An adult and a child are on a seesaw 14.5 ft long. The adult weighs 152 lb and the child weighs 80 lb.
How many feet from the child must the fulcrum be placed so that the seesaw balances?
To solve this problem, we need to understand the concept of torque. Torque is the measure of an object's tendency to rotate around a specific point due to the application of a force. In the case of a seesaw, the torque exerted on one side must be equal to the torque exerted on the other side for balance.
The general formula for torque is given by:
Torque = Force * Distance from the fulcrum
In this problem, the adult and child exert forces on opposite sides of the seesaw. Let's denote the distance from the fulcrum to the child as "x". The distance from the fulcrum to the adult is then (14.5 - x), as the seesaw is 14.5 ft long. The torque exerted by each person is equal to the force they exert multiplied by their respective distances from the fulcrum.
For the adult: Torque_adult = 152 lb * (14.5 - x) ft
For the child: Torque_child = 80 lb * x ft
Since the seesaw is balanced, the torques exerted by both the adult and the child must be equal.
Torque_adult = Torque_child
152 lb * (14.5 - x) ft = 80 lb * x ft
To solve for x, we can use algebraic methods. First, let's simplify the equation:
152 lb * 14.5 ft - 152 lb * x = 80 lb * x
152 lb * 14.5 ft = 232 lb * x
Dividing both sides by 232 lb:
14.5 ft = x
Therefore, the fulcrum must be placed at a distance of 14.5 feet from the child in order to balance the seesaw.
Thats a long seesaw. God they eat too much. The seesaw will break.
It should not be placed.
Nyan cat to the rescue.