## Step-by-Step Solution Let's break down the problem: - **Given:** - 50 N load at node 1 (downward) - Maximum vertical deflection at node 1: \(2\, \text{mm} = 0.002\, \text{m}\) - All elements are springs with stiffness \(k\) - Each spring is at a \(45^\circ\) angle You are to **determine the minimum value of \(k\)** so that the vertical deflection at node 1 does not exceed 2 mm. --- ### 1. **Identify the Structure** Node 1 is connected to two springs: - Both at 45° angles to the horizontal. --- ### 2. **Spring Stiffness in Vertical Direction** Each spring's force component in the vertical direction: - Vertical stiffness of each spring = \(k_y = k \cos^2(45^\circ)\) But, since there are two springs, both contribute to resist the vertical load. --- ### 3. **Total Vertical Stiffness at Node 1** For each spring: \[ \cos(45^\circ) = \frac{1}{\sqrt{2}} \] \[ k_y = k \cos^2(45^\circ) = k \left(\frac{1}{\sqrt{2}}\right)^2 = k \left(\frac{1}{2}\right) = \frac{k}{2} \] There are **two springs** acting together (both contribute): \[ k_{\text{total, vertical}} = 2 \times \frac{k}{2} = k \] --- ### 4. **Relate Force, Stiffness and Deflection** Hooke's Law (generalized for systems): \[ \delta = \frac{F}{k_{\text{total}}} \] Given: - \(\delta_{\text{max}} = 0.002\,\text{m}\) - \(F = 50\,\text{N}\) - \(k_{\text{total}} = k\) \[ 0.002 = \frac{50}{k} \] \[ k = \frac{50}{0.002} = 25{,}000\,\text{N/m} \] --- ## **Final Answer** \[ \boxed{ k = 25{,}000\,\text{N/m} } \] --- ### **Explanation Recap** - Both springs act together, each contributing half their stiffness in the vertical direction due to the 45° angle. - The total vertical stiffness at the loaded node is \(k\). - Use Hooke's Law to relate load, deflection, and stiffness. - Solve for \(k\) to ensure deflection does not exceed 2 mm. If you need the free-body diagram or more details on the vector components, let me know!