A uniform rod is 2.00 m long and has mass 1.80 kg. A 2.40-kg clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.20 m from the left-hand end of the rod?

Solution 3E Step 1 of 1: The location of center of gravity of the rod clamp system x = m 1 1m 2 2 cm m1+m 2 1.2 m = (1.8 kg×1 m)+2.4 k2×x 1.8 kg+2.4 kg x 2 1.2 m(1.8 kg+2.4 kg)(1.8 kg)(1 m) 2.4 kg = 1.35 m