Question

Ex1: A man of mass 84 kg stands
upright on the floor.
If the area of contact of his
shoes and floor is 420 cm?,
determine the average pressure
he exerts on the floor.
(take g=10 N/kg)​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Pressure=200\:pascal}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: man(m) = 84 \: kg \\ \\ \tt: \implies Area \: of \: contact = 420 \: cm^{2} \\ \\ \tt: \implies Acceleration \: due \: to \: gravity(g) = 10 { \: m/s}^{2} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Pressure \: exerted(p) =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies F = mg \\ \\ \tt: \implies F = 84 \times 10 \\ \\ \tt: \implies F= 840 \: N - - - - - (1) \\ \\ \tt \circ \: Area = \frac{420}{100} = 4.2 \: m^{2} \\ \\ \bold{As \: we \: know \: that } \\ \tt: \implies Pressure = \frac{Force}{Area} \\ \\ \tt: \implies Pressure = \frac{840}{4.2} \\ \\ \green{\tt: \implies Pressure = 200 \: pascal} \\ \\ \green{\tt \therefore Pressure \: exerted \: on \: floor \: is \: 200 \: pascal}$

Answer 2

Answer:

${\longrightarrow{\boxed{\rm{Pressure \: exerted = 200 \: Pascal}}}}$

Step-by-step explanation:

Given:-

⇢ A man of mass 84 kg stands upright on the floor. If the area of contact of his shoes and floor is 420 cm.

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Find:-

⇢ Determine the average pressure he exerts on the floor. (Take g = 10 N/kg).

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Using formula:-

${\longrightarrow{\boxed{\rm{Force = Mass \times Gravity}}}}$

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Calculations:-

$\longrightarrow\rm{84 \times 10}$

$\longrightarrow\rm{840 \: N}$ ---- Equation (1)

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Now finding for area:-

$\longrightarrow\rm{\dfrac{420}{100}}$

$\longrightarrow\rm{4.2 \: m^{2}}$ ---- Equation (2)

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Therefore, 4.2 m² is the required area.

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Using pressure formula:-

${\longrightarrow{\boxed{\rm{Pressure = \dfrac{Force}{Area}}}}}$

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Substitute the value from eq (1) and eq (2):-

$\longrightarrow\rm{\dfrac{840}{4.2}}$

${\longrightarrow{\boxed{\rm{200 \: P}}}}$

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Therefore, 200 pascal is the required pressure.