A60.0 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

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

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)

Learn more here:link

Explanation:

Given

Mass of dancer

Area of contact

(a)If dancer is stationary

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

(b)If dancer is leaping upwards with an acceleration of

So apparent weight of dancer is

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

increase masses, decrease separation