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6.4 HW 4

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6.4.1 Problem 1

The velocity in meter per second is $$\begin{align*} v & =228\left ( \frac{1000}{km}\right ) \left ( \frac{h}{3600}\right ) \\ & =\left ( 228\right ) \frac{1000}{3600}\\ & =63.333\text{ m/s} \end{align*}$$

In normal direction, the acceleration is $a_{n}=\frac{v^{2}}{\rho }=\frac{\left ( 63.333\right ) ^{2}}{115}=\allowbreak 34.879$ m/s$^{2}$ or in terms of $g$, it becomes $\frac{34.879}{9.81}=3.555$ g.

Now, force balance in vertical direction gives $$\begin{align*} L-mg & =ma_{n}\\ L & =m\left ( g+a_{n}\right ) \\ & =970\left ( 9.81+34.879\right ) \\ & =43348.33\text{ N} \end{align*}$$

6.4.2 Problem 2

For maximum speed, friction acts downwards as car assumed to be just about to slide upwards.  Resolving forces in normal and tangential gives $$\begin{align} \mu _{s}N\cos \theta +N\sin \theta & =m\frac{v_{\max }^{2}}{\rho }\tag{1}\\ -mg+N\cos \theta -\mu _{s}N\sin \theta & =0\tag{2} \end{align}$$

From (2) $$N=\frac{mg}{\cos \theta -\mu _{s}\sin \theta }$$ From (1) $$\begin{align*} v_{\max }^{2} & =\frac{\rho }{m}\left ( \mu _{s}N\cos \theta +N\sin \theta \right ) \\ & =\frac{\rho }{m}\left ( \mu _{s}\left ( \frac{mg}{\cos \theta -\mu _{s}\sin \theta }\right ) \cos \theta +\left ( \frac{mg}{\cos \theta -\mu _{s}\sin \theta }\right ) \sin \theta \right ) \\ & =\rho \left ( \mu _{s}\left ( \frac{g}{\cos \theta -\mu _{s}\sin \theta }\right ) \cos \theta +\left ( \frac{g}{\cos \theta -\mu _{s}\sin \theta }\right ) \sin \theta \right ) \\ & =\rho g\left ( \frac{\mu _{s}}{1-\mu _{s}\tan \theta }+\frac{\tan \theta }{1-\mu _{s}\tan \theta }\right ) \\ & =\rho g\left ( \frac{\mu _{s}+\tan \theta }{1-\mu _{s}\tan \theta }\right ) \\ & =324\left ( 9.81\right ) \left ( \frac{0.2+\tan \left ( 31\frac{\pi }{180}\right ) }{1-0.2\tan \left ( 31\frac{\pi }{180}\right ) }\right ) \\ & =2893.165\text{ m/s} \end{align*}$$

Hence $$\begin{align*} v_{\max } & =53.788\text{ m/s}\\ & =53.788\left ( 0.001\right ) \left ( 3600\right ) \\ & =193.637\text{ km/hr} \end{align*}$$

For minimum speed, friction acts upwards as car assumed to be just about to slide downwards.  Resolving forces in normal and tangential gives $$\begin{align} -\mu _{s}N\cos \theta +N\sin \theta & =m\frac{v_{\min }^{2}}{\rho }\tag{3}\\ -mg+N\cos \theta +\mu _{s}N\sin \theta & =0\tag{4} \end{align}$$

From (4) $$N=\frac{mg}{\cos \theta +\mu _{s}\sin \theta }$$ From (3) $$\begin{align*} v_{\min }^{2} & =\frac{\rho }{m}\left ( -\mu _{s}N\cos \theta +N\sin \theta \right ) \\ & =\frac{\rho }{m}\left ( -\mu _{s}\left ( \frac{mg}{\cos \theta +\mu _{s}\sin \theta }\right ) \cos \theta +\left ( \frac{mg}{\cos \theta +\mu _{s}\sin \theta }\right ) \sin \theta \right ) \\ & =\rho g\left ( -\mu _{s}\left ( \frac{1}{\cos \theta +\mu _{s}\sin \theta }\right ) \cos \theta +\left ( \frac{1}{\cos \theta +\mu _{s}\sin \theta }\right ) \sin \theta \right ) \\ & =\rho g\left ( \frac{-\mu _{s}}{1+\mu _{s}\tan \theta }+\frac{\tan \theta }{1+\mu _{s}\tan \theta }\right ) \\ & =\rho g\left ( \frac{\tan \theta -\mu _{s}}{1+\mu _{s}\tan \theta }\right ) \\ & =324\left ( 9.81\right ) \left ( \frac{\tan \left ( 31\frac{\pi }{180}\right ) -0.2}{1+0.2\tan \left ( 31\frac{\pi }{180}\right ) }\right ) \\ & =1137.425\text{ m/s} \end{align*}$$

Hence $$\begin{align*} v_{\min } & =33.726\text{ m/s}\\ & =33.726\left ( 0.001\right ) \left ( 3600\right ) \\ & =121.414\text{ km/hr} \end{align*}$$

Hence $$121.414\leq v\leq 193.637$$

6.4.3 Problem 3

6.4.3.1 part (1)

Since acceleration is constant along barrel, then using $$v_{f}^{2}=v_{0}^{2}+2\ddot{r}L$$ We will solve for $\ddot{r}$. Assuming $v_{0}=0$ then $$\begin{align*} 1681^{2} & =2\ddot{r}\left ( 4.2\right ) \\ \ddot{r} & =\frac{1681^{2}}{2\left ( 4.2\right ) }\\ & =336400.1\text{ m/s}^{2} \end{align*}$$

But $$\bar{a}=\left ( \ddot{r}-L\dot{\theta }^{2}\right ) \hat{u}_{r}+\left ( L\ddot{\theta }+2\dot{r}\dot{\theta }\right ) \hat{u}_{\theta }$$ But $\ddot{\theta }=0$, hence the above becomes $$\begin{align*} \bar{a} & =\left ( 336400.1-\left ( 4.2\right ) \left ( 0.18\right ) ^{2}\right ) \hat{u}_{r}+\left ( 2\left ( 1681\right ) \left ( 0.18\right ) \right ) \hat{u}_{\theta }\\ & =336400\ \hat{u}_{r}+605.16\ \hat{u}_{\theta } \end{align*}$$

6.4.3.2 part (2)

Free body diagram for bullet gives $$P=ma_{r}+mg\sin \theta$$ but $a_{r}=336400$ m/s$^{2}$ and $m=16.9\ $kg and $\theta =23^{0}$, hence $$\begin{align*} P & =\left ( 16.9\right ) \left ( 336400\right ) +\left ( 16.9\right ) \left ( 9.81\right ) \sin \left ( \left ( 23\right ) \frac{\pi }{180}\right ) \\ & =5685225\text{ N}\\ & =5.685\times 10^{6}\text{ N} \end{align*}$$

6.4.3.3 Part (3)

Free body diagram for bullet gives $$\begin{align*} N & =ma_{\theta }+mg\cos \theta \\ & =\left ( 16.9\right ) \left ( 605.16\right ) +\left ( 16.9\right ) \left ( 9.81\right ) \cos \left ( \left ( 23\right ) \frac{\pi }{180}\right ) \\ & =10379.81\text{ N}\\ & =10.379\times 10^{3}\text{ N} \end{align*}$$

6.4.4 Problem 4

The free body diagram is

For $A$, $$\begin{align} \sum F_{x} & =m_{A}a_{A_{x}}\nonumber \\ F_{A} & =m_{A}a_{A_{x}}\tag{2} \end{align}$$

$$\begin{align} \sum F_{y} & =m_{A}a_{A_{y}}\nonumber \\ N_{A}-W_{A} & =0\tag{3} \end{align}$$

For $B$ $$\begin{align} \sum F_{x} & =m_{B}a_{B_{x}}\nonumber \\ F_{o}-F_{A}-F_{B} & =m_{B}a_{B_{x}}\tag{4} \end{align}$$

$$\begin{align} \sum F_{y} & =m_{B}a_{B_{y}}\nonumber \\ N_{B}-W_{B}-N_{A} & =0\tag{5} \end{align}$$

Hence (3) becomes $$\begin{align} N_{A} & =W_{A}\tag{6A}\\ & =m_{A}g\nonumber \end{align}$$

And (5) becomes $$\begin{align} N_{B} & =W_{B}+N_{A}\nonumber \\ & =m_{B}g+m_{A}g\nonumber \\ & =\left ( m_{B}+m_{A}\right ) g\tag{6B} \end{align}$$

But  $F_{A}=N_{A}\mu _{1}$ then (2) becomes $$N_{A}\mu _{1}=m_{A}a_{A_{x}}$$ But from (6) the above reduces to (since $N_{A}=m_{A}g$) $$m_{A}g\mu _{1}=m_{A}a_{A_{x}}$$ Hence $$\begin{align*} a_{A_{x}} & =g\mu _{1}\\ & =\left ( 32.2\right ) \left ( 0.26\right ) \\ & =8.372\text{ ft/s}^{2} \end{align*}$$

Similarly $F_{B}=N_{B}\mu _{2}$ hence (4) becomes $$F_{o}-N_{A}\mu _{1}-N_{B}\mu _{2}=m_{B}a_{B_{x}}$$ But $N_{B}=\left ( m_{A}+m_{B}\right ) g$, then above becomes $$\begin{align*} F_{o}-m_{A}g\mu _{1}-\left ( m_{A}+m_{B}\right ) g\mu _{2} & =m_{B}a_{B_{x}}\\ a_{B_{x}} & =\frac{F_{o}-m_{A}g\mu _{1}-\left ( m_{A}+m_{B}\right ) g\mu _{2}}{m_{B}}\\ a_{B_{x}} & =\frac{438-\left ( 50\right ) \left ( 0.26\right ) -\left ( 50+71\right ) \left ( 0.46\right ) }{\frac{71}{32.2}}\\ & =167.504\text{ ft/sec}^{2} \end{align*}$$

6.4.5 Problem 5

Free body diagram for block $B$ results in $$2g-\frac{8T}{m_{B}}=a_{A}$$ Free body diagram for block $A$ resuts in $$-m_{A}g\sin \theta -\mu m_{A}g\cos \theta +2T=m_{A}a_{A}$$ In addision, since rope length is fixed, we find that $a_{A}=2a_{B}$. The above are 3 equations in 3 unknowns $T,a_{A},a_{B}$. Solving gives $$\begin{align*} a_{A} & =4.439\text{ m/s}^{2}\\ a_{B} & =2.2197\text{ m/s}^{2}\\ T & =37.95\text{ N} \end{align*}$$

6.4.6 Problem 6

maximum displacement is $2$ ft. Maximum speed is $9.404$ ft/sec.