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(a) if she stands on two feet

(b) if she stands on one feet

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The pressure \[P\] acting on an object is given by

\[P = \dfrac{F}{A}\] …… (1)

Here, \[F\] is the force acting on the object and \[A\] is the surface area on which the force is acting.

We have given that the weight of the girl is \[550\,{\text{N}}\].

\[mg = 550\,{\text{N}}\]

Here, \[m\] is the mass of the girl.

The area of contact of one flat shoe with the ground is \[160\,{\text{c}}{{\text{m}}^2}\].

\[A = 160\,{\text{c}}{{\text{m}}^2}\]

Convert the unit of the area of the one flat show in contact with the ground to the SI system of units.

\[A = \left( {160\,{\text{c}}{{\text{m}}^2}} \right)\left( {\dfrac{{{{10}^{ - 4}}\,{{\text{m}}^2}}}{{1\,{\text{c}}{{\text{m}}^2}}}} \right)\]

\[ \Rightarrow A = 160 \times {10^{ - 4}}\,{{\text{m}}^2}\]

Hence, the area of the one flat shoe in contact with the ground is \[160 \times {10^{ - 4}}\,{{\text{m}}^2}\].

(a) In the first case, the girl is standing on the ground with her two feet.

Hence, the area \[A'\] of the shoe in contact with the ground is twice the area of contact of the one shoe with the ground.

\[A' = 2A\]

The force acting on the two feet of the girl is her own weight. Let us calculate the pressure \[{P_1}\] exerted by the girl on the ground when she is standing on her two feet.

Substitute \[{P_1}\] for \[P\], \[mg\] for \[F\] and \[A'\] for \[A\] in equation (1).

\[{P_1} = \dfrac{{mg}}{{A'}}\]

Substitute \[2A\] for \[A'\] in the above equation.

\[{P_1} = \dfrac{{mg}}{{2A}}\]

Substitute \[550\,{\text{N}}\] for \[mg\] and \[160 \times {10^{ - 4}}\,{{\text{m}}^2}\] for \[A\] in the above equation.

\[{P_1} = \dfrac{{550\,{\text{N}}}}{{2\left( {160 \times {{10}^{ - 4}}\,{{\text{m}}^2}} \right)}}\]

\[ \Rightarrow {P_1} = 1.7187 \times {10^4}\,{\text{N/}}{{\text{m}}^2}\]

(b) In the second case, the girl is standing on the ground with her one foot.Hence, the area \[A\] of the shoe in contact with the ground is the area of contact of the one shoe with the ground.The force acting on the one feet of the girl is her own weight.Let us calculate the pressure \[{P_2}\] exerted by the girl on the ground when she is standing on her one foot.Substitute \[{P_2}\] for \[P\] and \[mg\] for \[F\] in equation (1).

\[{P_2} = \dfrac{{mg}}{A}\]

Substitute \[550\,{\text{N}}\] for \[mg\] and \[160 \times {10^{ - 4}}\,{{\text{m}}^2}\] for \[A\] in the above equation.

\[{P_2} = \dfrac{{550\,{\text{N}}}}{{160 \times {{10}^{ - 4}}\,{{\text{m}}^2}}}\]

\[ \Rightarrow {P_2} = 3.4375 \times {10^4}\,{\text{N/}}{{\text{m}}^2}\]