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A uniform rod of length 1m is placed on a table so that 0.2m projects over the edge. if the weight of the rod is 100, find the greatest weight which can be attached to the free end of the rod without causing it to topple over.......solve .........
Since the unit of mass is not given ,
let us consider the mass of the rod of uniform length 1 m as 100Kg
since it is uniform each unit length has a mass of 1Kg
the edge of the table acts as the pivot of the rod.
so the mass of the 0.8m part is 80Kg and the 0.2m part hanging out has mass 20Kg
the normal force exerted at pivoted point is mg = 1000N
R - F1 - F2 =0
where R is normal force at pivot , F1 is force at the 0.8m end on the table and F2 is the force that can be put on the free end
R = F1 + F2
1000 = 0.8 * 10 + 0.2 * M * g
200 = 20 M
M = 10Kg
therfore we can put 10Kg mass on the free end of the 0.2m part !
Please approve if it is right !
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