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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When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²