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 t
he mass of his friend is (A) 40 kg (B) 20 kg (C) 80 kg (D) Cannot be determined
1 answer:
Answer:
<h2>C. 80kg</h2>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
<em>Hence the mass of his friend that will keep the seesaw balanced horizontally is 80kg.</em>
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Answer:
C. less than 950 N.
Explanation:
Given that
Force in north direction F₁ = 500 N
Force in the northwest F₂ = 450 N
Lets take resultant force R
The angle between force = θ
θ = 45°
The resultant force R
R= 877.89 N
Therefore resultant force is less than 950 N.
C. less than 950 N
Note- When these two force will act in the same direction then the resultant force will be 950 N.
