What force must you exert on the 20 in. radius handle of a jackscrew if the pitch is 0.875 in. and the weight of the load is 8,500 lb.?

59.2

Explanation

For a jackscrew, resistance x pitch = effort force x 2πr. Solving for effort force:
\(F_{e}=\frac{R \times p}{2 \pi r}=\frac{8,500 l b . \times 0.875}{2 \pi \times 20 i n .}=\frac{7,438 l b . i n .}{125.66 i n .}=59.2 \mathrm{lb} .\)

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