Four acrobats of mass 70.0 kg, 68.0 kg, 62.0 kg, and 55.0 kg form a human tower,

Question

Four acrobats of mass 70.0 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 70.0-kg acrobat is at the bottom of the tower.

(a) What is the normal force acting on the 70.0-kg acrobat?
N

(b) If the area of each of the 70.0-kg acrobat's shoes is 430 cm2, what average pressure (not including atmospheric pressure) does the column of acrobats exert on the floor?
  
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. Pa

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

Explanation / Answer

a)

Normal force acting on the 70 kg acrobat

= M g , where M is the sum of mass of all other acrobats

=(68 + 62 + 55) x 9.8 N

=1813 N

b)

Area of two shoes = 430 x 2 = 860 cm^2 = 860 / (100 x 100) m^2 = 0.086 m^2

Force on the floor = M' g , where M' is sum of masses of all acrobats

=( 70 + 68 + 62 + 55 ) x 9.8 N = 2499 N

Thus pressure Force / Area = 2499 N / 0.086 m^2 = 29058.14 Pa

c) Yes , if area of shoes of all acrobats are same, pressure would be same bcoz Normal Force on floor will be same in all cases...

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