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

Step 1: Understanding the Question:
We have a system of two wires supporting masses. We need to find the maximum additional mass (\(M_{\text{pan}}\)) that can be added before one of the wires breaks. A wire breaks when the stress (Force/Area) on it exceeds the breaking stress.
Step 2: Key Formula or Approach:
1. Breaking Force: The maximum force a wire can withstand before breaking is \( F_{\text{break}} = (\text{Breaking Stress}) \times (\text{Area}) \).
2. Tension Calculation: Calculate the tension in each wire as a function of the unknown mass \(M_{\text{pan}}\).
3. Limiting Condition: The tension in each wire must be less than or equal to its breaking force. The most restrictive condition will give the maximum allowed mass.
Step 3: Detailed Explanation:
Given values:
- Breaking stress, \( \sigma_{\text{break}} = 1.25 \times 10^9 \) N/m\(^2\).
- Area of W\(_1\), \( A_1 = 8 \times 10^{-7} \) m\(^2\).
- Area of W\(_2\), \( A_2 = 4 \times 10^{-7} \) m\(^2\).
- \( g = 10 \) m/s\(^2\).
1. Calculate Breaking Force for each wire:
- For W\(_1\): \( F_{\text{break,1}} = \sigma_{\text{break}} \times A_1 = (1.25 \times 10^9) \times (8 \times 10^{-7}) = 10 \times 10^2 = 1000 \) N.
- For W\(_2\): \( F_{\text{break,2}} = \sigma_{\text{break}} \times A_2 = (1.25 \times 10^9) \times (4 \times 10^{-7}) = 5 \times 10^2 = 500 \) N.
2. Calculate Tension in each wire:
Let \(M_p\) be the mass placed in the pan.
- Tension in W\(_2\) (\(T_2\)): This wire supports the 10 kg mass and the pan's mass.
\( T_2 = (10 + M_p)g = (10 + M_p) \times 10 \) N.
- Tension in W\(_1\) (\(T_1\)): This wire supports everything below it: the 20 kg mass, the 10 kg mass, and the pan's mass.
\( T_1 = (20 + 10 + M_p)g = (30 + M_p) \times 10 \) N.
3. Apply Limiting Conditions:
For the system to be safe, both \( T_1 \le F_{\text{break,1}} \) and \( T_2 \le F_{\text{break,2}} \) must be true.
- Condition for W\(_2\):
\( T_2 \le F_{\text{break,2}} \)
\( 10(10 + M_p) \le 500 \)
\( 10 + M_p \le 50 \)
\( M_p \le 40 \) kg.
- Condition for W\(_1\):
\( T_1 \le F_{\text{break,1}} \)
\( 10(30 + M_p) \le 1000 \)
\( 30 + M_p \le 100 \)
\( M_p \le 70 \) kg.
The maximum mass \(M_p\) must satisfy both conditions. The most restrictive condition is \( M_p \le 40 \) kg. If we add more than 40 kg, wire W\(_2\) will break.
Step 4: Final Answer:
The maximum mass that can be placed in the pan is 40 kg.