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 - Fgrav)/m = (Fapplied - mg)/m => a = Fapplied/m - g a = 10 lbf/50 lb |32.2 ft • lb/s^2/lbf| - 32.2 ft/s^2 a = -25.8 ft/s^2 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/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
