(a) What is the mechanical advantage of a wheelbarrow, such as the one inFigure\({\rm{9}}{\rm{.24}}\), if the center of gravity of the wheelbarrow and its load has a perpendicular lever arm of \({\rm{5}}{\rm{.50}}\;{\rm{cm}}\), while the hands have a perpendicular lever arm of \({\rm{1}}{\rm{.02}}\;{\rm{m}}\)? (b) What upward force should you exert to support the wheelbarrow and its load if their combined mass is\({\rm{55}}{\rm{.0}}\;{\rm{kg}}\)? (c) What force does the wheel exert on the ground?

For the equilibrium under several forces the upward forces are equal to the downward forces.

The center of gravity of the wheelbarrow and its load has a perpendicular lever arm of \({\rm{5}}{\rm{.50}}\;{\rm{cm}}\), while the hands have a perpendicular lever arm of \({\rm{1}}{\rm{.02}}\;{\rm{m}}\).

The free body diagram is as shown below:

\(\begin{align}{\rm{MA = }}\frac{{{\rm{1}}{\rm{.02}}}}{{{\rm{5}}{\rm{.50 \times 1}}{{\rm{0}}^{{\rm{ - 2}}}}}}\\{\rm{ = 18}}{\rm{.5}}\end{align}\)

Hence, the value is \({\rm{18}}{\rm{.5}}\).

\({\rm{MA = }}\frac{{{\rm{mg}}}}{{{{\rm{F}}_{\rm{i}}}}}\)

For,\({\rm{mg = 55}}{\rm{.0 \times 9}}{\rm{.8}}\),

\(\begin{align}{{\rm{F}}_{\rm{i}}}{\rm{ = }}\frac{{{\rm{55}}{\rm{.0 \times 9}}{\rm{.8}}}}{{{\rm{18}}{\rm{.5}}}}\\{\rm{ = 29}}{\rm{.1}}\;{\rm{N}}\end{align}\)

Hence, the upward force is \({\rm{29}}{\rm{.1}}\;{\rm{N}}\).

\(\begin{align}{\rm{N}}{{\rm{l}}_{\rm{o}}}{\rm{ = }}{{\rm{F}}_{\rm{i}}}\left( {{{\rm{l}}_{\rm{i}}}{\rm{ - }}{{\rm{l}}_{\rm{o}}}} \right)\\{\rm{N = }}\frac{{{{\rm{F}}_{\rm{i}}}\left( {{{\rm{l}}_{\rm{i}}}{\rm{ - }}{{\rm{l}}_{\rm{o}}}} \right)}}{{{{\rm{l}}_{\rm{o}}}}}\\{\rm{N = }}{{\rm{F}}_{\rm{i}}}\left( {{\rm{MA - 1}}} \right)\end{align}\)

Substituting the values we get,

\(\begin{align}{\rm{N = 29}}{\rm{.1}}\left( {{\rm{18}}{\rm{.5 - 1}}} \right)\\{\rm{ = 509}}{\rm{.25}}\;{\rm{N}}\end{align}\)

Hence, the force is \({\rm{509}}{\rm{.25}}\;{\rm{N}}\).