91Ó°ÊÓ

According to the second condition of equilibrium, the torque about any point is zero.In this problem, the table is not tipping over, and the torque about any point is zero.

The mass of the table is \(m = 24.0\;{\rm{kg}}\).

The mass of the person is \(M = 66.0\;{\rm{kg}}\).

The length of the table is \(L = 2.20\;{\rm{m}}\).

You can assume that the mass of the table is at its middle point.

Let x be the distance of the person from the edge of the table.

The free-body diagram of the problem is given below.

Now, you know that the second condition for equilibrium is that the torque about any axis is zero. Let the right leg of the table be your rotational axis. Then, for stability, the torque about this point is zero.

Then,

\(\begin{array}{c}mg \times \left( {0.60\;{\rm{m}}} \right) - Mg \times \left( {0.50\;{\rm{m}} - x} \right) = 0\\M\left( {0.50\;{\rm{m}} - x} \right) = m\left( {0.60\;{\rm{m}}} \right)\\\left( {66.0\;{\rm{kg}}} \right) \times \left( {0.50\;{\rm{m}} - x} \right) = \left( {24.0\;{\rm{kg}}} \right) \times \left( {0.60\;{\rm{m}}} \right)\\x = 0.28\;{\rm{m}}\end{array}\).

Hence, the person can sit at 0.28 m from the edge without tipping it over.