A meter stick is balanced on knife edge as it's centre. when 2 coins each of mass 5kg are put one on top of other at 12.0cm mark the stick is found to be balance at 45.0cm. what is the mass of meter stick?
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Solution-
Let m be the mass of stick concentrated at C , the 50 cm mark.
For equilibrium about C, the 45cm mark,
10g (45-12) = mg (50-45)
10g × 33 = mg × 5
m = 10 × 33 / 5
= 66gm.
Let m be the mass of stick concentrated at C , the 50 cm mark.
For equilibrium about C, the 45cm mark,
10g (45-12) = mg (50-45)
10g × 33 = mg × 5
m = 10 × 33 / 5
= 66gm.
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