Figure 12-62 is an overhead view of a rigid rod that turns about a vertical axle until the identical rubber stoppers$\mathit{A}$and $\mathit{B}$are forced against rigid walls at distances${\mathit{r}}_{\mathbf{A}}\mathbf{=}\mathbf{7}\mathbf{.0}\mathbf{}\mathit{c}\mathit{m}$and${\mathit{r}}_{\mathbf{B}}\mathbf{=}\mathbf{4}\mathbf{.0}\mathbf{}\mathit{c}\mathit{m}$from the axle. Initially the stoppers touch the walls without being compressed. Then force$\mathit{F}$of magnitude $\mathbf{220}\mathbf{\text{N}}$is applied perpendicular to the rod at a distance $\mathit{R}\mathbf{=}\mathbf{5}\mathbf{.0}\mathit{c}\mathit{m}$from the axle. Find the magnitude of the force compressing (a) stopper$\mathit{A}$, and (b) stopper.$\mathit{B}$

The force applied perpendicular to the rod is,$\text{F}=220N$

The distance between F and axle is,$\text{R}=5.0cm\left(\frac{1\text{\hspace{0.17em}m}}{100\text{\hspace{0.17em}cm}}\right)=0.05\text{\hspace{0.17em}m}$

Distance between rubber stopper A and axle is,$r_A=7.0cm\left(\frac{1\text{\hspace{0.17em}m}}{100\text{\hspace{0.17em}cm}}\right)=0.07\text{\hspace{0.17em}m}$

Distance between rubber stopper A and axle is,$r_B=4.0cm\left(\frac{1\text{\hspace{0.17em}m}}{100\text{\hspace{0.17em}cm}}\right)=0.04\text{\hspace{0.17em}m}$

Using the free body diagram, condition for equilibrium, and assuming that two stoppers are identical,we can find the forces compressing the stopper.

Equation:

At equilibrium,${\mathit{\tau }}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathbf{0}$

The free body diagram:

At equilibrium, the torque about the axle is zero.

${\tau }_{net}=0$

$\text{FR}={\text{F}}_{\text{A}}{\text{r}}_{\text{A}}+{\text{F}}_{\text{B}}{\text{r}}_{\text{B}}$(1)

Let’s assume that two stoppers are identical. So force constant for both stoppers will be the same.

${\text{K}}_{\text{A}}={\text{K}}_{\text{B}}$

${\text{F}}_{\text{A}}{\text{y}}_{\text{A}}={\text{F}}_{\text{B}}{\text{y}}_{\text{B}}$(2)

If the stoppers are compressed by amounts${\text{y}}_{\text{A}}$and${\text{y}}_{\text{B}}$respectively, then they will rotate the rod by${\text{θ}}_{\text{A}}{\text{andθ}}_{\text{B}}\text{.}$

But,

$\begin{array}{l}{\text{y}}_{\text{A}}={\text{r}}_{\text{A}}{\text{θ}}_{\text{A}}\text{and}\\ {\text{y}}_{\text{A}}={\text{r}}_{\text{A}}{\text{θ}}_{\text{A}}\end{array}$

Therefore, equation 2 becomes,

$\frac{{\text{F}}_{\text{A}}}{{\text{r}}_{\text{A}}}=\frac{{\text{F}}_{\text{B}}}{{\text{r}}_{\text{B}}}$(3)

Putting equation 3 in equation 1 yields,

$\text{FR}={\text{F}}_{\text{A}}\text{r}+\left({\text{F}}_{\text{A}}\text{r}\right){\text{r}}_{\text{B}}$

$\begin{array}{l}\text{FR}={\text{F}}_{\text{A}}({\text{r}}_{\text{A}}+{\text{r}}_{\text{B}}^{\text{2}}/{\text{r}}_{\text{A}})\\ {\text{F}}_{\text{A}}=\text{FR}({\text{r}}_{\text{A}}+({\text{r}}_{\text{B}}^{\text{2}}/{\text{r}}_{\text{A}})\\ {\text{F}}_{\text{A}}=\frac{220N\left(0.05m\right)}{0.07m+\frac{{\left(0.04m\right)}^2}{0.07m}}\\ {\text{F}}_{\text{A}}=1.18\times {10}^{2}N\end{array}$

$\text{Therefore},\text{ThemagnitudeoftheforcecompressingstopperA}\left({\text{F}}_{\text{A}}\right)\text{is}1.2\times {10}^2\text{N}.$

.

$\begin{array}{c}{\text{F}}_B=\text{FR}({\text{r}}_B+({\text{r}}_A^{\text{2}}/{\text{r}}_{\text{B}})\\ =\frac{220N\left(0.05N\right)}{0.04m+\frac{0.07m^2}{0.04m}}\\ =68N\end{array}$

Therefore, the magnitude of the force compressing stopper$\text{A}\left({\text{F}}_{\text{B}}\right)$is$68N$