Chapter 12.1, Problem 12.72P

To determine

(a)

The initial tension in the cable.

Answer to Problem 12.72P

The initial tension: $T=142.7N$

Explanation of Solution

Given information:

The velocity of block A is $v_A=2m/s$ at the instant when $r=0.8m\text{ and }\theta {\text{=30}}^{\circ }$.

Mass of the pulley and the effect of the friction is negligible.

From results of part (b), the initial acceleration of block A: $a_A=6.18m/s^2$

Calculations:

From the free body and kinetic diagram of block A:

$\begin{array}{l}\text{Applying force balance in x-directi on}:\\ \stackrel{+}{\to }\sum F_x=m_A{a}_A:\\ T\cos \theta =m_A{a}_A\\ or\\ T=m_A{a}_A\sec \theta \\ Substitut\text{i}\text{n}gvalues,\\ T=\left(20\right)\left(6.18\right)\sec {30}^{\circ }\\ \Rightarrow T=142.7N\end{array}$

Conclusion:

The initial tension in the cable is $T=142.7N$

To determine

(b)

The initial acceleration of block A.

Answer to Problem 12.72P

The initial acceleration of block A: $a_A=6.18m/s^2.$

Explanation of Solution

Given information:

The figure:

The velocity of block A is $v_A=2m/s$ at the instant when $r=0.8m\text{ and }\theta {\text{=30}}^{\circ }$.

Mass of the pulley and the effect of the friction is negligible.

Calculation:

$\begin{array}{l}Considerrand\theta asthepolarcoord\text{i}\text{n}atesofblockA.\\ Represent\text{i}\text{n}gthecomponentsofvelocityofblockAasshown\text{i}\text{n}thebelowdiagram:\end{array}$

$\begin{array}{l}\text{The radial component of velocity of block A:}\\ \dot{r}=v_r=v_A\cdot e_r=-v_A\cos {30}^o\\ \dot{r}=-2\cos {30}^o=-1.73205m\text{/}s\\ \text{And, the transverse component of velocity of block A:}\\ \text{r}\dot{\theta }=v_{\theta }=v_A\cdot e_{\theta }=-v_A\sin {30}^o\\ \text{r}\dot{\theta }=2\sin {30}^o=1.000m\text{/}{s}^2\\ \dot{\theta }=\frac{v_{\theta }}{r}=\frac{1.000}{0.8}=1.25rad/s\\ \text{Since the length of the cable is constant}\text{.}\\ \therefore \text{Constraint of cable:}\\ r+y_B=\text{Constant,}\\ \dot{r}+v_B=0,\\ \ddot{r}+a_B=0\\ or\\ \ddot{r}=-a_B\mathrm{.............}\left(1\right)\end{array}$

Draw the free body and kinetic diagram of block A:

$\begin{array}{l}\text{Applying force balance in x-direction}:\\ \stackrel{+}{\to }\sum F_x=m_A{a}_A:\\ T\cos \theta =m_A{a}_A\\ or\\ T=m_A{a}_A\sec \theta \mathrm{..................}\left(2\right)\\ \end{array}$

Draw the free body and kinetic diagram of block B:

$\begin{array}{l}\text{Applying force balance in y-direction:}\\ +\downarrow \sum F_y=m_B{a}_B:m_{B}g-T=m_B{a}_B\mathrm{..........}\left(3\right)\\ Add\text{i}\text{n}gEq.\left(2\right)toEq.\left(3\right)to\text{ eliminate }T,\\ m_{B}g=m_A{a}_A\sec \theta +m_B{a}_B\mathrm{....................}\left(4\right)\end{array}$

Representing the components of acceleration of block A as shown in the below diagram:

$\begin{array}{l}Radialandtransverse{\text{components of a}}_A.\\ Useamethodsimilartothatusedforthecomponentsofvelocity.\\ \ddot{r}+r{\dot{\theta }}^2=a_r=a_A\cdot e_r=-a_A\cos \theta \mathrm{............}\left(5\right)\end{array}$

$\begin{array}{l}\text{Using }Eq.\left(1\right)toe\lim \text{i}\text{n}ate\ddot{r}andchang\text{i}\text{n}gsignsgives\\ a_B=a_A\cos \theta -r{\dot{\theta }}^2\mathrm{.......................}\left(6\right)\end{array}$

$\begin{array}{l}\\ Substitut\text{i}\text{n}g\text{ Eq}\text{. }\left(6\right)\text{ into}Eq.\left(4\right)\text{and solving for }{a}_A,\\ a_A=\frac{m_B\left(g+r{\dot{\theta }}^2\right)}{m_A\sec \theta +m_B\cos \theta }=\frac{\left(25\right)\left(9.81+\left(0.8\right){\left(1.25\right)}^2\right)}{20\sec {30}^o+25\cos {30}^o}\\ \Rightarrow a_A=6.18m\text{/}{s}^2\\ fromEq.\left(6\right),a_B=6.18\cos {30}^o-\left(0.8\right){\left(1.25\right)}^2=4.10m\text{/}{s}^2\\ \left(a\right)fromEq.\left(2\right),\\ T=\left(20\right)\left(6.18\right)\sec {30}^o=142.7\\ T=142.7N\\ \left(b\right)Acceleration\text{of block A}\text{.}\\ {\text{a}}_A=6.18m\text{/}{s}^2\\ \left(c\right)Acceleration\text{of block B}\text{.}\\ {\text{a}}_B=4.10m\text{/}{s}^2\end{array}$

Conclusion:

The initial acceleration of block A: $a_A=6.18m/s^2.$

To determine

(c)

The initial acceleration of block B.

Answer to Problem 12.72P

The initial acceleration of block B: $a_B=4.10m/s^2$

Explanation of Solution

Given information:

The velocity of block A is $v_A=2m/s$ at the instant when $r=0.8m\text{ and }\theta {\text{=30}}^{\circ }$.

Mass of the pulley and the effect of the friction is negligible.

From results of part (b), the initial acceleration of block A: $a_A=6.18m/s^2$

Calculations:

From eq. (6) defined in part (b):

$\begin{array}{l}{a}_B=a_A\cos \theta -r{\dot{\theta }}^2\\ substitut\text{i}\text{n}gvalues:a_A=6.18m/s^2,r=0.8m\theta ={30}^{\circ }and\dot{\theta }=1.25rad/s\\ a_B=\left(6.18\right)\cos {30}^{\circ }-\left(0.8\right){\left(1.25\right)}^2\\ \Rightarrow a_B=4.10m/s^2\end{array}$

Conclusion:

The initial acceleration of block B is $a_B=4.10m/s^2$