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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