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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Figure 7-27 shows an overhead view of three horizontal forces acting on a cargo canister that was initially stationary but now moves across a frictionless floor. The force magnitudes are ${{\mathbf{F}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{N}}$, ${{\mathbf{F}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{4}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{N}}$, and ${{\mathbf{F}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{10}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{N}}$, and the indicated angles are ${{\mathbf{\theta }}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{50}}{\mathbf{.}}{{\mathbf{0}}}^{{\mathbf{°}}}$and ${{\mathbf{\theta }}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{35}}{\mathbf{.}}{{\mathbf{0}}}^{{\mathbf{°}}}$. What is the net work done on the canister by the three forces during the first 4.00 m of displacement?

Work done on the canister by three forces during 4.0 m of the displacement will be 15.3 J.

See the step by step solution

## Step 1: Given data

Forces are given as, ${\mathrm{F}}_{1}=3.00\mathrm{N}$, ${\mathrm{F}}_{2}=4.00\mathrm{N}$, ${\mathrm{F}}_{3}=10.0\mathrm{N}$.

The angles are, ${\mathrm{\theta }}_{2}=50.{0}^{°}$, ${\mathrm{\theta }}_{3}=35.{0}^{°}$.

Displacement is, $\mathrm{d}=4.0\mathrm{m}$.

## Step 2: Understanding the concept

As the magnitude and the direction for each force are given, we can calculate the net force acting on the canister. We know the value for displacement, and we can find the net force acting on the canister; hence, from this data, we can find the work done.

Formula:

$\begin{array}{rcl}\sum {\mathrm{F}}_{\mathrm{x}}& =& 0\\ \sum {\mathrm{F}}_{\mathrm{y}}& =& 0\\ \mathrm{F}& =& \sqrt{{\mathrm{F}}_{\mathrm{x}}^{2}+{\mathrm{F}}_{\mathrm{y}}^{2}}\\ \mathrm{W}& =& \mathrm{Fd}\end{array}$

## Step 3: Calculate the net work done on the canister by the three forces during the first 4.00 m of displacement

The free Body Diagram for a canister is given as,

From this, we can find total force in the x direction as,

${\left({\mathrm{F}}_{\mathrm{x}}\right)}_{\mathrm{net}}=\left(10\mathrm{N}×\mathrm{cos}\left({35}^{°}\right)\right)-3\mathrm{N}-\left(4\mathrm{N}×\mathrm{sin}\left({50}^{°}\right)\right)\phantom{\rule{0ex}{0ex}}{\left({\mathrm{F}}_{\mathrm{x}}\right)}_{\mathrm{net}}=2.13\mathrm{N}$

The total force in the y direction is,

${\left({\mathrm{F}}_{\mathrm{y}}\right)}_{\mathrm{net}}=\left(10\mathrm{N}×\mathrm{sin}\left({35}^{°}\right)\right)-\left(4\mathrm{N}×\mathrm{cos}\left({50}^{°}\right)\right)\phantom{\rule{0ex}{0ex}}{\left({\mathrm{F}}_{\mathrm{y}}\right)}_{\mathrm{net}}=3.17\mathrm{N}$

Now, the total net force can be calculated as,

${\left(\mathrm{F}\right)}_{\mathrm{net}}=\sqrt{{\left({\mathrm{F}}_{\mathrm{x}}\right)}_{\mathrm{net}}^{2}+{\left({\mathrm{F}}_{\mathrm{y}}\right)}_{\mathrm{net}}^{2}}$

Substitute the values in the above expression, and we get,

${\left(\mathrm{F}\right)}_{\mathrm{net}}=\sqrt{{\left(2.13\mathrm{N}\right)}^{2}+{\left(3.16\mathrm{N}\right)}^{2}}\phantom{\rule{0ex}{0ex}}{\left(\mathrm{F}\right)}_{\mathrm{net}}=\sqrt{14.52{\mathrm{N}}^{2}}\phantom{\rule{0ex}{0ex}}{\left(\mathrm{F}\right)}_{\mathrm{net}}=3.82\mathrm{N}$

Work done can be calculated as,

$\mathrm{W}={\left(\mathrm{F}\right)}_{\mathrm{net}}\mathrm{d}$

Substitute the values in the above expression, and we get,

$\mathrm{W}=3.82\mathrm{N}×4.0\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{W}=15.3\mathrm{J}$

Thus, work done on the canister by three forces during 4.0 m of the displacement will be 15.3 J.