Chapter 16.2, Problem 16.116P

A 4-lb bar is attached to a 10-lb uniform cylinder by a square pin, P, as shown. Knowing that $r=16$ in., $h=8$ in., $\theta =20°$ , $L=20$ in., and $\omega =2$ rad/s at the instant shown, determine the reactions at P at this instant assuming that the cylinder rolls without sliding down the incline.

To determine

The reactions at P.

Answer to Problem 16.116P

The reactions at P is $P_x=1.1450Ib,P_y=3.2307Ib,P=3.43Ib,M_P=0.1550ft.Ib.$

Explanation of Solution

Given information:

$\begin{array}{l}r=16in\\ h=8in\\ \theta ={20}^0\\ L=20in\\ \omega =2rad/s\end{array}$

Find the mass of the cylinder (m_D):

$\begin{array}{l}{m}_D=\frac{W_D}{g}\\ m_D=\frac{10lb}{32.2ft/s^2}\\ m_D=0.3105slug\end{array}$

Find the moment of inertia of the cylinder.

$I_D=\frac{m_D{r}^2}{2}$

Substituting the values we get,

$\begin{array}{l}{I}_D=\frac{\left(0.3105\right){\left(16\frac{1ft}{12in.}\right)}^2}{2}\\ =0.2761slug.f{t}^2\end{array}$

Find the mass of the bar.

$\begin{array}{l}{m}_B=\frac{W_B}{g}\\ m_B=\frac{4lb}{32.2ft/s^2}\\ m_D=0.1242slug\end{array}$

Find the moment of inertia for bar.

$I_B=\frac{m_B{L}^2}{2}$

Substituting all the values we get,

$\begin{array}{l}{I}_B=\frac{\left(0.1242\right){\left(20\frac{1ft}{12in.}\right)}^2}{2}\\ =0.02875slug.f{t}^2\end{array}$

Here, the tangential component of acceleration of point C is zero.

Find the tangential acceleration at point C.

${\left(a_C\right)}_t=a_G-r\alpha$

Substituting the required values we get,

$\begin{array}{l}0=a_G-r\alpha \\ a_G=r\alpha \end{array}$

Then, the formula for tangential acceleration of point P is

${\left(a_P\right)}_t=a_G+h\alpha$

Substituting the values we get,

$\begin{array}{l}{\left(a_P\right)}_t=r\alpha +h\alpha \\ {\left(a_P\right)}_t=\left(r+h\right)\alpha ---(1)\end{array}$

Find the centripetal acceleration at point P.

${\left(a_P\right)}_n={\left(a_C\right)}_t+h{\omega }^2$

Substituting the values we get,

$\begin{array}{l}{\left(a_P\right)}_n=0+h{\omega }^2\\ =h{\omega }^2---(2)\end{array}$

Consider the free body diagram of cylinder.

First, take the moment about point C.

Here, the system of external forces is equal to system of effective forces.

$\begin{array}{l}{\sum M_C=\sum \left(M_C\right)}_{eff}---(3)\\ \sum M_C=\left(m_{D}g\sin \theta \right)r+\left(m_{B}g\sin \theta \right)\left(r+h\right)\end{array}$

Find the above values.

${\sum \left(M_C\right)}_{eff}=I_D\alpha +I_B\alpha +\left(m_D{a}_G\right)r+\left(m_B{\left(a_P\right)}_t\right)\left(r+h\right)$

We get,

$\begin{array}{l}\left(m_{D}g\sin \theta \right)r+\left(m_{B}g\sin \theta \right)\left(r+h\right)=I_D\alpha +I_B\alpha +\left(m_D{a}_G\right)r+\left(m_B{\left(a_P\right)}_t\right)\left(r+h\right)\\ \left(m_{D}g\sin \theta \right)r+\left(m_{B}g\sin \theta \right)\left(r+h\right)=\left\{I_D\alpha +I_B\alpha +\left(m_D\left(r\alpha \right)\right)r+\left(m_B\left(r+h\right)\right)\left(r+h\right)\alpha \right\}\\ \left(I_D+I_B+m_D{r}^2+m_B{\left(r+h\right)}^2\right)\alpha =\left(m_{D}gr+m_{B}g\left(r+h\right)\right)\sin \theta \\ \alpha =\frac{\left(m_{D}gr+m_{B}g\left(r+h\right)\right)\sin \theta }{\left(I_D+I_B+m_D{r}^2+m_B{\left(r+h\right)}^2\right)}---(4)\end{array}$

Substituting all the required values in equation 4 we get,

$\begin{array}{l}\alpha =\frac{\left(\left(0.3105slug\right)\left(32.2ft/s^2\right)\left(16in\frac{1ft}{12in.}\right)+\left(0.1242slug\right)\left(32.2ft/s^2\right)\left(16in\frac{1ft}{12in}\right)+8in\frac{1ft}{12in}\right)\sin {20}^0}{\left(\left(0.2761slug.f{t}^2\right)+\left(0.02875slug.f{t}^2\right)+\left(0.3105slug\right){\left(16in\frac{1ft}{12in}\right)}^2+\left(0.1242slug\right){\left(16in\frac{1ft}{12in}+8in\frac{1ft}{12in}\right)}^2\right)}\\ \alpha =\frac{7.295}{1.3536}\\ \alpha =5.3896rad/s^2\end{array}$

Consider the equation (1).

$\begin{array}{l}{\left(a_P\right)}_t=\left(16in\frac{1ft}{12in}+8in\frac{1ft}{12in}\right)5.3896rad/s^2\\ =10.779ft/s^2\end{array}$

Similarly, consider equation (2).

$\begin{array}{l}{\left(a_P\right)}_n=\left(8in\frac{1ft}{12in}\right)2rad/s^2\\ =2.6667ft/s^2\end{array}$

Take the free body diagram for bar.

Here, also the system of external forces is equivalent to system of effective forces.

$\sum F_x={\sum \left(F_x\right)}_{eff}---(5)$

And the force is considered to be positive.

Find the value of $\sum F_x.$

$\sum F_x=P_x$

Find the value of ${\sum \left(F_x\right)}_{eff}.$

$\begin{array}{l}\sum F_x=P_x\\ {\sum \left(F_x\right)}_{eff}=\left(m_B{\left(a_P\right)}_t\right)\cos {20}^0-\left(m_B{\left(a_P\right)}_n\right)\sin {20}^0\end{array}$

Substituting the values in equation (5) we get,

$\begin{array}{l}{P}_x=\left(m_B{\left(a_P\right)}_t\right)\cos {20}^0-\left(m_B{\left(a_P\right)}_n\right)\sin {20}^0\\ P_x=\left(0.1242slug\right)\left(10.7791ft/s^2\right)\cos {20}^0-\left(0.1242slug\right)\left(2.6667ft/s^2\right)\sin {20}^0\\ =1.1450Ib\end{array}$

Again the statement of system of external forces is equivalent to system of effective forces.

$\sum F_y={\sum \left(F_y\right)}_{eff}---(6)$

Find the value of $\sum F_y.$

$\sum F_y=P_y-m_{B}g$

Find the value of ${\sum \left(F_y\right)}_{eff}.$

${\sum \left(F_y\right)}_{eff}=-\left(m_B{\left(a_P\right)}_t\right)\sin {20}^0-\left(m_B{\left(a_P\right)}_n\right)\cos {20}^0$

Consider the equation 6 and substitute the values.

$\begin{array}{l}{P}_y-m_{B}g=-\left(m_B{\left(a_P\right)}_t\right)\sin {20}^0-\left(m_B{\left(a_P\right)}_n\right)\cos {20}^0\\ P_y-\left(0.1242slug\right)\left(32.2ft/s^2\right)=\left\{-\left(0.1242slug\right)\left(10.7791ft/s^2\right)\sin {20}^0-\left(0.1242slug\right)\left(2.6667ft/s^2\right)\cos {20}^0\right\}\\ P_y=3.9992-0.7691\\ P_y=3.2307Ib\end{array}$

Find the resultant reaction at point P.

$\begin{array}{l}P=\sqrt{P_x^2+P_y^2}\\ P=\sqrt{{\left(1.1450Ib\right)}^2+{\left(3.2307Ib\right)}^2}\\ P=3.43Ib\end{array}$

Find the resultant angle of reaction.

$\begin{array}{l}\tan \beta =\frac{P_y}{P_x}\\ \tan \beta =\frac{3.2307Ib}{1.1450Ib}\\ \beta ={\tan }^{-1}\left(\frac{3.2307Ib}{1.1450Ib}\right)\\ \beta ={70.5}^0\end{array}$

Take the moment about point P.

${\sum M_P=\sum \left(M_P\right)}_{eff}---(7)$

Take the moment as clockwise and positive.

Find the value of $\sum M_P.$

$\sum M_P=M_P$

Find the value of ${\sum \left(M_P\right)}_{eff}.$

$\begin{array}{l}{\sum \left(M_P\right)}_{eff}=I_B\alpha \\ M_P=I_B\alpha \end{array}$

Substituting the values in above equation we get,

$\begin{array}{l}{M}_P=I_B\alpha \\ M_P=\left(0.02875\right)\left(2.3896\right)\\ M_P=0.1550ft.Ib\end{array}$