Chapter 16, Problem 16.162RP

To determine

(a)

The angular acceleration of the plate.

Answer to Problem 16.162RP

The angular acceleration of the plate is $51.2346\text{rad}/{\text{s}}^2$ in clock -wise direction.

Explanation of Solution

Given information:

The side of the square plate is $150\text{mm}$, the mass is $2.5\text{kg}$

The following figure represents the general plane flow.

Figure-(1)

Write the expression for the distance between the point A and G.

$\begin{array}{c}AG=\frac{L\sqrt{2}}{2}\\ =\frac{L\sqrt{2}}{\sqrt{2}\times \sqrt{2}}\\ =\frac{L}{\sqrt{2}}\end{array}$ ...... (I)

Here, the distance between the point A and G is $AG$ and the side of the square plate is $L$.

Write the expression for the angular acceleration.

$\alpha =\frac{a_{G/A}}{AG}$ ...... (II)

Here, the vector position between the point A and G is $a_{G/A}$ and the angular acceleration is $\alpha$.

Substitute $\frac{L}{\sqrt{2}}$ for $AG$ in Equation (II).

$\begin{array}{l}\alpha =\frac{a_{G/A}}{\frac{L}{\sqrt{2}}}\\ a_{G/A}=\frac{L}{\sqrt{2}}\alpha \end{array}$

Write the expression for the acceleration at point A.

$\begin{array}{c}{a}_A=L\alpha \sin \left(30°\right)\\ =0.5L\alpha \end{array}$ ...... (III)

Here, the acceleration at point A is $a_A$.

Write the expression for the acceleration at point B.

$\begin{array}{c}{a}_B=L\alpha \cos \left(30°\right)\\ =0.8667L\alpha \end{array}$ ...... (IV)

Here, the acceleration at point B is $a_B$.

Write the expression for the direction of the acceleration.

$\begin{array}{c}a=\sqrt{{\left(a_A\right)}^2+{\left(a_{G/A}\right)}^2-2a_A{a}_{G/A}\cos \left(15°\right)}\\ =\sqrt{{\left(a_A\right)}^2+{\left(a_{G/A}\right)}^2-1.93185a_A{a}_{G/A}}\end{array}$ ...... (V)

Here, the direction of the acceleration is $a$.

Substitute $0.5L\alpha$ for $a_A$ and $\frac{L}{\sqrt{2}}\alpha$ for $a_{G/A}$ in Equation (V).

$\begin{array}{c}a=\sqrt{{\left(0.5L\alpha \right)}^2+{\left(\frac{L}{\sqrt{2}}\alpha \right)}^2-1.93185\left(0.5L\alpha \right)\left(\frac{L}{\sqrt{2}}\alpha \right)}\\ =\sqrt{\left(0.25L^2{\alpha }^2\right)+\left(\frac{L^2}{2}{\alpha }^2\right)-0.983012L^2{\alpha }^2}\\ =0.25882L\alpha \end{array}$

The following figure represents the acceleration diagram.

Figure-(2)

Write the expression for the angle.

$\begin{array}{l}\frac{a}{\sin \left(15°\right)}=\frac{a_{G/A}}{\sin \left(\beta \right)}\\ \sin \left(\beta \right)=a_{G/A}\times \frac{\sin \left(15°\right)}{a}\\ \sin \left(\beta \right)=a_{G/A}\times \frac{0.2588}{a}\\ \beta ={\sin }^{-1}\left(a_{G/A}\times \frac{0.2588}{a}\right)\end{array}$ ...... (VI)

He3re, the angle is $\beta$.

Write the expression for the distance between the point A and the point E.

$\begin{array}{c}AE=\left(AG\right)\cos \left(15°\right)\\ AE=0.9659\left(AG\right)\end{array}$ ...... (VII)

Here, distance between the point A and the point E is $AE$.

Write the expression for the distance between the point E and the point G along $x$ direction.

$\begin{array}{c}{\left(EG\right)}_x=\left(AG\right)\sin \left(15°\right)\\ {\left(EG\right)}_x=0.2588\left(AG\right)\end{array}$ ...... (VIII)

Here, the distance between the point E and the point G in $x$ direction is ${\left(EG\right)}_x$.

Write the expression for the distance between the point E and the point G.

$\begin{array}{c}{\left(EG\right)}_y=\left(AG\right)\sin \left(15°\right)\\ =0.2588\left(AG\right)\end{array}$ ...... (IX)

Here, the distance between the point E and the point G in $y$ direction is ${\left(EG\right)}_y$.

Write the expression for the resultant EG.

$EG=\sqrt{{\left({\left(EG\right)}_y\right)}^2+{\left({\left(EG\right)}_x\right)}^2}$ ....... (X)

Here, the resultant EG is $EG$.

Substitute $0.2588\left(AG\right)$ for ${\left(EG\right)}_y$ and $0.2588\left(AG\right)$ for ${\left(EG\right)}_x$ in Equation (X).

$\begin{array}{l}EG=\sqrt{{\left(0.2588\left(AG\right)\right)}^2+{\left(0.2588\left(AG\right)\right)}^2}\\ EG=\sqrt{0.066977{\left(AG\right)}^2+0.066977{\left(AG\right)}^2}\\ EG=0.365997\left(AG\right)\end{array}$

The following figure represents the free body diagram of the applied forces.

Figure-(3)

Write the expression for the acceleration at point G.

$a_G=a_{G/A}$ ....... (XI)

Here, the acceleration at point G is $a_G$.

Write the expression for the forces along $x$ direction.

$F_x=m{\left(a_G\right)}_x$ ...... (XII)

Here, the acceleration along $x$ direction about centre of gravity is ${\left(a_G\right)}_x$, the mass is $m$ and the forces along the $x$ direction is $F_x$.

Write the expression for the forces along $y$ direction.

$F_y=m{\left(a_G\right)}_y$ ...... (XIII)

Here, the acceleration along $y$ direction about centre of gravity is ${\left(a_G\right)}_y$ and the forces along the $y$ direction is $F_y$.

Write the expression for the moment of inertia.

$I=\frac{m{k}^2}{12}$ ...... (XIV)

Here, the radius of gyration is $k$ and the moment of inertia is $I$.

Write the expression for the radius of gyration.

$k^2=2L^2$ ...... (XV)

Substitute $2L^2$ for $k^2$ in equation (XV).

$\begin{array}{l}I=\frac{m\left(2L^2\right)}{12}\\ I=\frac{m\left(L^2\right)}{6}\end{array}$ ....... (XVI)

Write the expression for the moments about B.

$mg\left(EG\right)=I\alpha +ma\left(EG\right)$ ...... (XVII)

Substitute $0.365997\left(AG\right)$ for $EG$, $0.2588\left(AG\right)$ for ${\left(EG\right)}_x$, $\frac{m\left(L^2\right)}{6}$ for $I$ and $0.25882L\alpha$ for $a$ in equation (XVII).

$mg\times 0.2588\left(AG\right)=\left(\frac{m\left(L^2\right)}{6}\right)\alpha +m\left(0.25882L\alpha \right)\left(0.365997\left(AG\right)\right)$ …... (XVIII)

Substitute $\frac{L}{\sqrt{2}}$ for $AG$ in Equation (XVIII).

$\begin{array}{l}mg\times 0.2588\left(\frac{L}{\sqrt{2}}\right)=\left(\frac{m\left(L^2\right)}{6}\right)\alpha +m\left(0.25882L\alpha \right)\left(0.365997\left(\frac{L}{\sqrt{2}}\right)\right)\\ 0.183g=0.2336L\alpha \\ \alpha =\frac{0.183g}{0.2336L}\\ \alpha =0.7834\left(\frac{g}{L}\right)\end{array}$ ...... (XIX)

Calculation:

Substitute $150\text{mm}$ for $L$ in Equation (I).

$\begin{array}{c}AG=\frac{150\text{mm}}{\sqrt{2}}\\ =\frac{150\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\sqrt{2}}\\ =\frac{0.150\text{m}}{\sqrt{2}}\\ =0.106066\text{m}\end{array}$

Substitute $0.106066\text{m}$ for $AG$ in Equation (II).

$\begin{array}{l}\alpha =\frac{a_{G/A}}{0.106066\text{m}}\\ a_{G/A}=0.106066\text{m}\times \alpha \end{array}$

Substitute $150\text{mm}$ for $L$ in Equation (III).

$\begin{array}{c}{a}_A=0.5\left(150\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)\alpha \\ =0.5\left(0.150\text{m}\right)\alpha \\ =\left(0.25\alpha \right)\text{m}\end{array}$

Substitute $150\text{mm}$ for $L$ in Equation (IV).

$\begin{array}{c}{a}_B=0.8667\left(150\text{mm}\right)\alpha \\ a_B=0.8667\left(150\text{mm}\left(\frac{1\text{m}}{1000\text{m}}\right)\right)\alpha \\ a_B=\left(0.1299\text{m}\right)\alpha \end{array}$

Substitute $\left(0.25\alpha \right)\text{m}$ for $a_A$ and $0.106066\text{m}\times \alpha$ for $a_{G/A}$ in Equation (V).

$\begin{array}{l}a=\left[\sqrt{\begin{array}{l}{\left(0.25\alpha \right)}^2+{\left(0.106066\text{m}\times \alpha \right)}^2\\ -1.93185\left(0.25\alpha \right)\left(0.106066\text{m}\times \alpha \right)\end{array}}\right]\\ a=\left[\sqrt{\begin{array}{l}\left(0.0625{\alpha }^2\right)+\left(0.0112499{\text{m}}^2\times {\alpha }^2\right)\\ -\left(0.0512250\text{m}\times {\alpha }^2\right)\end{array}}\right]\\ a=\alpha \sqrt{\left(0.0625\right)+\left(0.0112499{\text{m}}^2\right)-\left(0.0512250\text{m}\right)}\\ a=0.150\alpha \end{array}$

Substitute $0.150\alpha$ for $a$ and $0.106066\text{m}\times \alpha$ for $a_{G/A}$ in Equation (VI).

$\begin{array}{c}\beta ={\sin }^{-1}\left(0.106066\text{m}\times \alpha \times \frac{0.2588}{\left(0.150\alpha \right)\text{m}}\right)\\ ={\sin }^{-1}\left(0.106066\times \frac{0.2588}{0.150}\right)\\ ={\sin }^{-1}\left(0.18299\right)\\ =10.54396°\end{array}$

Substitute $0.106066\text{m}$ for $AG$ in Equation (VII).

$\begin{array}{c}AE=0.9659\left(0.106066\text{m}\right)\\ =0.102449\text{m}\end{array}$

Substitute $0.106066\text{m}$ for $AG$ in Equation (VIII).

$\begin{array}{c}{\left(EG\right)}_x=0.2588\left(0.106066\text{m}\right)\\ =0.027449\text{m}\end{array}$

Substitute $0.106066\text{m}$ for $AG$ in Equation (IX).

$\begin{array}{c}{\left(EG\right)}_y=0.2588\left(0.106066\text{m}\right)\\ =0.0274498\text{m}\end{array}$

Substitute $0.0274498\text{m}$ for ${\left(EG\right)}_y$ and $0.027449\text{m}$ for ${\left(EG\right)}_x$ in Equation (X).

$\begin{array}{c}EG=\sqrt{{\left(0.0274498\text{m}\right)}^2+{\left(0.027449\text{m}\right)}^2}\\ =0.03881\text{m}\end{array}$

Substitute $150\text{mm}$ for $L$ in Equation (XV).

$\begin{array}{c}{k}^2=2{\left(150\text{mm}\right)}^2\\ k^2=2{\left(150\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2\\ k^2=0.045{\text{m}}^2\end{array}$

Substitute $0.045{\text{m}}^2$ for $k^2$ and $2.5\text{kg}$ for $m$ in Equation (XIV).

$\begin{array}{l}I=\frac{\left(2.5\text{kg}\right)\left(0.045{\text{m}}^2\right)}{12}\\ I=0.009375\text{kg}\cdot {\text{m}}^2\end{array}$

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$ and $150\text{mm}$ for $L$ in Equation (XIX).

$\begin{array}{c}\alpha =0.7834\left(\frac{9.81\text{m}/{\text{s}}^2}{150\text{mm}}\right)\\ =0.7834\left(\frac{9.81\text{m}/{\text{s}}^2}{150\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}\right)\\ =51.2346\text{rad}/{\text{s}}^2\end{array}$

Conclusion:

The angular acceleration of the plate is $51.2346\text{rad}/{\text{s}}^2$ in clock wise direction.

To determine

(b)

The reaction at the corner A.

Answer to Problem 16.162RP

The reaction at the corner A is $21.009\text{N}$.

Explanation of Solution

Given information:

The side of the square plate is $150\text{mm}$, the mass is $2.5\text{kg}$

The following figure represents the general plane flow.

Figure-(4)

Write the expression for the reaction at corner A.

$\begin{array}{l}{R}_A-mg=-ma\sin \left(45°\right)\\ R_A=-0.707ma+mg\end{array}$ ...... (XX)

Here, the reaction force at point A is $R_A$.

Substitute $0.25882L\alpha$ for $a$ in Equation (I).

$R_A=-0.707m\left(0.25882L\alpha \right)+mg$ ....... (XXI)

Substitute $0.7834\left(\frac{g}{L}\right)$ for $\alpha$ in Equation (XXI).

$\begin{array}{l}{R}_A=-0.707m\left(0.25882L\left(0.7834\left(\frac{g}{L}\right)\right)\right)+mg\\ R_A=-0.14335mg+mg\\ R_A=0.85664mg\end{array}$ ....... (XXII)

Calculation:

Substitute $2.5\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in Equation (XXII).

$\begin{array}{l}{R}_A=0.85664\left(2.5\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ R_A=0.85664\left(2.5\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)\\ R_A=21.009\text{kg}\cdot \text{m}/{\text{s}}^2\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\\ R_A=21.009\text{N}\end{array}$

Conclusion:

The reaction at the corner A is $21.009\text{N}$.