Two long metallic strips are joined together by two rivets each of radius `2 mm`. Each rivet can withstand a maximum shearing stress of `1.5 xx 10^(9) N//m^(2)`. Assuming that each rivet shares the stretching load equally, the maximum tensile force the strip can exert without rupture is

Explanation

To solve the problem, we need to find the maximum tensile force that the two rivets can withstand without rupture. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given data
- Radius of each rivet, \( r = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Maximum shearing stress, \( \tau_{\text{max}} = 1.5 \times 10^9 \, \text{N/m}^2 \)
- Number of rivets, \( n = 2 \)

### Step 2: Calculate the cross-sectional area of one rivet
The area \( A \) of one rivet can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (2 \times 10^{-3})^2 = \pi (4 \times 10^{-6}) \, \text{m}^2 = 4\pi \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the maximum shearing force that one rivet can withstand
The maximum shearing force \( F_{\text{max}} \) that one rivet can withstand is given by:
\[ F_{\text{max}} = \tau_{\text{max}} \times A \]
Substituting the values:
\[ F_{\text{max}} = 1.5 \times 10^9 \, \text{N/m}^2 \times 4\pi \times 10^{-6} \, \text{m}^2 \]
Calculating this:
\[ F_{\text{max}} = 1.5 \times 10^9 \times 4\pi \times 10^{-6} \approx 1.5 \times 4 \times 3.14 \times 10^{3} \approx 18,840 \, \text{N} \]

### Step 4: Calculate the total maximum tensile force for both rivets
Since there are two rivets sharing the load equally, the total maximum tensile force \( F_{\text{total}} \) is:
\[ F_{\text{total}} = 2 \times F_{\text{max}} = 2 \times 18,840 \, \text{N} = 37,680 \, \text{N} \]

### Final Answer
The maximum tensile force that the strip can exert without rupture is:
\[ F_{\text{total}} \approx 37,680 \, \text{N} \]

---