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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: gud prob ,rate assured
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[Post New]18 Apr 2008 20:37:03 IST
    Subject: Re:gud prob ,rate assured
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sarthak321 (62)

Cool goIITian

Olaaa!! Perrrfect answer. 10  [16 rates]
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ok..

net force acting on the rider are mg downwards and friction due to contact with the pole upwards...

now, mg - fr = ma
        72 - fr = 7.2 X 3 (taking g as 10m/s^2 ..m = 72/10 .. u may use 9.8 if required )
        fr = 72 - 21.6
       fr = 50.4N

also since every action has an equal and an opposite reaction friction will also act on the pole in downward direction and with magnitude 50.4N

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