lower wire has a length of $1 	ext{ m}$ and a linear mass density $\lambda = 0.2 	ext{ kg/m}$. 
The total mass of this wire is simply:
$$m_{	ext{wire}} = \lambda 	imes L = 0.2 	ext{ kg/m} 	imes 1 	ext{ m} = 0.2 	ext{ kg}$$

This might seem like a small number, but in the rigorous world of JEE and NEET, ignoring it is a fatal error.

### Part A: The Mid-Point of the Lower Wire

We need the tension $T_1$ exactly at the middle of the lower wire. Imagine taking a pair of conceptual scissors and snipping the wire right at that mid-point.

What happens to everything below the cut? It would fall! To prevent it from falling, the upper half of the wire must exert an upward force on the lower half. This force is precisely the tension $T_1$.

By isolating everything below the cut, we define our **first system**. What does this system contain?
1. The $1.9 	ext{ kg}$ block.
2. The lower half of the lower wire.

Since the whole wire is $0.2 	ext{ kg}$, half of it is $0.1 	ext{ kg}$. 
The total mass of our isolated system is:
$$m_{	ext{sys1}} = 1.9 	ext{ kg} + 0.1 	ext{ kg} = 2.0 	ext{ kg}$$

Now, we draw a Free Body Diagram (FBD) for this $2.0 	ext{ kg}$ system. 
There are only two forces acting on it:
- The upward tension $T_1$.
- The downward gravitational pull $W_1 = m_{	ext{sys1}}g = 2.0g$.

Crucially, this entire system is accelerating upwards at $a = 0.2 	ext{ m/s}^2$. 
Applying Newton's Second Law ($F_{	ext{net}} = ma$), we write the master equation:
$$T_1 - 2.0g = 2.0a$$

Rearranging to solve for $T_1$:
$$T_1 = 2.0(g + a)$$

Substituting the given values ($g = 9.8 	ext{ m/s}^2$ and $a = 0.2 	ext{ m/s}^2$):
$$T_1 = 2.0(9.8 + 0.2) = 2.0(10.0) = 20 	ext{ N}$$

Boom! The first part is conquered. Notice how elegantly the numbers align when we define our system correctly.

### Part B: The Mid-Point of the Upper Wire

Now, we shift our focus to the upper wire. We need the tension $T_2$ at its mid-point. We apply the exact same conceptual tool: we "cut" the upper wire in the middle and define everything below it as our **second system**.

This new system is much larger. It must support:
1. The lower half of the upper wire.
2. The $2.9 	ext{ kg}$ block.
3. The *entire* lower wire ($0.2 	ext{ kg}$).
4. The $1.9 	ext{ kg}$ block.

Wait, what about the mass of the upper wire? The problem states it has "negligible mass". Therefore, its mass is $0 	ext{ kg}$.

Let's calculate the total mass of this massive system:
$$m_{	ext{sys2}} = 0 	ext{ kg} + 2.9 	ext{ kg} + 0.2 	ext{ kg} + 1.9 	ext{ kg} = 5.0 	ext{ kg}$$

Once again, we draw an FBD. The forces are:
- The upward tension $T_2$.
- The downward gravitational pull $W_2 = m_{	ext{sys2}}g = 5.0g$.

The acceleration remains $a = 0.2 	ext{ m/s}^2$ upwards. 
Applying Newton's Second Law:
$$T_2 - 5.0g = 5.0a$$

Rearranging for $T_2$:
$$T_2 = 5.0(g + a)$$

Substituting the values:
$$T_2 = 5.0(9.8 + 0.2) = 5.0(10.0) = 50 	ext{ N}$$

### The Grand Takeaway

This problem beautifully illustrates that you don't need to write a dozen simultaneous equations for every single component. By strategically choosing your "system boundary", you can collapse a complex multi-body problem into a single, elegant application of $F = ma$.

Whether you are dealing with accelerating trains, stacked blocks, or heavy ropes, remember this technique. Cut the system where you need the force, sum up the mass below it, and let Newton's laws guide you to the solution!

Visualized Solution

$System Overview$

$Mass of the Lower Wire$

$System for T_{1}$

$m_{sys1} Calculation$

$Equation of Motion for T_{1}$

$Calculating T_{1}$

$System for T_{2}$

$m_{sys2} Calculation$

$Equation of Motion for T_{2}$

$Calculating T_{2}$

The Sigma Insight: Newton's Laws of Motion

The Trap of the Heavy Wire

Part A

Part B

The Grand Takeaway

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Overview

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of the Lower Wire

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System for T1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } msys1​ Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Motion for T1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating T1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System for T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } msys2​ Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Motion for T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating T2​