Let's break down the problem into the requested steps: --- ### **Step 1: Conceptual Introduction** This problem involves a 70-kg mass resting in equilibrium on a smooth, inclined ramp at \(25^\circ\) to the horizontal. Various springs and cables are attached to the mass and other points, and their constants and arrangements are specified. The system also involves a force moment at a pole. The problem asks for the force in a specific spring, the normal force from the ramp, tensions in several cables, and a specific angle. **Explanation Block:** In this setup, the 70-kg mass is held stationary by a combination of spring forces, cable tensions, and the normal force from the ramp. The equilibrium of the mass and the structure is governed by forces and moments in both horizontal and vertical directions. To solve the problem, we'll apply principles from statics: resolving forces, using Hooke's Law for springs, and considering moments for rotational equilibrium. --- ### **Step 2: Provide Formulae in Above Solution** Key formulae involved: 1. **Hooke’s Law for Springs:** \( F = kx \) where \( F \) is the force, \( k \) is the spring constant, \( x \) is the deformation. 2. **Force Equilibrium (Newton’s first law):** \( \sum F_x = 0 \), \( \sum F_y = 0 \) 3. **Component Forces on Ramp:** - Perpendicular: \( F_{N} = mg \cos(\theta) \) - Parallel: \( F_{\text{parallel}} = mg \sin(\theta) \) 4. **Moment (Torque) Equilibrium:** \( \sum M = 0 \) **Explanation Block:** The above formulae allow us to calculate individual forces and moments acting on the mass and the structure. Hooke's law tells us how much force a spring exerts based on its extension or compression. Summing forces in the \(x\) and \(y\) directions ensures the system is in equilibrium. Decomposing the weight of the mass along and perpendicular to the ramp helps in evaluating the normal and parallel forces, and the moment equilibrium is essential for the stability of the structure. --- ### **Step 3: Calculation for 1st Question (Force in Spring \( k_1 \))** **Step-by-Step:** 1. The weight component down the ramp: \( W = mg = 70 \times 9.81 = 686.7 \) N \( W_{\text{parallel}} = W \sin(25^\circ) = 686.7 \times 0.4226 = 290.1 \) N 2. The force in spring \( k_1 \) must balance the parallel component (since it's the only spring directly attached along the incline). **Explanation Block:** Here, the spring \( k_1 \) alone (assuming it's the only resistance along the ramp) must provide a force equal to the downslope weight component to keep the 70-kg mass stationary on the frictionless ramp. We find this by resolving the gravitational force into a component parallel to the incline and equating it to the spring force. --- ### **Step 4: Calculation for 2nd Question (Normal Force by Ramp)** **Step-by-Step:** 1. Normal force is from the component of weight perpendicular to the ramp: \( F_N = W \cos(25^\circ) = 686.7 \times 0.9063 = 622.2 \) N **Explanation Block:** The normal force is the reaction from the ramp surface and is equal to the component of the gravitational force perpendicular to the ramp. Since there are no vertical forces except the weight, this is a straightforward calculation using the cosine of the ramp angle. --- ### **Step 5: Calculation for 3rd Question (Tensions in Cables AB, BC, CD, EF)** **Step-by-Step:** 1. **Cable AB:** Supports 70-kg mass along its direction. 2. **Cable BC:** Supports the 50-kg mass: \( T_{BC} = 50 \times 9.81 = 490.5 \) N 3. **Cable CD:** Same as BC since they are joined. 4. **Cable EF:** Attached to the spring system, force equals the total force from parallel springs \( k_3, k_4, k_5 \). - Equivalent spring constant: \( k_{eq} = k_3 + k_4 + k_5 = 50 + 25 + 50 = 125 \) N/m (if in parallel). - Tension in EF is the total extension force, which can be inferred from the balance of moments (full calculation may depend on more data). **Explanation Block:** Cable tensions are determined by the weights or forces they support. BC and CD support the 50-kg mass, so their tensions are simply the weight. EF's tension is governed by the total force generated by the springs it connects, calculated using their combined spring constant and the extension produced by the applied load. --- ### **Step 6: Calculation for 4th Question (Angle \( \varphi \))** **Step-by-Step:** 1. The maximum moment at the base is given as 250 N·m (counterclockwise). 2. The force from the spring system (50 N) acts at the end of a 3.5 m bar: \( M = F \cdot l \cdot \sin(\varphi) \) \( 250 = 50 \times 3.5 \times \sin(\varphi) \) \( \sin(\varphi) = \frac{250}{50 \times 3.5} = \frac{250}{175} = 1.428 \) But since \( \sin(\varphi) > 1 \) is not possible, re-examine: Maybe only a component of 50 N is used; check the setup for the correct lever arm or component, but this is the process. **Explanation Block:** The angle \( \varphi \) is found from the moment equation, relating the perpendicular force component to the distance from the pivot. Setting the moment to the given maximum value and solving for the angle shows whether the force applied at the given distance can generate the required moment. --- ### **Final Answers:** a) **Force in spring \( k_1 \):** **290.1 N** b) **Normal force by the ramp:** **622.2 N** c) **Tensions:**  - AB: (Calculation depends on direction; if along ramp, same as spring \( k_1 \): **290.1 N**)  - BC: **490.5 N**  - CD: **490.5 N**  - EF: (Depends on spring extension; if 50 N acting, then **50 N**) d) **Angle \( \varphi \):** Not possible as calculated (\( \sin(\varphi) > 1 \)), needs re-evaluation with correct geometry or force value. **(If the maximum moment is less than the moment generated, the full 50 N may not be used, or the lever arm is less than 3.5 m.)**