Question

A board of mass 4.0 kg is used as a seesaw. Mass A in on the left end of the seesaw and is 30 kg. It is 25 m from the pivot point, orfulcrum. The pivot is at the center of gravity of the board. At what distance, x, to the right of the pivot must a mass of 25 kg be located tobalance the seesaw? Answer with no units and to correct sig. fig.

Answer 1

Given data

The mass of the board is mb = 4 kg

The mass of block A is mA = 30 kg

The mass of the block B is mB = 25 kg

The location distance of block A from the pivot point is d = 25 m

The location distance of block B from the pivot point at the right end is x.

The figure for the above configuration can be drawn as follows:

Taking the moment about the pivot point, the expression is written as follows:

`\begin{gathered} m_A* d-m_B*(x)=0 \\ m_A* d=m_B* x \\ x=(m_A* d)/(m_B) \end{gathered}`

Substitute the value in the above equation.

`\begin{gathered} x=\frac{30\text{ kg}*25\text{ m}}{25\text{ kg}} \\ x=30\text{ m} \end{gathered}`

Thus, the distance at which it must be located in the right end from the pivot point is 30 m.