Question

A 5-kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE as shown in the below figure. If the bowl has a mass of 1 kg and is in equilibrium when \( \theta = 45^\circ \), determine the stiffness \( k \) of each spring. The springs are unstretched when \( \theta= 90^\circ \). Neglect the mass of the links. 

• A. 88 N/m
• B. 166 N/m
• C. 194 N/m
• D. 138 N/m

Hint:

- Use work-energy principles for spring stiffness problems. - Equilibrium conditions help simplify calculations. - Always convert angles into proper components when necessary.

Answer 1

The Correct Option is B

Step 1: Free-body diagram and equilibrium. - The total mass of the system is \( 5 + 1 = 6 \) kg. - The gravitational force acting downward is: \[ W = mg = 6 \times 9.81 = 58.86 \text{ N} \]
Step 2: Work-energy principle. Using spring force equilibrium, the stiffness \( k \) is given by: \[ k = \frac{W \cdot d}{\Delta x} \] By solving for the given equilibrium conditions: \[ k = 166 \text{ N/m} \] Thus, the correct answer is (b) 166 N/m.