Relating Effort and the Weight Lifted in Screw Jacks
Let’s assume a few of the elements associated with the above two parameters and assign them in the following manner:
p = pitch of the screw,
d = mean diameter of the screw and,
α = helix angle.
Also a cross-sectional analysis of the produced angular displacement can be illustrated with the diagram shown alongside and the following expression:
tan α = P/πd
P = the effort applied for lifting the load,
W = Weight of the body being lifted and
µ = Coefficient of friction, between the screw threads and the stand threads.
As discussed earlier, since the calculations involved with a screw jack is similar and can be compared to that of an inclined plane, we can consider the external applied force to be horizontal.
Also, the weight which is being elevated generates a friction F acting downwards equal to:
F = µR
Where F = force of friction,
µ = coefficient of friction,
And R = normal reaction developed between the interacting surfaces.
Resolving the forces over the horizontal plane gives:
P cos α = W sin α + µR
Similarly, resolving the perpendicular forces gives:
R = P sin α + W cos α
Substituting the value of R in equation ( i ), we get:
P cos α = W sin α + µR = W sin α + µ(P sin α + W cos α)
= W sin α + µP sin α + µW cos α
Or P cos α – µP sin α = W sin α + µW cos α
Or P(cos α – µ sin α) = W(sin α + µ cos α)
Or P = W × (sin α + µ cos α) / (cos α – µ sin α)
Replacing the value of µ = tan φ,
P = W × (sin α + tan φ cos α) / (cos α – tan φ sin α)
Multiplying the numerator and the denominator by a common factor cos φ, we get
P = W × (sin α + cos φ + sin φ cos α) / (cos α cos φ – sin α sin φ),
= W × sin (α + φ) / cos (α + φ),
Finally, we write,
P = W tan (α + φ)
The above expression presents a clear relation between the weight P which is to be raised and the effort W needed to be applied externally using screw jacks.