A uniform seesaw is balanced at its center of mass, as seen below. The smaller boy on the right has a mass of 40.0 kg. What is the mass of his friend is (A) 40 kg (B) 20 kg (C) 80 kg (D) Cannot be determined

Explanation:

The question lacks the appropriate diagram. Find the diagram attached below:

Let the mass of his friend be M.

Using the principle of moment which states that the sum of clockwise moment is equal to the sum of its anticlockwise moment to solve the problem,.

Since Moment = Force * perpendicular distance.

The smaller boy will move in the clockwise direction and his friend will move in the anti clockwise direction.

Clockwise moment = 40 * 4 = 160kgm

anticlockwise moment = M * 2 = 2Mkgm

Equating both moments to get the mass M of his friend

160 = 2M

Divide both sides by 2

2M/2 = 160/2

M = 80kg

Hence the mass of his friend that will keep the seesaw balanced horizontally is 80kg.

this isn't the answer but they're the graphs

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.