A 79 kg diver stands at the end of a 21  kg springboard, as shown in the Figure. The board is attached to a hinge at the left end but simply rests on the right support. What is the magnitude of the vertical force exerted by the hinge on the board in Newton?

To solve this problem, we need to understand that the total force acting on the springboard must be zero when the system is in equilibrium. This means that the force exerted by the hinge and the force exerted by the support must balance the weight of the diver and the weight of the board.

1. First, calculate the weight of the diver and the board. Weight is mass times gravity. Assuming the acceleration due to gravity is 9.8 m/s², the weight of the diver is 79 kg * 9.8 m/s² = 774.2 N and the weight of the board is 21 kg * 9.8 m/s² = 205.8 N.

2. The total weight acting on the board is the sum of these two weights, which is 774.2 N + 205.8 N = 980 N.

3. The total weight is distributed between the hinge and the support. The diver is standing at the end of the board, so the weight of the diver acts at the center of gravity of the diver which is at the end of the board. The weight of the board acts at the center of the board, which is half the length of the board from the hinge.

4. If we assume the length of the board is L, the total torque acting on the board about the hinge is zero. This gives us the equation: (L/2) * 205.8 N + L * 774.2 N = L * F_support, where F_support is the force exerted by the support.

5. Solving this equation for F_support gives us F_support = ((L/2) * 205.8 N + L * 774.2 N) / L = 774.2 N + 205.8 N / 2 = 877.1 N.

6. The total force acting on the board is zero, so the force exerted by the hinge is the total weight minus the force exerted by the support, which is 980 N - 877.1 N = 102.9 N.

So, the magnitude of the vertical force exerted by the hinge on the board is approximately 102.9 N.