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From the markings 20 cm, 40 cm, 60 cm, 8...

From the markings 20 cm, 40 cm, 60 cm, 80 cm and 100 cm on a light metre ruler, masses of 1 g, 2 g, 3g , 4 g, 5 g, respectively are suspended. At which mark should the ruler be pivoted so that it remains horizontal?

Text Solution

Verified by Experts

The gravitational forces on 1 g, 2 g, 3 g, 4 g, 5 g, masses are downward parallel forces. Suppose their resultant acts at the x cm mark on the scale. Hence,
`x=(oversetundersetsum5(i=1)F_(i)x_(i))/(oversetundersetsum5(i=1)F_(i))`
`=(1xx20+2xx40+3xx60+4xx80+5xx100)/(1+2+3+4+5) =(1100)/(15)=73.3`
When the downward resultant and the upward normal force of the support, act at the same point the metre ruler remains in equilibrium. Hence the ruler should be pivoted at the 73.3 cm mark.
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