Four acrobats of mass 67.0 kg, 68.0 kg, 62.0 kg, and 55.0 kg form a human tower,

Question

Four acrobats of mass 67.0 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 67.0-kg acrobat is at the bottom of the tower (a) What is the normal force acting on the 67.0-kg acrobat? (Enter the magnitude only.) N (b) If the area of each of the 67.0 kg acrobat's shoes is 460 cm^2 what average pressure (not inducing 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? Calculate the mass of a solid gold rectangular bar that has dimensions = 4 00 cm times 13.0 cm times 21.0 cm. (The density of gold is 19.3 times 10^3 kg/m^3.) kg

Explanation / Answer

(1)normal force= weight of 4 acrobats

Fn= (67+68+62+55)(9.8)

Fn=2469.6N

(b)pressure= force /area

area= 460cm2 =0.046m2

P= 2469.6/0.046

P=53686.96 pa

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