Question: need answer asap with work PART 2: PIVOT POINT AT 40 CM MARK (10 CM TO THE LEFT OF THE CENTER OF GRAVITY) X 20
need answer asap with work

PART 2: PIVOT POINT AT 40 CM MARK (10 CM TO THE LEFT OF THE CENTER OF GRAVITY) X 20 cm -10 cm (@ 40em mark) Weight of Stick M2 = 100kg M1 = 200kg 1. Set the pivot point at the mark Ipin = .4 (or 40cm) and choose the stick mass M.= 100 kg 2. Place M1= 200 kg mass at 21 = 0.2 (20cm) 3. Find the position where M2 = 100 kg will produce equilibrium (zero net torque) 4. Determine the distance X between the M2 and the pivot point. X CID = 5 538 words English (U.S.) Text Predictions: On + 100% ! Fit Give Feedback 2:35 e 5/18/
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