Question

Bart, mass 32.4 kilograms, and Milhouse, mass 27.6 kilograms, play on the schoolyard seesaw. If Bart and Milhouse want to sit 4.0 meters apart, how far from the center of the seesaw should Bart sit? Include units in your answer. Answer must be in 3 significant digits.

Answer 1

Given data:

* The mass of the Bart is m_1 = 32.4 kg.

* The mass of the Milhouse is m_2 = 27.6 kg.

* The distance between the Millhouse and Bart is d = 4 m.

Solution:

To balance the seesaw, the net moment about the center should be zero.

The diagrammatic representation of the given system is,

The distance between the Bart and Milhouse can be written as,

`\begin{gathered} d=d_1+d_2 \\ 4=d_1+d_2 \\ d_2=4-d_1 \end{gathered}`

where d_2 is the distance of Milhouse from the center and d_1 is the distance of Bart from the center,

Consider the moment as positive if it is in an anticlockwise direction and negative if it is a clockwise direction.

Thus, the net moment about the center is,

`M=m_1d_1-m_2d_2`

Substituting the known values,

`\begin{gathered} 0=32.4* d_1-27.6*(4-d_1) \\ 0=32.4* d_1-110.4+27.6d_1 \\ 0=60d_1-110.4 \end{gathered}`

By simplifying,

`\begin{gathered} 60d_1=110.4 \\ d_1=(110.4)/(60) \\ d_1=1.84\text{ m} \end{gathered}`

Thus, the distance of the Bart from the center is 1.84 meters.