Answer:
Explanation:
Given
Mass of dancer [tex]m=60\ kg[/tex]
Area of contact [tex]A=26\ cm^2[/tex]
(a)If dancer is stationary
Pressure exerted by Floor is given by load per unit area of contact
[tex]P=\frac{weight}{A}[/tex]
[tex]P=\frac{mg}{A}[/tex]
[tex]P=\frac{60\times 9.8}{26\times 10^{-4}}[/tex]
[tex]P=226.153 kN[/tex]
(b)If dancer is leaping upwards with an acceleration of [tex]a=5 m/s^2[/tex]
So apparent weight of dancer is [tex]W'=mg+ma[/tex]
[tex]W'=m(g+a)=60\cdot (9.8+5)[/tex]
[tex]W'=888\ N[/tex]
[tex]Pressure=\frac{W'}{A}[/tex]
[tex]Pressure=\frac{888}{26\times 10^{-4}}[/tex]
[tex]=341.53\ kN[/tex]
The pressure exerted by the dancer on the floor when the dancer is stationary is 2.26 x 10⁵ N/m².
The pressure exerted by the dancer on the floor when the dancer is leaping up is 3.42 x 10⁵ N/m².
The given parameters;
The pressure exerted by the dancer on the floor when the dancer is stationary is calculated as follows;
F = ma + mg
a = 0 (stationary)
F = mg
[tex]P = \frac{F}{A} = \frac{mg}{A} = \frac{60 \times 9.8}{26\times 10^{-4}} = 2.26\times 10^5 \ N/m^2[/tex]
The pressure exerted by the dancer on the floor when the dancer is leaping up is calculated as follows;
F = ma + mg
F = m(a + g)
[tex]P = \frac{m(a + g)}{A} = \frac{60(5 + 9.8)}{26\times 10^{-4}} = 3.42 \times 10^5 \ N/m^2[/tex]
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