Question

Wires W\(_1\) and W\(_2\) are made of same material having the breaking stress of 1.25\(\times\)10\(^9\) N/m\(^2\). W\(_1\) and W\(_2\) have cross-sectional area of 8\(\times\)10\(^{-7}\) m\(^2\) and 4\(\times\)10\(^{-7}\) 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 m/s\(^2\)) 

 

Hint:

In problems with multiple constraints, you must check the limit for each component separately. The overall limit for the system is determined by the weakest link—the component that fails first. In this case, wire W\(_2\) is the weaker link under these loading conditions.

Answer 1

Correct Answer: 40

To solve this problem, we need to ensure neither wire exceeds its breaking stress. The breaking force \( F_{\text{break}} \) for each wire can be calculated using the formula:

\( F_{\text{break}} = \text{breaking stress} \times \text{cross-sectional area} \)

Given breaking stress = \( 1.25 \times 10^9 \, \text{N/m}^2 \).

For wire W\(_1\):
Cross-sectional area = \( 8 \times 10^{-7} \, \text{m}^2 \)
\( F_{1,\text{break}} = 1.25 \times 10^9 \times 8 \times 10^{-7} = 1000 \, \text{N} \)

For wire W\(_2\):
Cross-sectional area = \( 4 \times 10^{-7} \, \text{m}^2 \)
\( F_{2,\text{break}} = 1.25 \times 10^9 \times 4 \times 10^{-7} = 500 \, \text{N} \)

Let \( x \) be the maximum mass that can be placed in the pan.

For wire W\(_2\):
Total weight = weight of pan mass + 10 kg mass
\( F_{2,\text{total}} = (x + 10) \times 10 \leq 500 \)
\( x + 10 \leq 50 \)
\( x \leq 40 \)

For wire W\(_1\):
Total weight = weight of pan mass + 10 kg mass + 20 kg mass
\( F_{1,\text{total}} = (x + 10 + 20) \times 10 \leq 1000 \)
\( x + 30 \leq 100 \)
\( x \leq 70 \)

Since wire W\(_2\) imposes the stricter condition, the maximum mass in the pan is determined by it:
Maximum mass \( x = 40 \, \text{kg} \)

This value fits within the given range: 40,40.