Hiten whose mass is $75 k g$ is on a summer trip to goa. The cross-sectional area of Aryan's foot is $75 c m^{2}$ .
a) Find the pressure at the bottom of his feet.

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Solution

Force applied by Hiten on the sand is equal to his weight.
$F = m g$
$F = 75 \times 10 = 750 N$

$A r e a (A) = 2 \times 75 c m^{2}$
$A = 150 \times 10^{- 4} m^{2} = 0.015 m^{2}$

From definition of pressure,
$P r e s s u r e = \frac{T h r u s t}{A r e a}$
$P = \frac{m g}{A}$
$P = \frac{750}{0.015} = 50000 P a$

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Hiten whose mass is 75 kg is on a summer trip to goa. The cross-sectional area of Aryan's foot is 75 cm2. a) Find the pressure at the bottom of his feet.

Force applied by Hiten on the sand is equal to his weight. F=mg F=75×10=750N Area(A)=2×75 cm2 A=150×10−4 m2=0.015 m2 From definition of pressure, Pressure=ThrustArea P=mgA P=7500.015=50000 Pa

Hiten whose mass is $75 k g$ is on a summer trip to goa. The cross-sectional area of Aryan's foot is $75 c m^{2}$ .
a) Find the pressure at the bottom of his feet.