What force must you exert on the 14 in. radius handle of a jackscrew if the pitch is 0.25 in. and the weight of the load is 9,000 lb.?

25.6

Explanation

For a jackscrew, resistance x pitch = effort force x 2πr. Solving for effort force:
\(F_{e}=\frac{R \times p}{2 \pi r}=\frac{9,000 l b . \times 0.25}{2 \pi \times 14 i n .}=\frac{2,250 l b . i n .}{87.96 i n .}=25.6 l b\)

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