Determine the horizontal force FE, perpendicular to the 600 mm long handle of a simple screw jack,necessary to start lifting a 500 kg load. The square-threaded screw has a lead

of 8 mm and a mean diameter of 90 mm. The coefficient of static friction for the screw is 0.25. F_{E}=\frac{F_{L} D}{2 d} \tan (\emptyset+\theta) \quad F_{L}=F_{\mathrm{w}}=m g \quad T=F_{E} d

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