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
60 * 3 = 45 * ?
X=4 m
What formula did you use
To solve this problem, we can use the principle of moments, which states that the sum of the clockwise moments is equal to the sum of the anticlockwise moments. In this case, we want the seesaw to stay balanced and parallel to the ground, so the sum of the clockwise moments should be equal to the sum of the anticlockwise moments.
Let's calculate the moments for both sides of the seesaw:
Clockwise moments:
Student on the left: Moment = Weight × Distance = 60kg × 3m = 180kg*m
Student on the right: Moment = Weight × Distance
Anticlockwise moments:
Board: The center of mass of the board is over the fulcrum, so there is no moment contribution from the board.
Now let's set up the equation:
180kg*m = Weight of the student on the right × Distance
To find the distance from the center for the student on the right, we need to rearrange the equation to solve for it:
Distance = 180kg*m / Weight of the student on the right
Now we substitute the values:
Distance = 180kg*m / 45kg = 4m
Therefore, the student on the right should be 4 meters away from the center for the seesaw to stay parallel to the ground.
So, the correct option is a. 4m.