Question

Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is sitting three meters away from the fulcrum at the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. The mass of the board is evenly distributed so that its center of mass is over the fulcrum. What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground? 
a. 4m
b. 5m
c. 2m
d. 3m
NO LINKS. ​

Answer 1

Answer:

x = 4 m

Explanation:

For this exercise we must use the rotational equilibrium relationship, where we place zero at the turning point and counterclockwise rotations we will consider positive

as it indicates that the bar is in equilibrium, its center of mass coincides with the turning point, so the distance is zero and does not create torque on the system

        ∑τ  = 0

        W 3 - w x = 0

        x = 3W / w

        x = 3 Mg / mg

        x = 3 M / m

let's calculate

       x = 3 60/45

       x = 4 m