The question is..."A metre stick is balanced on a knife edge at its centre. When two coins each of 5 g are put on top of the other at 12 cm mark the stick is found to be balanced at the 45 cm mark. Mass of stick...?" And the solution is as attached. I could not understand the statement"The net torque will be conserved for rotational.." And the equation that follows it.

 

balance the torque about the point Plet the normal rxn be N at S, let the mass of the metre stick be mthen N=mg+10gbalance the torque12*10g+mg*50=N*45120g+mg*50=mg+10g*45120g+50mg=45mg+450g5mg=330gm=3305=66 gm Regards

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