Chapter 14, Problem 58PQ

To determine

The force balance and torque balance equations for the system.

Answer to Problem 58PQ

The equations (X) and (XI) represents the force balance equations, and equations (XII), (XIII), and (XIV) represents the torque balance equations of the system.

Explanation of Solution

Ten forces are acting on the fan, gravity ${\overrightarrow{F}}_g$ acts on the center of mass. The thrust ${\overrightarrow{F}}_{\text{thrust}}$ is a horizontal force whose point of contact to be the center of the fans blades. The eight other forces are due to the four points of contact between the floor and the fan’s feet. AT each point of contact, the floor exerts a normal force upward $\left({\overrightarrow{F}}_{N1},{\overrightarrow{F}}_{N2},{\overrightarrow{F}}_{N3},{\overrightarrow{F}}_{N4}\right)$ and static friction horizontally $\left({\overrightarrow{F}}_{s1},{\overrightarrow{F}}_{s2},{\overrightarrow{F}}_{s3},{\overrightarrow{F}}_{s4}\right)$.

Write the expression for the direction of the force of gravity.

    ${\overrightarrow{F}}_g=-F_g\widehat{j}$                                                                                                               (I)

Write the expression for the direction of the force of thrust.

    ${\overrightarrow{F}}_{\text{thrust}}=F_{\text{thrust}}\widehat{i}$                                                                                                        (II)

For each foot, the normal force is in the positive $y$ direction and static friction is in the negative $x$ direction.

The forces ${\overrightarrow{F}}_1,{\overrightarrow{F}}_2,{\overrightarrow{F}}_3,{\overrightarrow{F}}_4$ can be expressed as,

    ${\overrightarrow{F}}_1=-F_{s1}\widehat{i}+F_{N1}\widehat{j}$ (III)

The origin is at the center of the fan’s base.

The expression for the positions for the center of mass is given by,

    ${\overrightarrow{r}}_{\text{CM}}=r_{\text{CM}}\widehat{j}$ (IV)

The position where thrust acts is given by,

    ${\overrightarrow{r}}_{\text{thrust}}=h\widehat{j}$ (V)

The point of contact for the forces acting on each foot in terms of the leg length $b$ can be expressed as,

    $\begin{array}{c}{\overrightarrow{r}}_1=b\sin 45°\widehat{\text{i}}+b\cos 45°\widehat{\text{k}}\\ =\frac{b\sqrt{2}}{2}\widehat{\text{i}}\text{+}\frac{b\sqrt{2}}{2}\widehat{\text{k}}\end{array}$ (VI)

    ${\overrightarrow{r}}_2=\frac{b\sqrt{2}}{2}\widehat{\text{i}}-\frac{b\sqrt{2}}{2}\widehat{\text{k}}$ (VII)

    ${\overrightarrow{r}}_3=-\frac{b\sqrt{2}}{2}\widehat{\text{i}}-\frac{b\sqrt{2}}{2}\widehat{\text{k}}$ (VIII)

    ${\overrightarrow{r}}_4=-\frac{b\sqrt{2}}{2}\widehat{\text{i}}\text{+}\frac{b\sqrt{2}}{2}\widehat{\text{k}}$ (IX)

Since gravitational force and position only have $y$  components, the torque is zero

    $\begin{array}{c}{\overrightarrow{\tau }}_g={\overrightarrow{r}}_{\text{CM}}\times {\overrightarrow{F}}_g\\ =0\end{array}$

The torque due to the thrust using position vector (V) and force (II) can be written as,

    $\begin{array}{c}{\overrightarrow{\tau }}_{\text{thrust}}={\overrightarrow{r}}_{\text{thrust}}\times {\overrightarrow{F}}_{\text{thrust}}\\ =-\left(r_y{F}_x\right)\widehat{\text{k}}\\ \text{=}-h{F}_{\text{thrust}}\widehat{\text{k}}\end{array}$ (X)

The torques exerted by the each foot,

    $\begin{array}{c}{\overrightarrow{\tau }}_1={\overrightarrow{r}}_1\times {\overrightarrow{F}}_1\\ =\left(\frac{b\sqrt{2}}{2}\widehat{\text{i}}\text{+}\frac{b\sqrt{2}}{2}\widehat{\text{k}}\right)\times \left(-F_{s1}\widehat{i}+F_{N1}\widehat{j}\right)\\ =-\frac{b\sqrt{2}}{2}{F}_{N1}\widehat{\text{i}}-\frac{b\sqrt{2}}{2}{F}_{s1}\widehat{\text{j}}\text{+}\frac{b\sqrt{2}}{2}{F}_{N1}\widehat{\text{k}}\\ \end{array}$

Similarly,

    $\begin{array}{l}{\overrightarrow{\tau }}_2=-\frac{b\sqrt{2}}{2}{F}_{N2}\widehat{\text{i}}-\frac{b\sqrt{2}}{2}{F}_{s2}\widehat{\text{j}}\text{+}\frac{b\sqrt{2}}{2}{F}_{N2}\widehat{\text{k}}\\ {\tau }_3=-\frac{b\sqrt{2}}{2}{F}_{N3}\widehat{\text{i}}-\frac{b\sqrt{2}}{2}{F}_{s3}\widehat{\text{j}}\text{+}\frac{b\sqrt{2}}{2}{F}_{N3}\widehat{\text{k}}\\ {\tau }_4=-\frac{b\sqrt{2}}{2}{F}_{N4}\widehat{\text{i}}-\frac{b\sqrt{2}}{2}{F}_{s4}\widehat{\text{j}}\text{+}\frac{b\sqrt{2}}{2}{F}_{N4}\widehat{\text{k}}\end{array}$

Conclusion:

Apply force balance condition in component form,

    ${\displaystyle \sum F_x}=F_{\text{thrust}}-F_{s1}-F_{s2}-F_{s3}-F_{s4}=0$                                                         (X)

    ${\displaystyle \sum F_y=F_{N1}+F_{N2}+F_{N3}+F_{N4}-F_g}=0$ (XI)

Applying the torque balance condition in component form,

    ${\displaystyle \sum {\tau }_x=\frac{b\sqrt{2}}{2}\left(-F_{N1}+F_{N2}+F_{N3}-F_{N4}\right)}$ (XII)

    ${\displaystyle \sum {\tau }_x=\frac{b\sqrt{2}}{2}\left(-F_{S1}+F_{S2}+F_{S3}-F_{S4}\right)}$ (XIII)

    ${\displaystyle \sum {\tau }_x=\frac{b\sqrt{2}}{2}\left(F_{N1}+F_{N2}+F_{N3}-F_{N4}\right)}-h{F}_{\text{thrust}}$ (XIV)

To verify the normal forces 3 and 4 approach zero as the thrust reaches the maximum. Solve equation (XI) for $\left(F_{N1}+F_{N2}\right)$, use equation (XV) to eliminate $\left(F_{N1}+F_{N2}\right)=F_g-\left(F_{N3}+F_{N4}\right)$ (XV)

Substitute each values,

    $\begin{array}{c}\frac{b\sqrt{2}}{2}\left(F_g-\left(F_{N3}+F_{N4}\right)-F_{N3}-F_{N4}\right)-h{F}_{\text{thrust}}=0\\ \frac{b\sqrt{2}}{2}{F}_g-b\sqrt{2}\left(F_{N3}+F_{N4}\right)-h{F}_{\text{thrust}}=0\\ h{F}_{\text{thrust}}=\frac{b\sqrt{2}}{2}{F}_g-b\sqrt{2}\left(F_{N3}+F_{N4}\right)\end{array}$    (XVI)

Therefore, the equations (X) and (XI) represents the force balance equations, and equations (XII), (XIII), and (XIV) represents the torque balance equations of the system.