Two boys and their father are balance on a seesaw. The father, 70kg, sits on one end 0.65m from the center of the seesaw while the two boys sit together on the other end. If each boy weighs 30kg, where should they sit?
Take moments about the centre.
On father's side:
M=70*0.65=45.5 kg-m
On children's side:
M=(30+30)*x
Equate moments and solve for x.
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To determine where the boys should sit on the seesaw, we need to find the distance from the center where they should be balanced.
Let's assume that one of the boys is sitting closer to the center. Let's call the distance from the center point x (in meters). Since the father's weight is uniform (all concentrated at one point), we can consider the seesaw and the individuals on it as a system in equilibrium.
The principle of moments states that for the system to be in equilibrium, the clockwise moment must be equal to the counterclockwise moment. The moment of each object can be calculated by multiplying the weight by the distance from the center.
In this case, the moment of the father is (70 kg) * (0.65 m), and the moment of the two boys combined is (2 * 30 kg) * (x m). These moments should be equal, so we have:
(70 kg) * (0.65 m) = (2 * 30 kg) * (x m)
Simplifying this equation gives:
45.5 kg * m = 60 kg * x
Now we can solve for x:
x = (45.5 kg * m) / (60 kg)
x ≈ 0.7583 m
Therefore, the boys should sit approximately 0.7583 meters from the center of the seesaw in order to balance it with the father's weight.