Question

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?

Short answer

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

Step-by-step solution

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

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:

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

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,

$$\begin{gathered} {\left({\mathrm{F}}_{\mathrm{x}}\right)}_{\mathrm{net}}=\left(10\mathrm{N}\times \cos \left({35}^{°}\right)\right)-3\mathrm{N}-\left(4\mathrm{N}\times \sin \left({50}^{°}\right)\right) \\ {\left({\mathrm{F}}_{\mathrm{x}}\right)}_{\mathrm{net}}=2.13\mathrm{N} \end{gathered}$$

The total force in the y direction is,

$$\begin{gathered} {\left({\mathrm{F}}_{\mathrm{y}}\right)}_{\mathrm{net}}=\left(10\mathrm{N}\times \sin \left({35}^{°}\right)\right)-\left(4\mathrm{N}\times \cos \left({50}^{°}\right)\right) \\ {\left({\mathrm{F}}_{\mathrm{y}}\right)}_{\mathrm{net}}=3.17\mathrm{N} \end{gathered}$$

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,

$$\begin{gathered} {\left(\mathrm{F}\right)}_{\mathrm{net}}=\sqrt{{\left(2.13\mathrm{N}\right)}^2+{\left(3.16\mathrm{N}\right)}^2} \\ {\left(\mathrm{F}\right)}_{\mathrm{net}}=\sqrt{14.52{\mathrm{N}}^2} \\ {\left(\mathrm{F}\right)}_{\mathrm{net}}=3.82\mathrm{N} \end{gathered}$$

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,

$$\begin{gathered} \mathrm{W}=3.82\mathrm{N}\times 4.0\mathrm{m} \\ \mathrm{W}=15.3\mathrm{J} \end{gathered}$$

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