Question

A 60.0 kg ballet dancer stands on her toes during a performance with four square inches (26.0 cm2) in contact with the floor. What is the pressure exerted by the floor over the area of contact in the following cases?
(a) if the dancer is stationary
(b) if the dancer is leaping upwards with an acceleration of 5.00 m/s2

Answer 1

Answer:

Explanation:

Given

Mass of dancer $m=60\ kg$

Area of contact $A=26\ cm^2$

(a)If dancer is stationary

Pressure exerted by Floor is given by load per unit area of contact

$P=\frac{weight}{A}$

$P=\frac{mg}{A}$

$P=\frac{60\times 9.8}{26\times 10^{-4}}$

$P=226.153 kN$

(b)If dancer is leaping upwards with an acceleration of $a=5 m/s^2$

So apparent weight of dancer is $W'=mg+ma$

$W'=m(g+a)=60\cdot (9.8+5)$

$W'=888\ N$

$Pressure=\frac{W'}{A}$

$Pressure=\frac{888}{26\times 10^{-4}}$

$=341.53\ kN$