A woman whose weight is 530 Newtons is standing on the outer edge (right) of a diving board of 3.9 meters length. She is over the water of the pool. The left edge of the board is bolted down to the concrete. A support spring (shown as a fulcrum) is centered 1.4 meters from the left edge. The spring is somewhat compressed by the weight of the woman on the board. Determine the force in the support spring. The system is in static equilibrium
To determine the force in the support spring, we can analyze the forces acting on the diving board and the woman. Since the system is in static equilibrium, the sum of the forces acting on the system must be zero.
1. Start by drawing a free-body diagram of the system. On the diagram, indicate the forces acting on the diving board and the woman. The weight of the woman, 530 N, acts downward from a point located 1.4 meters from the left edge. Label this force as F1.
2. Since the woman exerts a force on the diving board, there must be a reaction force exerted by the diving board on the woman. This reaction force acts upward and has the same magnitude as the weight of the woman (530 N), but in the opposite direction.
3. The support spring exerts an upward force on the diving board, counteracting the weight of the woman and maintaining the static equilibrium. Label this force as F2.
4. The diving board is bolted down to the concrete at the left edge, creating a horizontal reaction force that prevents the board from moving horizontally. Label this force as F3.
5. Apply the condition of static equilibrium: the sum of the forces in both the horizontal and vertical directions must be zero.
In the vertical direction:
- F2 - F1 = 0
- F2 = F1 (since the forces are equal in magnitude but opposite in direction)
In the horizontal direction:
- F3 = 0 (since the diving board is prevented from moving horizontally)
Therefore, the force in the support spring (F2) is equal to 530 N.