Using the American Engineering system of units (AES), a) Calculate the weight of a 170.5 lbm person on the surface of the earth, where the local acceleration due to gravity is 32.174 ft/s2 . Report your answer in pound-force, lbf. b) What would be the weight of a 170.5 lbm astronaut on the moon, where the local acceleration due to gravity is 5.32 ft/s2 . Report your answer in pound-force, lbf.

a) the weight of the person is 170.5 lbf

b) weight of the astronaut on the moon is 28.2 lbf

Explanation:

Given the data in the question;

a)

we know that;

weight on the surface of the earth = m

given that m = 170.5 lbm and g = 32.174 ft/s²

we substitute

weight on the surface of the earth =  170.5 lbm × 32.174 ft/s²

= 5485.667 lbm-ft/s²

1 lbf = 32.174 lbm-ft/s²

so

weight on the surface of the earth = (5485.667 / 32.174) lbf

weight on the surface of the earth = 170.5 lbf

Therefore, the weight of the person is 170.5 lbf

b)

given that;

weight on the surface of the earth = m

m = 170.5 lbm and g = 5.32 ft/s²

weight on the surface of the earth = 170.5 lbm × 5.32 ft/s²

= 907.06 lbm-ft/s²

1 lbf = 32.174 lbm.ft/s²

weight on the surface of the earth = ( 907.06 / 32.174 ) lbf

weight on the surface of the earth = 28.2 lbf

Therefore, weight of the astronaut on the moon is 28.2 lbf

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explanation: