Question:
A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is
(a) constant and equal to mg in magnitude
(b) constant and greater than mg in magnitude
(c) variable but always greater than mg
(d) at first greater than mg and later becomes equal to mg
Solution:
(d) In the process of squatting on the ground he gets straight up and stand. Then he is tilted somewhat, the man exerts a variable force on the
ground to balance his weight, hence he also has to balance frictional force besides his weight in this case.
N = Normal reaction force = friction + mg => N > mg
Once the man gets straight up that variable force = 0 =>
Normal reaction force = mg