## Illustration 13.4: The Diving Board Problem

A 2-kg box sits 0.3 m from the right end of a board of negligible weight in Animation 1 and a board of weight 10 kg in Animation 2. Two supports (support 1 and support 2) exert forces on the left and right ends of the board, as indicated by the two force vectors (position is given in meters). The arrows represent the relative sizes of the force vectors, but their length does not represent their actual magnitudes (the actual value of the forces, as well as the separation between the supports, is shown in the table). The board is 6 m long and support 1 is 0.3 m in from the left edge. Restart.

Consider the situation in Animation 1 in which the board has a negligible weight. How does the force of each support on the board (F1 corresponds to the right support and F2 corresponds to the left support) depend on where the box is located? You can drag the second support from left to right to view the forces of the supports on the board. When the board is of a negligible mass, there are three forces that act on the board: the weight of the box, and the forces of the supports. As you move support 2 around, what do you notice? When the movable support is as far right as it can go, the force of the first support is zero and the force of the second support cancels the force of the box. This seems logical, but why this force? An F1 = 10 N and an F2 = 9.6 N, would work too, right? Well, yes and no. It would certainly make the sum of the forces on the board equal to zero, but what about the sum of the torques? This sum would not be equal to zero no matter where you measured the torques from. Only an equal and opposite force acting at the same position as the weight of the box will keep the board in equilibrium. As you move the second support to the left note that both forces exerted by the supports get bigger, and that the force that the first support exerts is negative. We can understand this effect by measuring the torque on the board from the first support. Since the force is in the y direction and the radius arm is in the x direction, the magnitude of the torque is rF. The radius arm for the weight of the box is always 5.4 m. The radius arm for the second support changes as you drag it to the left. Therefore, as you drag it to the left the force the support exerts must increase in order to keep rF the same magnitude as (and in the opposite direction of) that of the box. Once the right force is determined from the torque, the force due to the first support can be determined by requiring the sum of the forces on the board to be zero.

Now, consider the situation in Animation 2 in which the board has a weight of 10 kg. How does the force of each support on the board depend on where the box is located? You can drag the second support from left to right to view the forces of the supports on the board. When the board has a mass, there are four forces that act on the board: the weight of the box, the weight of the board, and the forces of the supports. Much of the same analysis described above still is valid here; you just need to factor in one more force and therefore one more torque. Again drag the support around and pay close attention to the forces. What happens to the forces when the separation between the supports is 3.15 m? How about 2.7 m?

Illustratron authored by Aaron Titus.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.

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