Wires W1 and W2 are made of same material having the breaking stress of 1.25 x 10^9 N/m^2. W1 and W2

Q: Why do we use the same breaking stress for both wires?

The problem explicitly states that both wires are made of the 'same material'. Breaking stress is a material property, so it remains constant regardless of the wire's thickness or length.

Q: Why does tension $T_1$ depend on the 10 kg block and the pan?

Wire $W_1$ is at the very top, supporting the entire chain below it. If you cut $W_1$, the 20 kg block, the 10 kg block, and the pan will all fall. Therefore, it must support the sum of all their weights.

Q: Can we just add the maximum tensions together?

No, tensions in a series system don't add up like that. Each wire experiences a specific tension based on the weight hanging directly below it. You must check each wire's safety limit individually.

Q: Why did we choose 40 kg instead of 70 kg as the final answer?

If we chose 70 kg, wire $W_1$ would be perfectly fine, but wire $W_2$ would experience a tension of $(10+70)\times 10 = 800$ N, which is way above its 500 N limit. It would snap! We must choose the smaller mass to ensure *both* wires survive.

Q: What if the wires had their own mass?

If the wires were massive, the tension wouldn't be constant throughout the wire. It would be highest at the top of each wire because the top point has to support the blocks *plus* the weight of the wire itself.

Question

Wires $W_1$ and $W_2$ are made of same material having the breaking stress of $1.25 	imes 10^9 	ext{ N/m}^2$. $W_1$ and $W_2$ have cross-sectional area of $8 	imes 10^{-7} 	ext{ m}^2$ and $4 	imes 10^{-7} 	ext{ m}^2$, respectively. Masses of 20 kg and 10 kg hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is .......... kg.
(Use, $g = 10 	ext{ m/s}^2$)

Sigma Percentile · Accepted Answer

Imagine you are an engineer tasked with designing a suspension system. You have two wires, $W_1$ and $W_2$, supporting a series of heavy blocks and a pan.

Your ultimate goal is to find the maximum mass, $m$, that can be safely placed in the pan without snapping either of the wires.

This is a classic problem of stress, strain, and structural integrity. Let's break it down step by step and ensure our system doesn't collapse!

### Analyzing the Setup

First, let's look at the properties of our materials. Both wires are made of the exact same material.

This means they share the same **breaking stress**, denoted by $\sigma$, which is given as $1.25 	imes 10^9 	ext{ N/m}^2$.

However, they have different thicknesses. Wire $W_1$ has a cross-sectional area $A_1 = 8 	imes 10^{-7} 	ext{ m}^2$, while wire $W_2$ is thinner with an area $A_2 = 4 	imes 10^{-7} 	ext{ m}^2$.

### The Master Equation

To ensure a wire doesn't break, the tension inside it must never exceed its maximum capacity.

We know from the fundamental principles of elasticity that stress is defined as the internal restoring force (tension) per unit area.

$$ 	ext{Stress} = \frac{	ext{Tension}}{	ext{Area}} $$

Therefore, the maximum tension a wire can handle before snapping is simply its breaking stress multiplied by its cross-sectional area.

$$ T_{	ext{max}} = \sigma 	imes A $$

### Calculating Maximum Tensions

Let's calculate the breaking point for each wire individually.

For the thicker upper wire, $W_1$, we substitute its specific area into our master equation.

$$ T_{1,	ext{max}} = (1.25 	imes 10^9) 	imes (8 	imes 10^{-7}) $$

When we compute this, $1.25 	imes 8 = 10$, and the powers of 10 combine to $10^2$. This gives us a maximum tension of $1000 	ext{ N}$.

$$ T_{1,	ext{max}} = 1000 	ext{ N} $$

Now, let's look at the thinner lower wire, $W_2$. Since its area is exactly half of $W_1$, it makes intuitive sense that it can only withstand half the tension.

$$ T_{2,	ext{max}} = (1.25 	imes 10^9) 	imes (4 	imes 10^{-7}) $$

Calculating this yields a maximum tension of $500 	ext{ N}$.

$$ T_{2,	ext{max}} = 500 	ext{ N} $$

### Free Body Diagrams and Constraints

Now that we know the limits, we need to find out how much actual tension is being applied to each wire by the hanging masses.

Let's visualize the forces by drawing free body diagrams.

The tension $T_2$ in the lower wire must support everything hanging below it. This includes the 10 kg block and the unknown mass $m$ in the pan.

$$ T_2 = (10 + m)g $$

On the other hand, the tension $T_1$ in the upper wire must support the entire system's weight. This includes the 20 kg block, the 10 kg block, and the mass $m$.

$$ T_1 = (20 + 10 + m)g = (30 + m)g $$

### Final Calculation and The Catch

Here is where we apply our safety constraints. For the system to survive, neither wire can exceed its maximum tension.

Let's solve the constraint for the lower wire, $W_2$, first. We know $T_2$ must be less than or equal to $500 	ext{ N}$.

$$ (10 + m) 	imes 10 \le 500 $$

Dividing both sides by 10, we get $10 + m \le 50$. Subtracting 10 gives us our first limit.

$$ m \le 40 	ext{ kg} $$

Now, let's check the constraint for the upper wire, $W_1$. Its tension $T_1$ must be less than or equal to $1000 	ext{ N}$.

$$ (30 + m) 	imes 10 \le 1000 $$

Dividing by 10 gives $30 + m \le 100$. Subtracting 30 gives us our second limit.

$$ m \le 70 	ext{ kg} $$

**There is a catch here.** We have two different limits: 40 kg and 70 kg.

If we were to place 50 kg in the pan, the upper wire would be perfectly fine, but the lower wire would snap instantly!

To keep the entire system intact, we must satisfy both conditions simultaneously. This means we must choose the stricter, smaller limit.

Therefore, the maximum mass we can safely place in the pan is **40 kg**.

Enter Numerical Value:

Visualized Solution

$System Setup$

$Maximum Tension Formula$

$Substituting Values for T_{max}$

$Calculating Maximum Tensions$

$Free Body Diagrams$

$Tension Equations$

$Setting up Constraints$

$Solving for m from W_{2}$

$Solving for m from W_{1}$

$Final Answer$

$The Way Forward$

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

The Master Equation

Calculating Maximum Tensions

Free Body Diagrams and Constraints

Final Calculation and The Catch

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for m from W2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for m from W1​

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

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