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PROBLEM 1.16 The FBD of the object is as shown with an upward applied force of 10 lbf and the force downward due to gravity where Fgrav = mg and g is given as 32.2 ft/s². Summing forces yields the following equation that can be rearranged to solve for acceleration. It is assumed that up is positive. Fapplied = 10 lbf m = 50 lb g = 32.2 ft/s² a = ?, g = T Fapplied - Fgrav = ma Fgrav = mg a = Fapplied / m - Fgrav / m = Fapplied / m - mg / m = Fapplied / m - g a = 10 lbf / 50 lb / 32.2 ft•lb / s² / 1 lbf / 32.2 ft = -25.8 ft/s² downward PROBLEM 1.17 Fgrav,E = m ge (m Earth) Fgrav,S = m gs (m Space Station) Mass remains the same. So, Fgrav,S / Fgrav,E = gs / ge => Fgrav,S = Fgrav,E ( gs / ge ) = 700 N ( 6 m/s² / 9.81 m/s² ) = 428.1 N Fgrav,S Also, Fgrav,E = 700 N | 0.22481 lbf / 1 N | = 157.4 lbf Fgrav,S = 428.1 N | 0.22481 lbf / 1 N | = 96.2 lbf lbf.