A0.120-kg, 50.0-cm-long uniform bar has a small 0.055-kg mass glued to its left end and a small 0.110-kg mass glued to the other end. the two small masses can each be treated as point masses. you want to balance this system horizontally on a fulcrum placed just under its center of gravity. how far from the left end should the fulcrum be placed?

29.8 cm

Explanation:

For the system to be at equilibrium, the counterclockwise moments must be equal to clockwise moments. Let the fulcrum be placed x cm from the left end. If the fulcrum is placed in the middle of the bar, the clockwise moments are the net moments so the fulcrum must be moved closer to the right hand side to decrease the effects of the moment due to the mass on the right.  Let the fulcrum be placed x cm from the left end.

counterclockwise moments = clockwise moments

(0.055 kg * x) + (0.120*(x-25)) = 0.110 kg * (50-x)

0.055x + 0.12x - 3 = 5.5 - 0.11x

                            x = 29.8 cm

answer: strength i guess

explanation: