A man doing push-ups pauses in the position shown in the figure(Figure 1) . His mass = 68 Determine the normal force exerted by the floor on each hand. Determine the normal force exerted by the floor on each foot. A=42cm B=95cm c=28cm

Total Weight = 68 kg

Let Rh be the total reaction from the ground on the man's hands.

Let Rl be the total reaction from the ground on the man's legs.

Rh/Rl= 95/42 (As ground reactions are in the inverse proportions of distances)

Rh + Rl = 68 kg (Total ground reactions = total weight)

Solving the above 2 equations, we get Rl=20.8 kg Rh=47.2 kg

Therefore, force exerted by the ground on each foot = Rl/2 = 20.8/2 = 10.4 kg

As the questions ask for force, so we convert 10.4 kg into force units.

10.4 kg = 10.4 x 9.8 m/s^2 = 101.9 N

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