If the handles of a wheelbarrow are 4 ft. from the wheel axle, what force must you exert to lift the handles if it's carrying a 340 lb. load concentrated at a point 0.5 ft. from the axle?

42.5 lbs.

Explanation

This problem describes a second class lever and, for a second class lever, the effort force multiplied by the effort distance equals the resistance force multipied by the resistance distance:

\(P_{e} d_{e}=F_{r} d_{r}\)

Plugging in the variables from this problem yields:

\(F_{e} \times 4 f t .=340 \mathrm{lbs} . \times 0.5 \mathrm{ft.}\)
\(F_{e}=170 \mathrm{ft} . \mathrm{lbs} . \div 4 \mathrm{ft} .=42.5 \mathrm{lbs}\)

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