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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/s^2. 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/s^2 a = ? Fapplied - Fgrav = ma Fgrav = mg a = Fapplied - Fm/m = Fapplied/m - mg/m = Fapplied/m - g a = 10 lbf/50 lb | 32.2 lb·ft/s^2 /1 lbf | -32.2 ft/s^2 a = -25.8 ft/s^2 downward PROBLEM 1.17 Fgrav,E = m * ge (on Earth) Fgrav,S = m * gs (on Space Station) Mass remains the same. So, Fgrav,S / Fgrav,E = gs / ge => Fgrav,S = Fgrav,E (gs/ge) = 700 N (6 m/s^2 / 9.81 m/s^2) = 428.1 N 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 In lbf