A 10 meter long seesaw has a boy of mass 20 kg sitting on one end . How far away from him should a girl of mass 30kg sit to balance the seesaw?
I am not sure where the fulcrum is but if it is in the middle then:
20 * 5 = 30 * x
x = 5(2/3) = 10/3 from center
which is 10/3 + 5 from the boy
which is
25/3 = 8 1/3 meters from the boy
To balance the seesaw, the torques on both sides of the fulcrum need to be equal. Torque is calculated by multiplying the force applied to an object by the distance from the fulcrum.
In this case, we have a boy of mass 20 kg sitting on one end of the seesaw. The force acting on the boy is his weight, which is the product of his mass and the acceleration due to gravity (9.8 m/s^2). So the force on the boy is 20 kg * 9.8 m/s^2 = 196 N.
Let's assume that the distance from the boy to the fulcrum is x meters. The torque on the boy's side would be the force (196 N) multiplied by the distance (x meters).
On the other side of the fulcrum, we have a girl of mass 30 kg. The force acting on the girl is her weight, which is 30 kg * 9.8 m/s^2 = 294 N.
For the seesaw to balance, the torques on both sides need to be equal. So we can set up an equation:
Torque on boy's side = Torque on girl's side
196 N * x meters = 294 N * (10 meters - x meters)
Simplifying the equation:
196x = 2940 - 294x
490x = 2940
x = 2940 / 490
x = 6 meters
Therefore, the girl should sit 6 meters away from the boy to balance the seesaw.