Quantity the unknown plenty m1, m2, m3 with m1 on the backside
Steadiness the torques
4cm*5g=2cm*m1 => m1 = 10g unknown mass on the backside
10g 5g = 15g
1cm*15g = 5cm*m2 => m2 = 3g unknown mass second from high
3g 15g = 18g
8cm*18g=4cm*m3 => m3 = 36g unknown mass at high
The underside left mass is 10g., = 15g. for the decrease section.
The center left mass is (15/5) = 3g., complete now (15 3) = 18g.
Mass M must be twice that, = 36g.
This post is last updated on hrtanswers.com at Date : 1st of September – 2022