Chapter 16.2, Problem 16.134P

The hatchback of a car is positioned as shown to help determine the appropriate size for a damping mechanism AB. The weight of the door is 40 lbs, and its mass moment of inertia about the center of gravity G is 15 lb·ft·s². The linkage DEFH controls the motion of the hatch and is shown in more detail in part (b) of the figure. Assume that the mass of the links DE, EF, and FH are negligible, compared to the mass of the door. With AB removed, determine (a) the initial angular acceleration of the 40-lb door as it is released from rest, (b) the force on link FH.

(a)

(b)

Fig. P16.134

To determine

The initial angular acceleration of the door.

Answer to Problem 16.134P

The initial angular acceleration of the door is $\underset{\_}{{\alpha }_{EF}=4.144\text{rad}/{\text{s}}^2}$.

Explanation of Solution

Given information:

The weight of the door is $W=40\text{lbs}$.

The mass moment of inertia of the center of gravity is $I_G=15\text{lb}\cdot \text{ft}\cdot {\text{s}}^2$.

Calculation:

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

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

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

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

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

Sketch the Free Body Diagram of the door as shown in Figure 1.

Refer to Figure 1.

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

$\begin{array}{c}a=2\cos 24°\\ =1.827\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =0.1523\text{ft}\end{array}$

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

$\begin{array}{c}b=2\sin 24°\\ =0.8135\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =0.06779\text{ft}\end{array}$

Calculate the position vectors as shown below.

The position of F with respect to H.

$\begin{array}{c}{\text{r}}_{F/H}=\left(4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\text{j}\\ =\frac{1}{3}\text{j}\end{array}$

The position of E with respect to F.

${\text{r}}_{E/F}=-a\text{i}+b\text{j}$

Substitute $0.1523\text{ft}$ for a and $0.06779\text{ft}$ for b.

$r_{E/F}=-0.1523\text{i}+0.06779\text{j}$

The position of E with respect to D.

$\begin{array}{c}{\text{r}}_{E/D}=\left(9.5\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\text{cos48}°\text{i}+\left(9.5\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\right)\text{sin48}°\text{j}\\ =0.5297\text{i}+0.5883\text{j}\end{array}$

The position of G with respect to E.

${\text{r}}_{G/E}=-2\text{i}$

Apply the Equations of Equilibrium as shown below.

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

$\begin{array}{l}{\displaystyle \sum F_x}=m{a}_x\\ F_E\cos 48°=m{\overline{a}}_x\end{array}$

Substitute $1.242\text{slugs}$ for m.

$\begin{array}{l}0.6691F_E=1.242{\overline{a}}_x\\ F_E=1.856{\overline{a}}_x\end{array}$ (1)

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

$\begin{array}{l}{\displaystyle \sum F_y}=m{a}_y\\ -mg+F_E\sin 48°+F_F=m{\overline{a}}_y\end{array}$

Substitute $1.242\text{slugs}$ for m and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{l}-1.242\times 32.2+F_E\sin 48°+F_F=1.242{\overline{a}}_y\\ -40+0.743F_E+F_F=1.242{\overline{a}}_y\end{array}$

Substitute $1.856{\overline{a}}_x$ for $F_E$.

$\begin{array}{l}-40+0.743\times 1.856{\overline{a}}_x+F_F=1.242{\overline{a}}_y\\ F_F=1.242{\overline{a}}_y-1.379{\overline{a}}_x+40\end{array}$ (2)

Apply the Equilibrium of moment about G as shown below.

$\begin{array}{l}{\displaystyle \sum M_G}=\overline{I}{\alpha }_{EF}\\ 2.1523F_F+0.06779F_E\cos 48°+2F_E\sin 48°=15{\alpha }_{EF}\\ 2.1523F_F+1.53165F_E=15{\alpha }_{EF}\end{array}$ (3)

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

${\text{a}}_F={\text{a}}_H-{\omega }_{FH}^2{\text{r}}_{F/H}+{\alpha }_{FH}\times {\text{r}}_{F/H}$

Substitute $0$ for ${\text{a}}_H$, $0$ for ${\omega }_{FH}$, and $\frac{1}{3}\text{j}$ for ${\text{r}}_{F/H}$.

$\begin{array}{c}{\text{a}}_F=0+{\alpha }_{FH}\text{k}\times \frac{1}{3}\text{j}\\ =-\frac{1}{3}{\alpha }_{FH}\text{i}\end{array}$

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

${\text{a}}_E={\text{a}}_F-{\omega }_{EF}^2{\text{r}}_{E/F}+{\alpha }_{EF}\times {\text{r}}_{E/F}$

Substitute $-\frac{1}{3}{\alpha }_{FH}\text{i}$ for ${\text{a}}_F$, $0$ for ${\omega }_{EF}$, and $-0.1523\text{i}+0.06779\text{j}$ for ${\text{r}}_{E/F}$.

$\begin{array}{c}{\text{a}}_E=-\frac{1}{3}{\alpha }_{FH}\text{i}+0+{\alpha }_{EF}\text{k}\times \left(-0.1523\text{i}+0.06779\text{j}\right)\\ =-\frac{1}{3}{\alpha }_{FH}\text{i}-0.1523{\alpha }_{EF}\text{j}-0.06779{\alpha }_{EF}\text{i}\\ \text{=}-\left(\frac{1}{3}{\alpha }_{FH}+0.06779{\alpha }_{EF}\right)\text{i}-0.1523{\alpha }_{EF}\text{j}\end{array}$ (4)

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

${\text{a}}_E={\text{a}}_D-{\omega }_{DE}^2{\text{r}}_{E/D}+{\alpha }_{DE}\times {\text{r}}_{E/D}$

Substitute $0$ for ${\text{a}}_D$, $0$ for ${\omega }_{DE}$, and $0.5297\text{i}+0.5883\text{j}$ for ${\text{r}}_{E/D}$.

$\begin{array}{c}{\text{a}}_E=0+0+{\alpha }_{DE}\text{k}\times \left(0.5297\text{i}+0.5883\text{j}\right)\\ =-0.5883{\alpha }_{DE}\text{i}+0.5297{\alpha }_{DE}\text{j}\end{array}$ (5)

Equating Equations (4) and (5) as shown below.

$-\left(\frac{1}{3}{\alpha }_{FH}+0.06779{\alpha }_{EF}\right)\text{i}-0.1523{\alpha }_{EF}\text{j}=-0.5883{\alpha }_{DE}\text{i}+0.5297{\alpha }_{DE}\text{j}$

Resolving i and j components as shown below.

For i component.

$\frac{1}{3}{\alpha }_{FH}+0.06779{\alpha }_{EF}=0.5883{\alpha }_{DE}$ (6)

For j component.

$\begin{array}{l}-0.1523{\alpha }_{EF}=0.5297{\alpha }_{DE}\\ {\alpha }_{DE}=-0.2875{\alpha }_{EF}\end{array}$

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

${\overline{\text{a}}}_G={\text{a}}_E-{\omega }_{EF}^2{\text{r}}_{G/E}+{\alpha }_{EF}\times {\text{r}}_{G/E}$

Substitute $-0.5883{\alpha }_{DE}\text{i}+0.5297{\alpha }_{DE}\text{j}$ for ${\text{a}}_E$, $0$ for ${\omega }_{EF}$, and $-2\text{i}$ for ${\text{r}}_{G/E}$.

$\begin{array}{l}{\overline{\text{a}}}_G=-0.5883{\alpha }_{DE}\text{i}+0.5297{\alpha }_{DE}\text{j}+0+{\alpha }_{EF}\text{k}\times \left(-2\text{i}\right)\\ {\overline{a}}_x\text{i}+{\overline{a}}_y\text{j}=-0.5883{\alpha }_{DE}\text{i}+0.5297{\alpha }_{DE}\text{j}-2{\alpha }_{EF}\text{j}\\ {\overline{a}}_x\text{i}+{\overline{a}}_y\text{j}=-0.5883{\alpha }_{DE}\text{i}+\left(0.5297{\alpha }_{DE}-2{\alpha }_{EF}\right)\text{j}\end{array}$

Resolving i and j components as shown below.

For i component.

${\overline{a}}_x=-0.5883{\alpha }_{DE}$

For j component.

${\overline{a}}_y=0.5297{\alpha }_{DE}-2{\alpha }_{EF}$

Calculate the force at F $\left(F_F\right)$ as shown below.

Substitute $-0.5883{\alpha }_{DE}$ for ${\overline{a}}_x$ and $0.5297{\alpha }_{DE}-2{\alpha }_{EF}$ for ${\overline{a}}_y$ in Equation (2).

$\begin{array}{c}{F}_F=1.242\left(0.5297{\alpha }_{DE}-2{\alpha }_{EF}\right)-1.379\left(-0.5883{\alpha }_{DE}\right)+40\\ =0.6579{\alpha }_{DE}-2.484{\alpha }_{EF}+0.8113{\alpha }_{DE}+40\\ =1.4692{\alpha }_{DE}-2.484{\alpha }_{EF}+40\end{array}$

Substitute $-0.2875{\alpha }_{EF}$ for ${\alpha }_{DE}$.

$\begin{array}{c}{F}_F=1.4692\times \left(-0.2875{\alpha }_{EF}\right)-2.484{\alpha }_{EF}+40\\ =-2.9064{\alpha }_{EF}+40\end{array}$ (7)

Calculate the force at E $\left(F_E\right)$ as shown below.

Substitute $-0.5883{\alpha }_{DE}$ for ${\overline{a}}_x$ in Equation (1).

$\begin{array}{c}{F}_E=1.856\times \left(-0.5883{\alpha }_{DE}\right)\\ =-1.0919{\alpha }_{DE}\end{array}$

Substitute $-0.2875{\alpha }_{EF}$ for ${\alpha }_{DE}$.

$\begin{array}{c}{F}_E=-1.0919\times \left(-0.2875{\alpha }_{EF}\right)\\ =0.3139{\alpha }_{EF}\end{array}$

Calculate the angular acceleration $\left({\alpha }_{EF}\right)$ as shown below.

Substitute $-2.9064{\alpha }_{EF}+40$ for $F_F$ and $0.3139{\alpha }_{EF}$ for $F_E$ in Equation (3).

$\begin{array}{l}2.1523\left(-2.9064{\alpha }_{EF}+40\right)+1.53165\left(0.3139{\alpha }_{EF}\right)=15{\alpha }_{EF}\\ -6.2554{\alpha }_{EF}+86.092+0.4808{\alpha }_{EF}-15{\alpha }_{EF}=0\\ 20.7746{\alpha }_{EF}=86.092\\ {\alpha }_{EF}=4.144\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Therefore, the initial angular acceleration of the door is $\underset{\_}{{\alpha }_{EF}=4.144\text{rad}/{\text{s}}^2}$.

To determine

The force on link FH.

Answer to Problem 16.134P

The force on link FH is $\underset{\_}{F_F=27.96\text{N}}$.

Explanation of Solution

Given information:

The weight of the door is $W=40\text{lbs}$.

The mass moment of inertia of the center of gravity is $I_G=15\text{lb}\cdot \text{ft}\cdot {\text{s}}^2$.

Calculation:

Refer to part (a).

The initial angular acceleration of the door is ${\alpha }_{EF}=4.144\text{rad}/{\text{s}}^2$.

Calculate the force at F $\left(F_F\right)$ as shown below.

Substitute $4.144\text{rad}/{\text{s}}^2$ for ${\alpha }_{EF}$ in Equation (7).

$\begin{array}{c}{F}_F=-2.9064\times 4.144+40\\ =27.96\text{N}\end{array}$

Therefore, the force on link FH is $\underset{\_}{F_F=27.96\text{N}}$.