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 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.
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- Use work-energy principles for spring stiffness problems.
- Equilibrium conditions help simplify calculations.
- Always convert angles into proper components when necessary.
Updated On: Feb 27, 2025
- 88 N/m
- 166 N/m
- 194 N/m
- 138 N/m
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The Correct Option is B
Solution and Explanation
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.
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.
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