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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: pl turn to page 133 in hc ver prob no11
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swati.07 (852)

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Olaaa!! Perrrfect answer. 156  [192 rates]

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Mass of man = 40 kg   , g = 10 m/s^2
Frictional force is developed b/w hands,  legs & block side with the wall by the weight of man. So he remains in equilibrium .
He gives equal force on both the walls so he gets equal reaction from both the walls. If he applies unequal forces R that should be different because he can't rest b/w the walls.
Frictional force = 2R to balance his weight.
   R + R = 40g
    2R = 40 * 10
        R = 400 / 2 * 0.8
           = 250 N
 
( b ) Normal force = 250 N
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