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:
P= 454.11 N
Explanation:
Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).
The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:
In order for the small cube to not slide down, the friction force must equal the weight of the small cube:
The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N