Chapter 16.2, Problem 16.142P

To determine

Find a couple of M.

Answer to Problem 16.142P

The couple of M=-7.19N.m.

Explanation of Solution

Given information:

The attached rod A mass is=0.8kg

And length is=160mm

Rod BP mass is=1kg

Length is=200mm

Explanation:

First consider the unit vector along the axis of X and Y direction.

So, we have to find the vertical distance from point P and point A(y)

$\begin{array}{l}y=\left[\left(200mm\right)\frac{1m}{1000mm}\right]\sin {30}^0\\ =0.1m\end{array}$

Here, we take the coordinate points as (0,0.1m)

Then the position of the vector $r_{P/A}=-\left(0.1m\right)j$

Coordinate points of the point P with the point B is $\left(-0.2\cos {30}^{0}m,-0.2\sin {30}^{0}m\right)$

Then the position vector is $r_{P/B}=-\left(0.1732m\right)i-\left(0.1m\right)j$

Then the coordinate points with the point P and point E as (0,-0.16m)

Position vector is $r_{E/A}=-\left(0.16m\right)j$

Here, the unit vector k as angular direction in clockwise is negative and it is positive when counter clockwise.

Angular velocity of the rod BP in vector form is

$-\left(20rad/s\right)k$

Velocity of the rod BP is

$\begin{array}{l}{v}_p={\omega }_{BP}\times r_{P/B}\\ v_p=\left[-\left(20rad/s\right)k\right]\left[-\left(0.1732m\right)i-\left(0.1m\right)j\right]\\ =\left(3.464m/s\right)\left(k\times i\right)+\left(2m/s\right)\left(k\times j\right)\\ =\left(3.464m/s\right)j+\left(2m/s\right)\left(-i\right)\\ =\left(3.464m/s\right)j-\left(2m/s\right)i----(1)\end{array}$

Formulate the velocity equation of point P

$\begin{array}{l}{v}_p={\omega }_{AE}\times r_{P/A}+v_{P/AE}\\ v_p=\left({\omega }_{AE}k\right)\left[-\left(0.1m\right)j\right]+uj\\ =-\left(0.1{\omega }_{AE}\right)\left(k\times j\right)+uj\\ =-\left(0.1{\omega }_{AE}\right)\left(-i\right)+uj\\ =\left(0.1{\omega }_{AE}\right)i+uj---(2)\end{array}$

Compare the i terms from equation (1) And (2)

$\begin{array}{l}-2m/s=0.1{\omega }_{AE}\\ {\omega }_{AE}=-\frac{2}{0.1}\\ =-20rad/s\end{array}$

Compare the equation (1) and (2) for j terms. We get,

$u=3.464m/s$

So, the angular acceleration is zero.

Formulate the acceleration of the rod BP

$\begin{array}{l}{a}_p=\left[\left({\alpha }_{BP}\right)\left(r_{P/B}\right)\right]-{\omega }_{BP}^2{r}_{P/B}\\ a_p=\left\{\left(80rad/s^2\right)k\times \left(-0.1732m\right)i-\left(0.1m\right)j-{\left(-\left(20rad/s\right)k\right)}^2\left(-\left(0.1732\right)i-\left(0.1m\right)j\right)\right\}\\ =\left(13.856\right)\left(k\times i\right)+\left(8\right)\left(k\times j\right)-\left(400\right)\left(-\left(0.1732m\right)i-\left(0.1m\right)j\right)\\ =\left(13.856m/s^2\right)j+\left(8m/s^2\right)\left(-i\right)+\left(69.32m/s^2\right)i+\left(40m/s^2\right)j\\ =\left(13.856m/s^2\right)j-\left(8m/s^2\right)\left(i\right)+\left(69.32m/s^2\right)i+\left(40m/s^2\right)j----(3)\end{array}$

Find the acceleration of the point P to the point E

$\begin{array}{l}{a}_{p^1}=\left[\left({\alpha }_{AE}\right)\left(r_{P/A}\right)\right]-{\omega }_{AE}^2{r}_{P/A}\\ a_{p^{\text{'}}}=\left\{\left({\alpha }_{AE}k\right)\times \left(-0.1m\right)j-{\left(-\left(20rad/s\right)k\right)}^2\left[-\left(0.1m\right)j\right]\right\}\\ =-\left(0.1{\alpha }_{AE}\right)\left(k\times j\right)-\left(400\right)\left[-\left(0.1m\right)j\right]\\ =-\left(0.1{\alpha }_{AE}\right)\left(-i\right)+\left(40m/s^2\right)j\\ =0.1{\alpha }_{AE}i+\left(40m/s^2\right)j\end{array}$

Formulate the acceleration at the point P

$a_p=a_{p^{\text{'}}}+a_{p/AE}+2{\omega }_{BP}\times v_{p/AE}$

Substitute the above values

$\begin{array}{l}{a}_P=0.1{\alpha }_{AE}i+\left(40\right)j+0+2\left[-\left(20rad/s\right)k\right]\left((3.464m/s)j\right)\\ =0.1{\alpha }_{AE}i+\left(40\right)j-\left(138.56m/s^2\right)\left(k\times j\right)\\ =0.1{\alpha }_{AE}i+\left(40\right)j-\left(138.56\right)\left(-i\right)\\ =0.1{\alpha }_{AE}i+\left(40\right)j+\left(138.56\right)i----(4)\end{array}$

Based on equations (3) and (4), we get,

$\begin{array}{l}-\left(8m/s^2\right)+\left(69.32m/s^2\right)=0.1{\alpha }_{AE}+\left(138.56m/s^2\right)\\ {\alpha }_{AE}=-\frac{77.282}{0.1}\\ {\alpha }_{AE}=-772.82rad/s^2\end{array}$

Find the weight of the rod AE is get by

$\begin{array}{l}{W}_{AE}=m_{AE}\times g\\ W_{AE}=0.8kg\times 9.81m/s^2\\ =7.848N\end{array}$

Then the moment of inertial of the rod AE is

$\begin{array}{l}{I}_{AE}=\frac{m_{AE}{I}_{AE}^2}{12}\\ I_{AE}=\frac{0.8\times {\left(0.16m\right)}^2}{12}\\ =1.7066\times {10}^{-3}kg.m^2\end{array}$

Similarly, find the weight of the rod BP is

$\begin{array}{l}{W}_{AE}=m_{AE}\times g\\ W_{AE}=1kg\times 9.81m/s^2\\ =9.81N\end{array}$

Then the moment of inertial of the rod BP

$\begin{array}{l}{I}_{BP}=\frac{m_{BP}{I}_{BP}^2}{12}\\ I_{BP}=\frac{1kg\times {\left(0.2m\right)}^2}{12}\\ =3.333\times {10}^{-3}kg.m^2\end{array}$

Here, the mass of the centers of rod AE and BP are to be considered as G and H.

Then the reference for point G with point A as (0,-0.08m).

Here the position vector is −(0.08m)j.

Similarly, the point H with point B as the reference as (-0.0866m,-0.05m). Position vector is −(0.0866m)i-(0.05m)j

Acceleration of the point G is

$a_G={\alpha }_{AE}{r}_{G/A}-{\omega }_{AE}^2{r}_{G/A}$

$\begin{array}{l}{a}_G=\left[\left(-772.82rad/s^2\right)k\right]\left[-\left(0.08m\right)j\right]-{\left[\left(-20rad/s\right)k\right]}^2\left[-\left(0.08m\right)j\right]\\ =\left(61.8256m/s^2\right)\left(k\times j\right)+\left(32m/s^2\right)j\\ =\left(61.8256m/s^2\right)\left(-i\right)+\left(32m/s^2\right)j\\ =-\left(61.8256m/s^2\right)\left(i\right)+\left(32m/s^2\right)j\end{array}$

Find the acceleration at point H

$\begin{array}{l}{a}_H={\alpha }_{BP}{r}_{H/B}-{\omega }_{BP}^2{r}_{H/B}\\ a_H=\left\{\left(-\left(80rad/s^2\right)k\right)\left[-\left(0.0866m\right)i-\left(0.05m\right)j\right]-{\left[\left(-20rad/s\right)k\right]}^2\left[-\left(0.0866m\right)i-\left(0.05m\right)j\right]\right\}\\ =\left(6.928\right)\left(k\times i\right)+\left(4m/s^2\right)\left(k\times j\right)-\left(400\right)\left(-\left(0.0866m\right)i\right)-\left(0.05m\right)j\\ =\left(6.928\right)\left(j\right)+\left(4\right)\left(-i\right)\left(34.64\right)i+\left(20m/s^2\right)j\\ =\left(6.928\right)\left(j\right)-\left(4m/s^2\right)i\left(34.64m/s^2\right)+\left(20m/s^2\right)j\\ =\left(30.64m/s^2\right)i+\left(26.928m/s^2\right)j\end{array}$

Based on above figure, consider the moment about point B.

$\left\{\left(-\left(\frac{5}{12}\right)j\right)\times \overline{P}i+\left(-\left(0.3608\right)i-\left(0.2083\right)j\right)\times \left(-2j\right)+Mk\right\}=\left\{\left(-\left(0.3608\right)i-\left(0.2083\right)j\right)\times \left[\left(0.0621\right)\left(\left(288.6\right)i+\left(166.64\right)j\right)+\left(3.5944\times {10}^{-3}\right)\left(0\right)\right]\right\}$

$\begin{array}{l}\left[-0.4166\overline{P}\left(j\times i\right)+\left(0.7216\right)\left(i\times j\right)+Mk\right]=\left\{\left(-\left(0.3608\right)i-\left(0.2083\right)j\right)\times \left[\left(\left(17.930\right)i+\left(10.352\right)j\right)+0\right]\right\}\\ \left\{0.4166\overline{P}k+\left(0.7216\right)k+Mk\right\}=-3.7354k+3.735k---(5)\end{array}$ Substitute the values

$\begin{array}{l}\left\{0.1\left(52.757\right)k+\left(0.8495\right)k+Mk\right\}=-0.8k-0.2666k\\ 6.1252+M=-1.0666\\ M=-7.19N.m\end{array}$

The magnitude of the couple applied on the rod is BP. It is considered as -7.19N.m and acts in a clockwise direction.

To determine

Find the force exerted on rod AE.

Answer to Problem 16.142P

The force exerted by rod AE is F=52.757N.

Explanation of Solution

Considering the above diagram of free body of the rod AE

$\begin{array}{l}{r}_{P/A}\times \left(-\overline{P}i\right)=I_{AE}{\alpha }_{AE}+r_{G/A}\times \left(m_{AE}{a}_G\right)\\ -\left(0.1m\right)j\times \left(-\overline{P}i\right)=\left\{\left(1.7066\times {10}^{-3}\right)\left(\left(-772.82\right)k\right)+\left(-\left(0.08m\right)j\right)\times \left(0.8\right)\left[-\left(61.8256\right)i+\left(32\right)j\right]\right\}\\ (0.1\overline{P})(j\times i)=-1.3188k+(3.9568)(j\times i)-(2.048)(j\times j)\\ -0.1\overline{P}k=-1.3188k+3.8249(-k)-(2.048)(0)\\ -0.1\overline{P}k=-1.3188k-3.9568k\\ -0.1\overline{P}=-1.3188-3.9568\\ \overline{P}=52.757N\end{array}$

The force exerted by rod AE by block P is 52.75N. It is acted upon the left direction.