A see-saw 3.76 meters long perfectly balances on its own at its center. Two children get on the see-saw. One, weighing 345 N, sits at the left hand end. The other, sits 25.0 cm in from the right hand end. The see-saw is perfectly balanced with the children sitting in these locations. What is the weight of the child on the right?
sum moments about the pivot
W*3.76/2-345(3.76/2-.25)=0
solve for Weight W
F1 * d1 = F2 * d2.
345 * 3.76/2 = F2 * (3.76/2-0.25).
F2 = ?.
To find the weight of the child on the right, we need to consider the principle of moments or torques. Moments or torques are the rotational forces that cause an object to rotate.
In this case, the see-saw is perfectly balanced, which means the total clockwise moments are equal to the total counterclockwise moments.
To calculate the moments, we can use the equation:
Moment = Force × Distance
First, let's calculate the moment caused by the child on the left.
The child on the left has a weight of 345 N, and they are sitting at the left hand end of the see-saw. The distance from the center to the left hand end is half of the total length of the see-saw.
So the moment caused by the child on the left is:
Moment_left = Weight_left × Distance_left
Since the see-saw is balanced, this moment must be equal to the moment caused by the child on the right. Let's call the weight of the child on the right "Weight_right" and the distance from the center to the right hand end "Distance_right".
Moment_left = Moment_right
Weight_left × Distance_left = Weight_right × Distance_right
Substituting the given values:
345 N × (3.76 m / 2) = Weight_right × (3.76 m - 0.25 m)
Now we can solve this equation to find the weight of the child on the right.
Weight_right = (345 N × (3.76 m / 2)) / (3.76 m - 0.25 m)
Weight_right ≈ 240.41 N
Therefore, the weight of the child on the right is approximately 240.41 N.