Question

Problem 2 Figure 1 PO stillin An object of mass 10 kg (m) is lifred and lowered through a vertical distance of 2.0m using a power screw similar to that of Figure I. Parameters are : Thread angle B: 30° Pitch diameter dp: 8.0 cm lead (pitch) P: 5.0 mm Frictional coef' h 0.2 Determine a) lead angle Trad & deg! b) normal angle (radd dey) c) torque to lift Tp (Nm) d) torque to lower Tu e) work done w (Nm) for one cycle f) is the system Self locking ? mg rc Thrust Bearing (bulltype) (Nm)

Answer 1

Soution given ka=9.81X10NW man lo L= 2m 20 dp: Sim o.08m HREO.N To find ) dead angle - tan Sxo (TXO-08) tan- Td 1.1h0 noimal amg &: tam (HA) =tam Coz) l1-3 in)Tootque to t w dp tomlo+ )taamam.as =9.81 X1otx e.08 tam T 865.clNm iv) Toau to lourer wdp Lo-a) tam 2 9.8(XLohKo.o tam C3 - /- I4) 2 R=703. 21NM herk don er y cle v) = 9.81x 10 x 2 wo 196- 2 kNm VSelf lodkny ar nat Sinu TR poliAive & also aly O.2 7 d.02 KR 7 Sely Nacking Souw