Question

A barbellbell is loaded with two 20 kg plates on its right side and two 20kg plates on its left side. The barbell is 2.2m long, and its unloaded mass is 20kg. The two 20kg plates on the right side are locked in place at 35cm and 40cm from the right end of the bar. The two 20kg plates on the left side have slipped one iss 30 cm from the left end of the bar and the other is 20 cm from the left end of th bar. where is the center of gravity of this bar?

Answer 1

Answer:

Center of gravity of the system of all masses is 105 cm from the left end of the rod

Explanation:

Let the position of Left end of the rod is our reference

So here we will have

$m_1 = 20 kg$

$x_1 = 20 cm$

$m_2 = 20 kg$

$x_2 = 30 cm$

$m_3 = 20 kg$

$x_3 = 110 cm$

$m_4 = 20 kg$

$x_4 = (220 - 40) = 180 cm$

$m_5 = 20 kg$

$x_5 = (220 - 35) = 185 cm$

so we will have

$x_{cm} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + m_4x_4 + m_5x_5}{m_1 + m_2 + m_3 + m_4 + m_5}$

$x_{cm} = \frac{20(20 + 30 + 110 + 180 + 185)}{20 + 20 + 20 + 20 + 20}$

$x_{cm} = 105 cm$