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6.6 HW 6

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6.6.1 Problem 1

Let state 1 be when the crate is thrown at the platform. Let the crate by body $A$ and the platform be body $B$. We will use work-energy to solve this.

$$T_{1}+V_{1}+U_{12}^{internal}+U_{12}^{external}=T_{2}+V_{2}$$ Where $U_{12}^{internal}$ is work done due to internal forces between the two bodies, which is the friction.  We will use notation $v_{A_{1}}$ to mean velocity of $A$ in state $1$ and $v_{A_{2}}$ to mean velocity of $A$ in state $2$ and the same for body $B$. Therefore the above equation becomes $$\frac{1}{2}m_{A}v_{A_{1}}^{2}+\frac{1}{2}m_{A}v_{B_{1}}^{2}-\int _{0}^{d}\mu _{k}m_{A}gdx=\frac{1}{2}m_{A}v_{A_{2}}^{2}+\frac{1}{2}m_{A}v_{B_{2}}^{2}$$ Notice that $V_{1}=V_{2}$ and hence they cancel. Also since in state 2 both body $A$ and $B$ move with same speed $v$, and also $v_{B_{1}}=0$, then the above simplifies to $$\frac{1}{2}m_{A}v_{A_{1}}^{2}-\mu _{k}m_{A}gd=\frac{1}{2}\left ( m_{A}+m_{B}\right ) v^{2}$$ We now solve for $d$, the distance that body $A$ (the crate) slides. The above is one equation with one unknown. $$\begin{align*} d & =\frac{\frac{1}{2}m_{A}v_{A_{1}}^{2}-\frac{1}{2}\left ( m_{A}+m_{B}\right ) v^{2}}{\mu _{k}m_{A}g}\\ & =\frac{1}{2}\frac{\left ( 210\right ) \left ( 15\right ) ^{2}-\left ( 210+741\right ) \left ( 2.652\right ) ^{2}}{\left ( 0.28\right ) \left ( 210\right ) \left ( 32.2\right ) }\\ & =10.712\text{ ft} \end{align*}$$

6.6.2 Problem 2

Let state 1 be just before release and state 2 after it moves by $1.5$ meter

Therefore

$$T_{1}+V_{1}+U_{12}^{internal}+U_{12}^{external}=T_{2}+V_{2}$$ $T_{1}=0,V_{1}=0$ and $U_{12}=0$ since there is no friction and no external force. $T_{2}=\frac{1}{2}\left ( m_{A}+m_{B}\right ) v^{2}$ since both bodies will have same speed. $V_{2}$ comes from spring and gravity. The distance $h=1.5$ meter. Therefore $V_{2}=\frac{1}{2}kh^{2}-m_{B}gh$ since spring extend by same amount $m_{B}$ and $m_{A}$ moves. The above becomes $$0=\frac{1}{2}\left ( m_{A}+m_{B}\right ) v^{2}+\frac{1}{2}kh^{2}-m_{B}gh$$ Where $v$ above is $1.3$ since both bodies move with same speed. We want to solve for $m_{B}$ the only unknown in this equation $$\begin{align*} 0 & =\frac{1}{2}m_{A}v^{2}+\frac{1}{2}m_{B}v^{2}+\frac{1}{2}kh^{2}-m_{B}gh\\ & =m_{B}\left ( \frac{1}{2}v^{2}-gh\right ) +\frac{1}{2}m_{A}v^{2}+\frac{1}{2}kh^{2}\\ m_{B} & =\frac{m_{A}v^{2}+kh^{2}}{2gh-v^{2}} \end{align*}$$

Plug-in numerical values gives

$$\begin{align*} m_{B} & =\frac{\left ( 4\right ) \left ( 1.3\right ) ^{2}+\left ( 9\right ) \left ( 1.5\right ) ^{2}}{2\left ( 9.81\right ) \left ( 1.5\right ) -\left ( 1.3\right ) ^{2}}\\ & =0.974\text{ kg} \end{align*}$$

6.6.3 Problem 3

Let state 1 be just before release and state 2 after it rotation.

$$T_{1}+V_{1}+U_{12}^{internal}+U_{12}^{external}=T_{2}+V_{2}$$ But $U_{12}=0$ since there is no friction and no external forces. Now $T_{1}=0$ since at rest. And $$V_{1}=-m_{B}gL_{1}\sin \alpha +m_{A}gL_{2}\sin \alpha$$ Where $\alpha =36^{o}$. In state 2 $$T_{2}=\frac{1}{2}m_{B}\left ( L_{1}\omega \right ) ^{2}+\frac{1}{2}m_{A}\left ( L_{2}\omega \right ) ^{2}$$ Where $\omega$ is the angular velocity, which we do not know, but will solve for. And $$V_{2}=m_{B}gL_{1}\sin \beta -m_{A}gL_{2}\sin \beta$$ Therefore (1) becomes $$\begin{align*} -m_{B}gL_{1}\sin \alpha +m_{A}gL_{2}\sin \alpha & =\frac{1}{2}m_{B}\left ( L_{1}\omega \right ) ^{2}+\frac{1}{2}m_{A}\left ( L_{2}\omega \right ) ^{2}+m_{B}gL_{1}\sin \beta -m_{A}gL_{2}\sin \beta \\ -m_{B}gL_{1}\sin \alpha +m_{A}gL_{2}\sin \alpha -m_{B}gL_{1}\sin \beta +m_{A}gL_{2}\sin \beta & =\omega ^{2}\left ( \frac{1}{2}m_{B}L_{1}^{2}+\frac{1}{2}m_{A}L_{2}^{2}\right ) \\ \omega ^{2} & =\frac{m_{A}gL_{2}\left ( \sin \alpha +\sin \beta \right ) -m_{B}gL_{1}\left ( \sin \alpha +\sin \beta \right ) }{\frac{1}{2}m_{B}L_{1}^{2}+\frac{1}{2}m_{A}L_{2}^{2}}\\ \omega & =\sqrt{\frac{2\left ( m_{A}gL_{2}-m_{B}gL_{1}\right ) \left ( \sin \alpha +\sin \beta \right ) }{m_{B}L_{1}^{2}+m_{A}L_{2}^{2}}} \end{align*}$$

Now we solve for $\omega$ and use it to find speed of $B$ from $v_{B}=L_{1}\omega$. Since $m_{A}=814,m_{B}=129,L_{1}=10,L_{2}=5,\alpha =36,\beta =110-36=74^{o}$ then

$$\begin{align*} \omega & =\sqrt{2\frac{\left ( \left ( 814\right ) \left ( 32.2\right ) \left ( 5\right ) -\left ( 129\right ) \left ( 32.2\right ) \left ( 10\right ) \right ) \left ( \sin \left ( 36\left ( \frac{\pi }{180}\right ) \right ) +\sin \left ( 74\left ( \frac{\pi }{180}\right ) \right ) \right ) }{\left ( 129\right ) 10^{2}+\left ( 814\right ) 5^{2}}}\\ & =2.888\,\text{\ rad/sec} \end{align*}$$

Hence

$$\begin{align*} v_{B} & =L_{1}\omega \\ & =10\left ( 2.888\right ) \\ & =28.88\text{ ft/sec} \end{align*}$$

6.6.4 Problem 4

Power $P$ is

$$P=Fv$$

Where $F$ is force generated by cyclist. From force balance we see that $F=mg\sin \theta$. Hence for constant power, we want

$$\left ( mg\sin 13^{0}\right ) v_{1}=\left ( mg\sin 20^{0}\right ) v_{2}$$

Solving for $v_{2}$

$$\begin{align*} v_{2} & =\frac{\left ( mg\sin 13^{0}\right ) v_{1}}{\left ( mg\sin 20^{0}\right ) }\\ & =\frac{\sin \left ( 13\left ( \frac{\pi }{180}\right ) \right ) }{\sin \left ( 20\left ( \frac{\pi }{180}\right ) \right ) }21\\ & =13.812\text{ mph} \end{align*}$$

6.6.5 Problem 5

$$\varepsilon =\frac{P_{out}}{P_{in}}$$

Where $\varepsilon$ is the efficiency and $P_{out}$ is power out and $P_{in}$ is power in. But $P_{out}=Fv_{c}$. So we just need to find force in the cable that the motor is pulling with. This force is $\frac{W}{4}$, since there are $4$ cables and hence the weight is distributed over them, and then the tension in the one cable attached to the motor is $\frac{W}{4}$. Now we have all the information to find $\varepsilon$

$$\begin{align*} P_{out} & =\left ( 4.5\right ) \left ( \frac{460}{4}\right ) \\ & =517.\,\allowbreak 5\text{ lb-ft/sec} \end{align*}$$

But $hp=550$ lb-ft/sec, therefore in hp the above is $\frac{517.\,\allowbreak 5}{550}=\allowbreak 0.941$, hence

$$\varepsilon =\frac{\allowbreak 0.941}{1.37}=0.687$$

6.6.6 Problem 6

$$\varepsilon =\frac{P_{out}}{P_{in}}$$

Mass of crate is $\frac{216}{32.2}$ slug. (note, number given $216$ is weight) These problem should make it more clear if $lb$ given is meant to be weight or mass.

We need to find $P_{in}$. We are given $\varepsilon$. We now calculate $P_{out}$ and then will be able to find $P_{in}$. But $P_{out}=Fv$, where $F$ is force given by motor to pull the crate. From free body diagram, we see that this force is $F=\mu mg\cos \theta +mg\sin \theta$. Hence

$$\begin{align*} P_{out} & =\left ( \mu mg\cos \theta +mg\sin \theta \right ) v\\ & =mg\left ( \mu \cos \theta +\sin \theta \right ) v\\ & =\left ( \frac{216}{32.2}\right ) \left ( 32.2\right ) \left ( 0.24\cos \left ( 28\left ( \frac{\pi }{180}\right ) \right ) +\sin \left ( 28\left ( \frac{\pi }{180}\right ) \right ) \right ) \left ( 6.6\right ) \\ & =971.374\text{ lb-ft/sec}\\ & =\frac{971.374}{550}=1.\,766\text{ hp} \end{align*}$$

Hence

$$\begin{align*} P_{in} & =\frac{P_{out}}{\varepsilon }\\ & =\frac{1.\,766}{0.78}\\ & =2.264\text{ hp} \end{align*}$$