Determine the force of gravity and body weight with a mass of: 50kg; 0.35g; 4t: what pressure is exerted on the arena

| July 24, 2021 | education

Determine the force of gravity and body weight with a mass of: 50kg; 0.35g; 4t: what pressure is exerted on the arena by a circus elephant standing on one leg. The mass of an elephant is 3.5t, the area of one sole is 700cm2

m1 = 50 kg.

m2 = 0.35 g = 0.00035 kg.

m3 = 4 t = 4000 kg.

g = 10 m / s2.

Fт1 -?

Ft2 -?

Ft3 -?

P1 -?

P2 -?

P3 -?

The force of gravity Fт is called the force with which the earth attracts bodies located on its surface. Gravity is applied to the body.

Ft1 = m1 * g, Ft1 = 50 kg * 10 m / s2 = 500 N.

Ft2 = m2 * g, Ft2 = 0.00035 kg * 10 m / s2 = 0.0035 N.

Ft3 = m3 * g, Ft3 = 4000 kg * 10 m / s2 = 40,000 N.

The weight of the body P is the force with which the body stretches the suspension or presses on the support. The body weight P is applied to the support or suspension.

P1 = m1 * g, P1 = 50 kg * 10 m / s2 = 500 N.

P2 = m2 * g, P2 = 0.00035 kg * 10 m / s2 = 0.0035 N.

P3 = m3 * g, P3 = 4000 kg * 10 m / s2 = 40,000 N.

Answer: Ft1 = P1 = 500 N, Ft2 = P2 = 0.0035 N, Ft3 = P3 = 40,000 N.

m = 3.5 t = 3500 kg.

g = 10 m / s2.

S = 700 cm2 = 0.07 m2.

R – ?

P = F / S = m * g / S.

P = 3500 kg * 10 m / s2 / 0.07 m2 = 500000 Pa.

Answer: the elephant’s pressure is P = 500,000 Pa.



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