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Found in: Page 317

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

(a) What is the mechanical advantage of a wheelbarrow, such as the one in Figure $${\rm{9}}{\rm{.24}}$$, 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?

1. The mechanical advantage is$${\rm{18}}{\rm{.5}}$$.
2. The upward force is$${\rm{29}}{\rm{.1}}\;{\rm{N}}$$.
3. The force exerted by the wheel on the ground is$${\rm{509}}{\rm{.25}}\;{\rm{N}}$$.
See the step by step solution

Step 1: Idea of equilibrium under Forces

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

Step 2: Diagram

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:

Step 3: Calculation of the mechanical advantage of the wheelbarrow

\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 $${\rm{18}}{\rm{.5}}$$.

Step 4: Calculation of the upward force

b. We know, mechanical advantage is,

$${\rm{MA = }}\frac{{{\rm{mg}}}}{{{{\rm{F}}_{\rm{i}}}}}$$

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 $${\rm{29}}{\rm{.1}}\;{\rm{N}}$$.

Step 5: Calculation of the force by the wheel on the ground

c. The force exerted by the wheel on the ground is,

\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 $${\rm{509}}{\rm{.25}}\;{\rm{N}}$$.