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6.11 HW 11

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6.11.1 Problem 1

Let us assume the center of mass of the overall system is at some distance $z$ from point $A$ somewhere between $A$ and $C$. It does not matter where it is. Therefore the rotational equation of motion for the hanging system is

$$M_{cg}=I_{A}\alpha$$ Where $M$ is the moment of external forces around this center of mass and $I_{A}$ is the mass moment of inertia around $A$. But since we want the system to be translating, then $\alpha =0$. Therefore $$\begin{align} M & =0\nonumber \\ F_{y}z\sin \theta -F_{x}z\cos \theta & =0\tag{1} \end{align}$$

Notice the weights do not come into play, since we are taking moments about center of mass of the overall system.

So we just need to find $F_{x},F_{y}$. These forces are the reactions on point $A$ where it is connected. These can be found by resolving forces in the horizontal and vertical direction. In horizontal direction

$$\begin{equation} F_{x}=\left ( m_{AB}+m_{C}\right ) a_{0}\tag{2} \end{equation}$$ In vertical direction (where these is no acceleration) $$\begin{align} F_{y}-W_{AB}-W_{C} & =0\nonumber \\ F_{y} & =W_{AB}+W_{C}\tag{3} \end{align}$$

Plugging (2,3) into (1) and canceling $z$ (as we see, we really did not need to find where $z$ is), gives

$$\begin{align*} \left ( W_{AB}+W_{C}\right ) \sin \theta -\left ( m_{AB}+m_{C}\right ) a_{0}\cos \theta & =0\\ \tan \theta & =\frac{\left ( m_{AB}+m_{C}\right ) a_{0}}{\left ( W_{AB}+W_{C}\right ) } \end{align*}$$

Plugging the numerical values

$$\begin{align*} \tan \theta & =\frac{\left ( \frac{128}{32.2}+\frac{462}{32.2}\right ) 9}{\left ( 128+462\right ) }\\ & =0.279 \end{align*}$$

Hence

$$\begin{align*} \theta & =\arctan \left ( 0.279\right ) \\ & =0.272\\ & =15.616^{0} \end{align*}$$

6.11.2 Problem 2

Taking moments about $G$ (and assuming no friction from the ground as problems says to neglect rotational inertia of wheels, which seems to imply this).

$$-Fh+N_{B}L_{B}-N_{A}L_{A}=I\alpha$$ For $\alpha =0$ $$-Fh+N_{B}L_{B}-N_{A}L_{A}=0$$ And when $N_{A}=0$ $$F=\frac{N_{B}L_{B}}{h}$$ But $N_{A}+N_{B}=mg$ or since $N_{A}=0$ then $N_{B}=mg$ and the above becomes $$\begin{align*} F_{\min } & =\frac{mgL_{B}}{h}\\ & =\frac{\left ( 37\right ) \left ( 9.81\right ) \left ( 0.35\right ) }{\left ( 0.71\right ) }\\ & =178.929\text{ N} \end{align*}$$

And the acceleration is

$$\begin{align*} F & =ma\\ 178.929 & =37a\\ a & =\frac{178.929}{37}\\ & =4.836\text{ m/s}^{2} \end{align*}$$

6.11.3 Problem 3

$$\begin{align*} F & =ma\\ \mu N & =ma\\ a & =\frac{\mu N}{m}\\ & =\frac{\left ( 0.51\right ) \left ( mg\right ) }{m}\\ & =\left ( 0.51\right ) \left ( 32.2\right ) \\ & =16.422\text{ ft/s}^{2} \end{align*}$$

Hence

$$\begin{align*} v & =at\\ t & =\frac{v}{a}\\ & =\frac{18.2}{16.422}\\ & =1.108\text{ sec} \end{align*}$$

6.11.4 Problem 4

Resolve forces in vertical direction for hanging mass

$$T-m_{B}g=m_{B}a_{y}$$ But $a_{y}=R\alpha$ where $\alpha$ is angular acceleration of spool. Hence $$\begin{equation} T-m_{B}g=m_{B}R\alpha \tag{1} \end{equation}$$ For the spool, the equation of motion is $M=I\alpha$ or $$\begin{equation} -TR=mr_{G}^{2}\alpha \tag{2} \end{equation}$$ Where $r_{G}$ is radius of gyration. We have two equations and two unknowns $\alpha ,T$., solving gives $$\begin{align*} \alpha & =\frac{-m_{B}gR}{mr_{G}^{2}+m_{B}R^{2}}\\ T & =m_{B}R\alpha +m_{B}g \end{align*}$$

Hence

$$\alpha =\frac{-\left ( 6\right ) \left ( 9.81\right ) \left ( 0.18\right ) }{\left ( 4\right ) \left ( 0.11\right ) ^{2}+\left ( 6\right ) \left ( 0.18\right ) ^{2}}=-43.636\text{ rad/sec}^{2}$$ And $$\begin{align*} T & =\left ( 6\right ) \left ( 0.18\right ) \left ( -43.636\right ) +\left ( 6\right ) \left ( 9.81\right ) \\ & =11.733\text{ N} \end{align*}$$

6.11.5 Problem 5

I will use $L$ for $w$ so not to confuse it with $\omega$. Resolving forces in $x$ direction

$$-O_{x}=ma_{Gx}$$ in the $y$ direction $$P+O_{y}=ma_{Gy}$$ But $a_{Gx}=\frac{L}{2}\omega ^{2}$ and $a_{G_{y}}=-\frac{L}{2}\alpha$ where $\alpha$ is angular acceleration of gate. Hence the above becomes $$\begin{align} -O_{x} & =m\frac{L}{2}\omega ^{2}\tag{1}\\ P+O_{y} & =-m\frac{L}{2}\alpha \tag{2} \end{align}$$

Now the angular acceleration equation for the gate is, taking moments around center of mass

$$\begin{align} O_{y}\frac{L}{2} & =I_{cg}\alpha \nonumber \\ & =\frac{mL^{2}}{12}\alpha \tag{3} \end{align}$$

From (2) $O_{y}=-m\frac{L}{2}\alpha -P$, plug this in (3) gives

$$\begin{align*} \left ( -m\frac{L}{2}\alpha -P\right ) \frac{L}{2} & =\frac{mL^{2}}{12}\alpha \\ -P\frac{L}{2} & =\frac{mL^{2}}{12}\alpha +m\frac{L^{2}}{4}\alpha \\ -P\frac{L}{2} & =\alpha \left ( \frac{mL^{2}}{12}+\frac{mL^{2}}{4}\right ) \\ \alpha & =-\frac{P\frac{L}{2}}{\frac{1}{3}L^{2}m}\\ & =-\frac{3}{2}\frac{P}{Lm} \end{align*}$$

Plug-in numerical values

$$\alpha =-\frac{3}{2}\frac{\left ( 21\right ) }{\left ( 14\right ) \left ( \frac{213}{32.2}\right ) }=-0.340$$ From (3) $$\begin{align*} O_{y} & =\frac{mL}{6}\alpha \\ & =\frac{\frac{213}{32.2}\left ( 14\right ) }{6}\left ( -0.3408\right ) \\ & =-5.25\text{ N} \end{align*}$$

To find $\omega$, from $\omega =\alpha t=-0.340\,\left ( 2.2\right ) =-0.748$ rad/sec, hence from  (1)

$$\begin{align*} -O_{x} & =m\frac{L}{2}\omega ^{2}\\ & =\frac{213}{32.2}\left ( \frac{14}{2}\right ) \left ( -0.748\,\right ) ^{2}\\ O_{x} & =-25.929\text{ N} \end{align*}$$

6.11.6 Problem 6

Resolving forces in $x$ direction, where $F_{x},F_{y}$ are forces in hinge

$$\begin{equation} F_{x}=-m\left ( \frac{3}{2}L\right ) \omega ^{2}\tag{1} \end{equation}$$ In $y$ direction $$\begin{equation} F_{y}-2mg=m\left ( \frac{3}{2}L\right ) \alpha \tag{2} \end{equation}$$ Taking moments about the hinge $O$ $$\begin{equation} \left ( -mg\frac{L}{2}-mgL\right ) =\left ( \left ( m\frac{L^{2}}{3}\right ) +\left ( \frac{1}{12}mL^{2}+mL^{2}\right ) \right ) \alpha \tag{3} \end{equation}$$ Solving (2,3) for $F_{y},\alpha$ gives $$\begin{align*} F_{y} & =40.394\text{ N}\\ \alpha & =-5.52503\text{ rad/sec}^{2} \end{align*}$$

We are given $\omega =6.9$ rad/sec., hence from (1)

$$F_{x}=-1342.6\text{ N}$$