# If a child of mass 40kg sits 1.5 meters from the middle of a seesaw, how far from the middle should his or her friend sit to balance it if the friend's mass is 30 kg?

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### 1 Answer

In order for the seesaw to be balanced, the vector sum of torques of the forces acting on it should be equal. The forces acting on the seesaw are the forces with which the children push on it, and they equal the weights, or the forces of gravity acting on the children. Since the children sit on the opposite sides of the seesaw, the torques will be in the opposite direction. So if the magnitudes of the torques are equal, the seesaw will be balanced.

Since the force from a child on the seesaw is perpendicular to the seesaw, the magnitude of the force torque will be the magnitude of the force times the distance between the point where the force is applied (that is, where the child is sitting), and the center of the seesaw.

For the first child, this torque will be `m_1*g*l_1` , where `m_1 = 40 kg` , `l_1 = 1.5 m` , and g is 9.8 m/s^2.

For the friend, the distance is unknown, so if it is denoted by `l_2` , the torque will be `m_2*g*l_2` , where `m_2 = 30 kg` and g is the same.

The magnitudes of the torques have to be equal, so

`40*g*1.5 = 30*g*l_2`

g cancels from both sides of the equation, and it can be solved for `l_2` :

`l_2 = (40*1.5)/30 = 2` meters.

**The friend should sit 2 meters away from the center of the seesaw in order for the seesaw to be balanced.**

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