biomechanics homework question I am stuck on

Question

A 2m barbell is loaded with a 10kg and 20kg plate on its left side and three 10kg plates on its right side. The mass of the unloaded barbell is 20kg. On the left side, the 10kg plate is 5cm from the end of the bar and the 20kg plate is 28cm from the end of the bar. On the right side, the three 10kg plates are 10, 15, and 20cm from the end of the bar. If the axis of rotation is the centre of the bar, where is the centre of gravity of the loaded barbell?

Answer 1

The center of gravity is calculated by... x_cog = ∑m_ix_i / ∑m_i

Since the axis of rotation for the bar is at the center of the bar, we'll take that center as the origin. therefore the positions of each weight are as follows:

Left:

10 kg position = -95 cm

20 kg position = -72 cm

Right:

10 kg position = +90 cm

10 kg position = +85 cm

10 kg position = +80 cm

Position of center of gravity for the unloaded barbell itself is at the center.

∑m_ix_i = 10(-95) + 20(-72) + 10(90) + 10(85) + 10(80) + 20(0) = -2390 + 2550 = 160 kg.cm

∑m_i = 10+20+10+10+10 +20 = 80 kg

Position of the COG for the loaded barbell = 160/80 = 2 cm.

So the position of the center of gravity for the loaded barbell is 2 cm to the right.