Four acrobats of mass 78.5 kg, 68.0 kg, 62.0 kg, and 55.0 kg form a human tower,

Question

Four acrobats of mass 78.5 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 89.5-kg acrobat is at the bottom of the tower.

a. I found the normal force to be 2582.3 which was marked correct
Then
(b) If the area of each of the 89.5-kg acrobat's shoes is 355 cm2, what average pressure (not including atmospheric pressure) does the column of acrobats exert on the floor?
I did
2582.3/2(355*1 m^2/10^4 cm^2) and it kept marking me wrong! Am i doing something wrong or if not what is the answer to compare to mine which was 363704.2 (was told i was off by a multiple of 10)

Explanation / Answer

The formula for pressure is P = F/A

The area is 355 cm2 which must be converted to m2. That is .0355 m2

Since that is one shoe, we can multiply by two to get the total area, which is .071 m2

Then, P = F/A

P = (2582.3)/(.071) = 36370.4 Pa so yes, you were off by a factor of 10.

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