Maximum distance physics forums, Here the fulcrum is the edge of the table. since the block is 30 cm on the table and 20 cm outside it, you can take the downward force acting on the left hand side as 3/5 x 7 = 4.2 kg. this acts at the cg which is 15 cm away. hence moments on the left side = 4.2 x .15 = 0.63 kg-m. similarly moment on the right side = (2/5 x 7) x .1 = 0.28.. Solved: thin 2.00kg box rests 6.50kg board , A thin 2.00kg box rests on a a 6.50kg board that hangs over the end of a table, as shown in the figure. how far can the center of the box from the end of the table before the board begins to tilt? answer: x=_____m.
Solved: thin 3.00kg box rests 6.50kg board han, Answer thin 3.00kg box rests 6.50kg board hangs table, shown figurehow cent skip navigation chegg home. A thin 2.50 kg box rests 6.00 kg board hangs, A thin 2.50 kg box rests edge 6.00 kg board length 2m hangs table. center board table board begins . A thin 2kg box rests 6kg board 50cm long 20cm, If assume cg board center, summing moments board table. = length box edge table. 2kg**+ 6kg**25cm-(8kg**30cm)=0.
