Chapter 16.2, Problem 16.119P

(a)

To determine

Find the angular acceleration of the ladder.

Answer to Problem 16.119P

The angular acceleration of the ladder is $\underset{\_}{\alpha =-0.786\text{rad}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

Given information:

The weight of the ladder is $W=40\text{lb}$.

The length of the ladder is $l=30\text{ft}$.

The coefficient of kinetic friction is ${\mu }_k=0.2$.

The angle is $\theta =40°$.

Calculation:

Consider the acceleration due to gravity $g=32.2\text{ft}/{\text{s}}^2$.

Calculate the mass of the ladder $\left(m\right)$ as shown below.

$m=\frac{W}{g}$

Substitute $32.2\text{ft}/{\text{s}}^2$ for g and $40\text{lb}$ for $W$.

$\begin{array}{c}m=\frac{40}{32.2}\\ =1.242\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\times \frac{1\text{slug}}{1\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}}\\ =1.242\text{slug}\end{array}$

Calculate the moment of inertia $\left(\overline{I}\right)$ as shown below.

$\overline{I}=\frac{1}{12}m{l}^2$

Substitute $30\text{ft}$ for $l$ and $1.242\text{slug}$ for $m$.

$\begin{array}{c}\overline{I}=\frac{1}{12}\times 1.242\times {30}^2\\ =93.15\text{slug}\cdot {\text{ft}}^2\end{array}$

Sketch the geometry of the ladder rests on the wall as shown in Figure 1.

Refer to Figure 1.

Calculate the distance $\left(c\right)$ as shown below.

$2c=l\cos \theta$

Substitute $30\text{ft}$ for $l$ and $40°$ for $\theta$.

$\begin{array}{l}2c=30\cos 40°\\ c=11.49\text{ft}\end{array}$

Calculate the distance $\left(d\right)$ as shown below.

$2d=l\sin \theta$

Substitute $30\text{ft}$ for $l$ and $40°$ for $\theta$.

$\begin{array}{l}2d=30\sin 40°\\ d=9.64\text{ft}\end{array}$

Sketch the Free Body Diagram of the ladder as shown in Figure 2.

Refer to Figure 2.

Apply the Equations of Equilibrium as shown below.

Apply the Equilibrium of force along x direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_x}=m{a}_x\\ {\mu }_k{N}_A-N_B=m{\overline{a}}_x\end{array}$ (1)

Apply the Equilibrium of force along y direction as shown below.

$\begin{array}{l}{\displaystyle \sum F_y}=m{a}_y\\ N_A+{\mu }_k{N}_B-mg=m{\overline{a}}_y\end{array}$ (2)

Apply the Equilibrium of moment about G as shown below.

$\begin{array}{l}{\displaystyle \sum M_G}=\overline{I}\alpha \\ {\mu }_k{N}_{B}c-N_{A}c+{\mu }_k{N}_{A}d+N_{B}d=\overline{I}\alpha \\ \left({\mu }_{k}d-c\right)N_A+\left({\mu }_{k}c+d\right)N_B=\overline{I}\alpha \end{array}$ (3)

Apply the kinematics as shown below.

Calculate the acceleration $\left(\overline{\text{a}}\right)$ as shown below.

$\overline{\text{a}}={\text{a}}_A-{\omega }^2{r}_{G/A}+\alpha \times r_{G/A}$

Substitute $0$ for $\omega$, $c\text{i}+d\text{j}$ for $r_{G/A}$, and modify the Equation in vector form as shown below.

$\begin{array}{c}\overline{\text{a}}={\text{a}}_A\text{i}-{\left(0\right)}^2{r}_{G/A}+\alpha \text{k}\times \left(c\text{i}+d\text{j}\right)\\ ={\text{a}}_A\text{i}+c\alpha \text{j}-d\alpha \text{i}\\ =\left({\text{a}}_A-d\alpha \right)\text{i}+c\alpha \text{j}\end{array}$ (4)

Calculate the acceleration $\left({\text{a}}_A\right)$ as shown below.

${\text{a}}_A={\text{a}}_B-{\omega }^2{r}_{A/B}+\alpha \times r_{A/B}$

Substitute $0$ for $\omega$, $-2c\text{i}-2d\text{j}$ for $r_{A/B}$, $a_A\text{j}$ for $a_B$ and modify the Equation in vector form as shown below.

$\begin{array}{c}{a}_A\text{i}=a_A\text{j}-{\left(0\right)}^2{r}_{A/B}+\alpha \text{k}\times \left(-2c\text{i}-2d\text{j}\right)\\ =a_A\text{j}-2c\alpha \text{j}+2d\alpha \text{i}\\ =2d\alpha \text{i}+\left(a_A-2c\alpha \right)\text{j}\end{array}$ (5)

Resolving the components of i and j in Equation (5) as shown below.

$a_A=2d\alpha$

Calculate the acceleration $\left(\overline{a}\right)$ as shown below.

Substitute $2d\alpha$ for $a_A$ in Equation (4).

$\begin{array}{c}\overline{a}=\left(2d\alpha -d\alpha \right)\text{i}+c\alpha \text{j}\\ =d\alpha \text{i}+c\alpha \text{j}\end{array}$

Hence, ${\overline{a}}_x=d\alpha$ and ${\overline{a}}_y=c\alpha$.

Calculate the reaction $\left(N_B\right)$ as shown below.

Substitute $d\alpha$ for ${\overline{a}}_x$ in Equation (1).

$\begin{array}{l}{\mu }_k{N}_A-N_B=md\alpha \\ N_B={\mu }_k{N}_A-md\alpha \end{array}$ (6)

Calculate the reaction $\left(N_A\right)$ as shown below.

Substitute $c\alpha$ for ${\overline{a}}_y$ and ${\mu }_k{N}_A-md\alpha$ for $N_B$ in Equation (2).

$\begin{array}{l}{\displaystyle \sum F_y}=m{a}_y\\ N_A+{\mu }_k\left({\mu }_k{N}_A-md\alpha \right)-mg=mc\alpha \\ N_A+{\mu }_k^2{N}_A-{\mu }_{k}md\alpha -mg=mc\alpha \\ N_A\left(1+{\mu }_k^2\right)=mc\alpha +{\mu }_{k}md\alpha +mg\end{array}$

$N_A=\frac{m\left(c\alpha +{\mu }_{k}d\alpha +g\right)}{\left(1+{\mu }_k^2\right)}$ (7)

Calculate the reaction $\left(N_B\right)$ as shown below.

Substitute $\frac{m\left(c\alpha +{\mu }_{k}d\alpha +g\right)}{\left(1+{\mu }_k^2\right)}$ for $N_A$ in Equation (6).

$\begin{array}{c}{N}_B={\mu }_k\frac{m\left(c\alpha +{\mu }_{k}d\alpha +g\right)}{\left(1+{\mu }_k^2\right)}-md\alpha \\ =m\left[\frac{{\mu }_k\left(c\alpha +{\mu }_{k}d\alpha +g\right)}{\left(1+{\mu }_k^2\right)}-d\alpha \right]\end{array}$ (8)

Calculate the angular acceleration $\left(\alpha \right)$ as shown below.

Substitute $\frac{m\left(c\alpha +{\mu }_{k}d\alpha +g\right)}{\left(1+{\mu }_k^2\right)}$ for $N_A$ and $m\left[\frac{{\mu }_k\left(c\alpha +{\mu }_{k}d\alpha +g\right)}{\left(1+{\mu }_k^2\right)}-d\alpha \right]$ for $N_B$ in Equation (3).

$\left[\begin{array}{l}\left({\mu }_{k}d-c\right)\frac{m\left(c\alpha +{\mu }_{k}d\alpha +g\right)}{\left(1+{\mu }_k^2\right)}\\ +\left({\mu }_{k}c+d\right)m\left[\frac{{\mu }_k\left(c\alpha +{\mu }_{k}d\alpha +g\right)}{\left(1+{\mu }_k^2\right)}-d\alpha \right]\end{array}\right]=\overline{I}\alpha$

Substitute $0.2$ for ${\mu }_k$, $9.64\text{ft}$ for $d$, $11.49\text{ft}$ for $c$, $1.242\text{slug}$ for $m$, $93.15\text{slug}\cdot {\text{ft}}^2$ for $\overline{I}$, and $32.2\text{ft}/{\text{s}}^2$ for $g$.

$\begin{array}{l}\left[\begin{array}{l}\left(0.2\times 9.64-11.49\right)\frac{1.242\left(11.49\alpha +0.2\times 9.64\alpha +32.2\right)}{\left(1+{0.2}^2\right)}\\ +\left(0.2\times 11.49+9.64\right)\times 1.242\\ \times \left[\frac{0.2\left(11.49\alpha +0.2\times 9.64\alpha +32.2\right)}{\left(1+{0.2}^2\right)}-9.64\alpha \right]\end{array}\right]=93.15\alpha \\ -11.419\left(13.418\alpha +32.2\right)+2.851\left(13.418\alpha +32.2\right)-142.932\alpha =93.15\alpha \\ -153.22\alpha -367.6918+38.255\alpha +91.8022-236.082\alpha =0\\ -351.047\alpha -275.8896=0\end{array}$

$\begin{array}{l}-351.047\alpha =275.8896\\ \alpha =-0.786\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Hence, the angular acceleration of the ladder is $\underset{\_}{\alpha =-0.786\text{rad}/{\text{s}}^{\text{2}}}$.

(b)

To determine

Find the forces at A and B

Answer to Problem 16.119P

The force at A is $\underset{\_}{F_A=25.86\text{lb}}$.

The force at B is $\underset{\_}{F_B=14.58\text{lb}}$.

Explanation of Solution

Given information:

The weight of the ladder is $W=40\text{lb}$.

The length of the ladder is $l=30\text{ft}$.

The coefficient of kinetic friction is ${\mu }_k=0.2$.

The angle is $\theta =40°$.

Calculation:

Refer to part (a).

The angular acceleration of the ladder is $\alpha =-0.786\text{rad}/{\text{s}}^2$.

Calculate the reaction $\left(N_A\right)$ as shown below.

Substitute $0.2$ for ${\mu }_k$, $9.64\text{ft}$ for $d$, $11.49\text{ft}$ for $c$, $1.242\text{slug}$ for $m$, $-0.786\text{rad}/{\text{s}}^2$ for $\alpha$, and $32.2\text{ft}/{\text{s}}^2$ for $g$ in Equation (7).

$\begin{array}{c}{N}_A=\frac{1.242\left(11.49\times \left(-0.786\right)+0.2\times 9.64\times \left(-0.786\right)+32.2\right)}{\left(1+{0.2}^2\right)}\\ =25.86\text{lb}\end{array}$

Calculate the force at A $\left({\mu }_k{N}_A\right)$ as shown below.

${\mu }_k{N}_A$

Substitute $0.2$ for ${\mu }_k$ and $25.86\text{lb}$ for $N_A$.

$\begin{array}{c}{\mu }_k{N}_A=0.2\times 25.86\\ =5.172\text{lb}\end{array}$

Hence, the force at A is $\underset{\_}{F_A=25.86\text{lb}}$.

Calculate the reaction $\left(N_B\right)$ as shown below.

Substitute $0.2$ for ${\mu }_k$, $9.64\text{ft}$ for $d$, $11.49\text{ft}$ for $c$, $1.242\text{slug}$ for $m$, $-0.786\text{rad}/{\text{s}}^2$ for $\alpha$, and $32.2\text{ft}/{\text{s}}^2$ for $g$ in Equation (7).

$\begin{array}{c}{N}_B=1.242\times \left[\begin{array}{l}\frac{0.2\left(11.49\left(-0.786\right)+0.2\times 9.64\left(-0.786\right)+32.2\right)}{\left(1+{0.2}^2\right)}\\ -9.64\left(-0.786\right)\end{array}\right]\\ =1.242\left[4.164+7.577\right]\\ =14.58\text{lb}\end{array}$

Calculate the force at B $\left({\mu }_k{N}_B\right)$ as shown below.

${\mu }_k{N}_B$

Substitute $0.2$ for ${\mu }_k$ and $14.58\text{lb}$ for $N_A$.

$\begin{array}{c}{\mu }_k{N}_B=0.2\times 14.58\\ =2.916\text{lb}\end{array}$

Therefore, the force at B is $\underset{\_}{F_B=14.58\text{lb}}$.