The meterstick shown is 100 cm long. It is free to pivotaround its center of gravity (CG), which is at the 50 cm mark.There is a 20.0 N block hanging from the 80 cm mark.Decide where each of the other blocks should be placed, oneat a time, to balance out the 20.0 N block.At what mark on the meter stick would you place a 19.0 Nblock to balance the 20.0 N block?mark:cmAt what mark on the meter stick would you place a 44.0 Nblock to balance the 20.0 N block?mark:cm10320230WOR405060708090

We are given the meter scale. We are given a weight hanging on one side. We are also given distance where it is hanged. We need to find the position where other weight is to be hanged to balance the stick. We apply equilibrium of torque about center of mass.First part: Weight hanged W'=20 N Distance from center is d'=80-50d'=30cm Other weight W=19 N Let its distance be d from center. Hence applying equilibrium of torque we get Wd=W'd'd=W'd'Wd=20×3019d=31.58cm Hence the mark is x=50-dx=50-31.58x=18.42 cm The mark is 18.42 cm. First part: Other weight W=44 N Let its distance be d from center. Hence applying equilibrium of torque we get Wd=W'd'd=W'd'Wd=20×3044d=13.64cm Hence the mark is x=50-13.64x=50-31.58x=36.36 cm The mark is 36.36 cm.

Answered: The meterstick shown is 100 cm long. It…