Answer
The center of gravity of the clamp should be a distance of 1.35 m from the left-hand end of the rod.
Work Step by Step
Let $m_1$ be the mass of the rod and let $m_2$ be the mass of the clamp. Let $r_2$ be the distance of the clamp's center of gravity from the left end of the rod. Note that the center of gravity of the rod is located at the midpoint of the rod.
$x_{cog} = \frac{m_1~r_1+m_2~r_2}{m_1+m_2}$
$m_2~r_2 = x_{cog}~(m_1+m_2)-m_1~r_1$
$r_2 = \frac{x_{cog}~(m_1+m_2)-m_1~r_1}{m_2}$
$r_2 = \frac{(1.20~m)(1.80~kg+2.40~kg)-(1.80~kg)(1.00~m)}{2.40~kg}$
$r_2 = 1.35~m$
The center of gravity of the clamp should be a distance of 1.35 m from the left-hand end of the rod.
