How much pressure will a man of weight 80 kgf exert on the ground if (1) he is lying and (2)he is standing on his feet? Given that the area of the body of the man is 0.6m² and that of his foot is 80cm².?

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Answer 1 · 2017-12-01T14:14:15

Lying: $1309 Pa$ $"1309 Pa"$
Standing: $98 062.5 Pa$ $"98 062.5 Pa"$

When the man is lying:

Pressure $= \frac{Force}{Area}$ $= "Force"/"Area"$

$Force = m g = (80 kg \cdot {9.81 m/s}^{2}) = 784.8 N$ $"Force" = mg = ("80 kg" * "9.81 m/s"^2)= "784.8 N"$

$P = \frac{784.8 N}{{0.6 m}^{2}} = 1308 Pa$ $P = "784.8 N"/"0.6 m"^2 = "1308 Pa"$

When the man is standing:

$Pressure = \frac{Force}{Area}$ $"Pressure" = "Force"/"Area"$

The same force as above because of the same weight:

$Force = 784.8 N$ $"Force = 784.8 N"$

So

$P = \frac{784.8 N}{80 \times 10^{- 4} m^{2}} = 98 062.5 Pa$ $P = "784.8 N"/(80 xx 10^-4"m"^2) = "98 062.5 Pa"$

Note: Do not forget to convert ${cm}^{2}$ $"cm"^2$ to $m^{2}$ $"m"^2$ !