Let's solve the problem **step by step**: --- ## **Step 1: Understanding the Problem** - The building has 4 frames on its perimeter, forming a square \(200 \text{ ft} \times 200 \text{ ft}\). - Frames 1, 2, and 3 have the **same stiffness** \(k\). - Frame 4 has **twice the stiffness** (\(2k\)). - We are to find the **center of rigidity** (CR). --- ## **Step 2: Assign Coordinates** Let’s assign coordinates to each frame (based on the diagram): - **Frame 1:** at \((, 200)\) - **Frame 2:** at \((200, 200)\) - **Frame 3:** at \((200, )\) - **Frame 4:** at \((, 200)\) to \((, )\) (but we treat its effect at \((, 200)\) for calculation) --- ## **Step 3: List Stiffness and Coordinates** | Frame | \(x\) (ft) | \(y\) (ft) | Stiffness | |-------|-----------|-----------|-----------| | 1 | | 200 | \(k\) | | 2 | 200 | 200 | \(k\) | | 3 | 200 | | \(k\) | | 4 | | 200 | \(2k\) | --- ## **Step 4: Center of Rigidity Formula** The **center of rigidity** \((x_{CR}, y_{CR})\) is given by: \[ x_{CR} = \frac{\sum k_i x_i}{\sum k_i} \] \[ y_{CR} = \frac{\sum k_i y_i}{\sum k_i} \] where \(k_i\) and \((x_i, y_i)\) are the stiffness and coordinates of each frame. --- ## **Step 5: Calculate \(\sum k_i\)** \[ \sum k_i = k + k + k + 2k = 5k \] --- ## **Step 6: Calculate \(\sum k_i x_i\) and \(\sum k_i y_i\)** ### \(x\)-coordinate: \[ \sum k_i x_i = (k \times ) + (k \times 200) + (k \times 200) + (2k \times ) = + 200k + 200k + = 400k \] ### \(y\)-coordinate: \[ \sum k_i y_i = (k \times 200) + (k \times 200) + (k \times ) + (2k \times 200) = 200k + 200k + + 400k = 800k \] --- ## **Step 7: Calculate \(x_{CR}\) and \(y_{CR}\)** \[ x_{CR} = \frac{400k}{5k} = 80\ \text{ft} \] \[ y_{CR} = \frac{800k}{5k} = 160\ \text{ft} \] --- ## **Step 8: Choose the Closest Answer** None of the provided options exactly match (80, 160) ft, but let's check the closest: - (a) Cannot be determined - (b) (,) - (c) (100, 100) - (d) (100, 133) **(d) (100, 133 ft)** is the closest to our computed value. --- ## **Final Answer** \[ \boxed{(d)\ (100\ \text{ft},\ 133\ \text{ft})} \] > **Explanation:** The center of rigidity shifts toward the stiffer frame (Frame 4), so the answer closest to our calculation is (100, 133 ft).