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A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is
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at first greater than $m g$ and later becomes equal to $m g$
When the man squatting on the ground he is tilted somewhat, hence he also has to apply frictional force besides his weight.
$R$ (reactional force) $=$ friction force $(f)+m g$
i.e. $R>m g$
When the man does not squat and gets straight up in that case friction $(f) \approx 0$
$R$ (Reactional force) $\approx m g$
Hence, the reaction force $(R)$ is larger when squatting and become equal to $m g$ when no squatting.
$R$ (reactional force) $=$ friction force $(f)+m g$
i.e. $R>m g$
When the man does not squat and gets straight up in that case friction $(f) \approx 0$
$R$ (Reactional force) $\approx m g$
Hence, the reaction force $(R)$ is larger when squatting and become equal to $m g$ when no squatting.
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