Question

The piano stool consists of a circular seat supported by a square threaded screw that goes through the frame mechanism sitting on the floor. The frame mechanism has a threaded hole. The radius of the square threaded screw is 0.380 in. It has a lead of 0.510 in. The coefficient of friction between the threads on the screw and the frame is Hs 0.38 he screw andthe 220lb Determine the magnitude of the moment in ft. Ib necessary to raise the stool ft lb Determine the magnitude of the moment in ft lb necessary to lower the stool ft lb

Answer 1

RADIUS OF SQUARE TenpeD SCREW, = 0.38 in LEAD OF SCREW, L = 0.5l in HEUX ANGLE, ) o = tant ( 22 ) tas' ( 0-51" (27x0-38 = 12.06° FRICTWN, P = 0.38 Øs = tan (0.38) = 20.810 COEFFICIENT of STATIC MOMENT NECESSARY TO RAISE THE STOOL - Wotan (Ps+0) – 220 10 * 0.38 tan (20-21.6) = 54.021 in.lb = 4.502 ft-lb [ift = 12in] +12.06) MOMENT NECESSARY TO LOWER TITE STOOL = wrtan ( 8 0 ) = 22026 x 0.38 in tans (20.8) - 12.06 = 12.867 in.lbl = 1.072 ft. lb