Engineering Mechanics, Lecture notes of Engineering

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Fg = Py, sin15° = 120.35 N ee teiocal force ax the rp Hence, friction force = 120.35 N [Ans] 7.4.4 Screw Friction In every enginccring application, use of screw is very common. Like wedges, screws have application of frictional forces aiding the lifting/lowering the heavy loads. z bent) Figure 7.3 Screw jack and irs FBD. Figure 7.3(a) shows a typical right-hand threaded screw-jack. Figure 7.3(b) shows a development of 2 typical screw thread with helix angle a Clearly, tno= es 2ar, Jar, where L is the lead, p is the pitch, mis che numberof scar _ of the screw thread and ris the mean radius of thread. Finally, Figure 7.3(c) shows FBD of a representa- tive thread when the load W7 is lifted by turning the screw with torque M. Forces are shown in radial and axial directions. F and N here represent the sum of reactions of all threads. Therefore, TIF =0= Nosa—Fsina-W=0 => Neos a-Fsina= W F=uN = Neos @—- ys sin @) = W Ww cosa = fi, sina [@, static angle of friction] SNe cost +,) Finally, taking moment of all these forces about the screw-axis (i.c., z-axis) M-—r_(Nsin a+ F cos a) =0 M=nNisina+4, cove) = = sina) Bur =Ne= But “i, = tang, GA Guilt ce