Let's break down the problem step by step: ## **Step 1: Understand the Problem** - There is a crate of mass \( m = 59 \) kg hanging from point \( D \). - The spring (with stiffness \( k \)) connects from \( D \) to \( B \). - The spring has an **unstretched length** of \( 1.7 \) m, but the actual segment \( DB \) is longer. - We are to find the **stiffness \( k \) of the spring** so that the crate is held in the position shown. --- ## **Step 2: Find the Geometry** Let's find the coordinates of points \( D \) and \( B \): - Point \( D \) is vertically below \( C \), so its coordinates: \( (0, 0) \). - The crate is hanging directly below \( D \), so we focus on \( D \) for forces. - Point \( B \): From the top view, \( B \) is 3 meters to the right of the vertical line through \( D \), and 2 meters up. - So, \( B = (3, 2) \). **Length \( DB \):** \[ DB = \sqrt{(3 - 0)^2 + (2 - 0)^2} = \sqrt{9 + 4} = \sqrt{13} = 3.606 \text{ m} \] --- ## **Step 3: Spring Stretch** - **Unstretched length** of spring: \( L_0 = 1.7 \) m - **Stretched length** (actual \( DB \)): \( L = 3.606 \) m - **Extension** (\( x \)): \[ x = L - L_0 = 3.606 - 1.7 = 1.906 \text{ m} \] --- ## **Step 4: Free Body Diagram and Equilibrium** The crate is in equilibrium, so the **vertical forces** must balance the weight: - The forces acting at \( D \) are: - Tension in \( DC \) (let's call it \( T_{DC} \)), acting along \( DC \) - Spring force \( F_s \) along \( DB \) - Weight \( W = mg \) acting downward Sum of vertical forces at \( D \) must be zero. --- ## **Step 5: Find the Angle of Spring** We need the vertical component of the spring force. - **Direction vector of \( DB \):** \( (3, 2) \) - **Magnitude:** \( 3.606 \) - **Unit vector along \( DB \):** \( \frac{(3, 2)}{3.606} = (0.832, 0.555) \) - **Vertical component is the \( y \)-component:** \( 0.555 \) times the total spring force. --- ## **Step 6: Write the Force Balance Equation** Let \( F_s \) be the spring force. Vertically, components must add up to weight: \[ \text{Upward force from spring} = F_s \cdot \sin(\theta) \] Where \( \theta \) is the angle between \( DB \) and the horizontal. But from the unit vector, vertical component is \( 0.555 \): \[ F_s \cdot 0.555 = mg \] \[ F_s = \frac{mg}{0.555} \] But \( F_s = kx \): \[ kx = \frac{mg}{0.555} \] \[ k = \frac{mg}{0.555 \times x} \] Recall \( x = 1.906 \) m, \( m = 59 \) kg, \( g = 9.81 \) m/s\(^2\): --- ## **Step 7: Plug in the Numbers** \[ k = \frac{59 \times 9.81}{0.555 \times 1.906} \] \[ k = \frac{578.79}{1.0588} \] \[ k = 546.7 \, \text{N/m} \] **To three significant figures:** \[ \boxed{547 \, \text{N/m}} \] --- ## **Summary Table:** | Quantity | Value | |-------------------|---------------| | \( DB \) length | 3.606 m | | Extension \( x \) | 1.906 m | | \( mg \) | 578.79 N | | Vertical fraction | 0.555 | | \( k \) | 547 N/m | --- ### **Final Answer** \[ \boxed{ k = 547 \ \text{N/m} } \] Let me know if you'd like a breakdown of the forces or vector calculations!