A 20-kg child sits on a see saw on the playground, 3.0 m from the pivot. A 2nd child is seated on the other side of the pivot and 2.00 m from the pivot. The mass of the second child needed to lift the first child off the ground is
To find the mass of the second child needed to lift the first child off the ground, we can use the principle of torque balance.
The torque is the measure of the turning force or moment. In this case, the torque exerted by each child is given by the product of their respective masses, distances from the pivot, and the acceleration due to gravity (9.8 m/s^2).
According to the principle of torque balance, the sum of the torques on one side of the pivot should be equal to the sum of the torques on the other side of the pivot when the see-saw is in equilibrium.
Let's denote the mass of the second child as M2.
The torque exerted by the first child is given by the product of their mass (20 kg), distance (3.0 m), and acceleration due to gravity (9.8 m/s^2): 20 kg × 3.0 m × 9.8 m/s^2 = 588 N·m.
For the see-saw to be in equilibrium, the torque exerted by the second child must be equal and opposite to that of the first child.
The torque exerted by the second child is given by the product of their mass (M2), distance (2.00 m), and acceleration due to gravity (9.8 m/s^2): M2 × 2.00 m × 9.8 m/s^2 = 19.6 M2 N·m.
By setting the torques equal to each other, we have:
588 N·m = 19.6 M2 N·m.
To solve for M2, divide both sides of the equation by 19.6 N·m:
M2 = 588 N·m / 19.6 N·m = 30 kg.
Therefore, the mass of the second child needed to lift the first child off the ground is 30 kg.