A table has a heavy circular top of radius 1   m and mass 20   kg placed on four light (considered massless) legs placed symmetrically on its circumference. The maximum mass that can be kept anywhere on the table without toppling it is close to

Step-by-step solution

From the above figure, we can see that due to mass placed at edge of table, it tends to rotate about a line joining two legs. Hence, we take pivot in balancing of weight of mass m and weight of mass M of table at point B . In critical condition, Moment of weight of mass m about B = Moment of weight of table about B ⇒ m g · A B = M g · B C ⇒ m = M · B C A B ........(i) From geometry of table, B C = R cos 45 ° = R 2 So, A B = R - R 2 = R 1 - 1 2 Hence, from Eq. (i), we have m = M × R / 2 R 1 - 1 2 Here, M = 20 kg ∴ m