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?
To be balanced the angular acceleration must be equal (but opposite).
α = mg/r
So: m_b/r_b = m_f/r_f
.: r_b = (60)(0.65)/(70) ≈ 0.56m
actually, it is angular torque, not acceleation that is balanced. Hopefully, the acceleration is zero.
To find out where the two boys should sit on the seesaw, we need to consider the torque acting on each end of the seesaw.
Torque is calculated by multiplying the force applied by the distance from the pivot point (in this case, the center of the seesaw). If the seesaw is balanced, the total torque acting on both ends should be equal.
Let's determine the torque acting on the father's side first. The force exerted by the father is his weight, which is 70 kg. The distance from the center of the seesaw is given as 0.65 m. So, the torque on the father's side is:
Torque = Force * Distance
Torque = 70 kg * 0.65 m
Torque = 45.5 Nm
Since the seesaw is balanced, the torque on the boys' side should be the same. Let's denote the distance the boys sit from the center of the seesaw as "x".
The total weight of the boys is 30 kg + 30 kg = 60 kg. So, the force exerted by the boys is 60 kg.
Now we can set up an equation to find the value of "x":
Torque on Boys' side = Torque on Father's side
Force * Distance = 45.5 Nm
60 kg * x = 45.5 Nm
60x = 45.5
x = 45.5 / 60
x = 0.758 m
Therefore, the two boys should sit together at a distance of approximately 0.758 meters from the center of the seesaw in order to balance the seesaw with their father.