Four acrobats of mass 86.5 kg, 68.0 kg, 62.0 kg, and 55.0 kg form a human tower,

Question

Four acrobats of mass 86.5 kg, 68.0 kg, 62.0 kg, and 55.0 kg form a human tower, with each acrobat standing on the shoulders of another acrobat. The 86.5-kg acrobat is at the bottom of the tower.
(a) What is the normal force acting on the 86.5-kg acrobat?
N

(b) If the area of each of the 86.5-kg acrobat's shoes is 445 cm2, what average pressure (not including atmospheric pressure) does the column of acrobats exert on the floor?
Pa

(c) Will the pressure be the same if a different acrobat is on the bottom?

Explanation / Answer

(a) the acrobat of mass 86.5-kg feel others weight acting on him AND HIS WEIGHT!

(68.0 + 62.0 +55.0+86.5) * g =
Ans : 2660.7 N (if g = 9.8)

(b)

including himself ( Pa = N/ m2) (1m2=10000cm2)

(86.5 + 68.0 + 62.0 + 55.0) * g / (0.0445 * 2) 3050.5618 * g

Ans : 29895.5056 Pa (if g = 9.8)

(c)

depending on the area of each of th person's shoes .

plz rate asap!,

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