Final answer:
The magnitudes of the forces exerted by the supports on the plank can be found by considering the torques acting on the plank. The torque due to the person's weight and the torque due to the weight of the plank can be calculated and set equal to zero, since the plank is in equilibrium. Solving for the forces, we find that one support exerts a force of -490 N and the other support exerts a force of 490 N.
Step-by-step explanation:
To find the magnitudes of the forces exerted by the supports on the plank, we need to consider the torques acting on the plank. The torque due to the person's weight can be calculated as the product of their weight and the distance from the support to their center of mass. The torque due to the weight of the plank can be calculated as the product of its weight and the distance from its center of mass to the support. The magnitudes of the forces exerted by the supports can then be found by summing up the torques and setting them equal to zero, since the plank is in equilibrium.
Let's calculate the torques. The person's weight is 50 kg times the acceleration due to gravity, which is approximately 9.8 m/s². The distance from the support to their center of mass is 1.0 m. Therefore, the torque due to the person's weight is 50 kg × 9.8 m/s² × 1.0 m = 490 Nm.
The weight of the plank is 20 kg times the acceleration due to gravity, which is again approximately 9.8 m/s². The distance from the center of mass to the support is half the length of the plank, which is 1.5 m. Therefore, the torque due to the weight of the plank is 20 kg × 9.8 m/s² × 1.5 m = 294 Nm.
Since the plank is in equilibrium, the sum of the torques must be zero. So, we can write the equation 490 Nm + 294 Nm = 0. Solving for the forces, we find that the magnitude of the force exerted by one support is -490 N and the magnitude of the force exerted by the other support is 490 N.