Chapter 12.1, Problem 12.71P

The two blocks are released from rest when $r=0.8$ m and $\theta =30°$ . Neglecting the mass of the pulley and the effect of friction in the pulley and between block A and the horizontal surface, determine (a) the initial tension in the cable, (b) the initial acceleration of block A, (c) the initial acceleration of block B.

    

To determine

(a)

The initial tension in the cable.

Answer to Problem 12.71P

The initial tension: $T=126.6N$

Explanation of Solution

Given information:

Blocks are released from rest when $r=0.8m\text{ and }\theta {\text{=30}}^{\circ }$.

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

From part (b), The initial acceleration of block A: $a_A=5.48m/s^2$

Calculations:

From the free body and kinetic diagram of block A:

$\begin{array}{l}\text{Applying force balance in x-direction:}\\ \stackrel{+}{\to }\sum F_x=m_A{a}_A:\\ T\cos \theta =m_A{a}_A\\ or\\ T=m_A{a}_A\sec \theta \\ substitut\text{i}\text{n}gvalues,\\ T=\left(20\right)\left(5.48\right)\sec {30}^{\circ }\\ \Rightarrow T=126.6N\end{array}$

Conclusion:The initial tension in the cable is $T=126.6N$

To determine

(b)

The initial acceleration of block A.

Answer to Problem 12.71P

The initial acceleration of block A: $a_A=5.48m/s^2$

Explanation of Solution

Given information:

The figure:

Blocks are released from rest when $r=0.8m\text{ and }\theta {\text{=30}}^{\circ }$.

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

Calculations:

$\begin{array}{l}Considerrand\theta asthepolarcoord\text{i}\text{n}atesofblockA,\\ \text{Since the length of the cable is constant}\text{.}\\ \therefore \text{Constraint of the cable:}\\ r+y_B=\text{constant,}\\ \left(\begin{array}{l}\text{where,}\\ r\to \text{length of the cable between the block }A\text{ and the pulley}\text{.}\\ y_B\to \text{position coordinate of block B}\end{array}\right)\\ \text{<mover accent="’true'"><mi>r</mi><mo>˙</mo></mover>}+v_B=0,\\ \text{<mover accent="’true'"><mi>r</mi><mo>¨</mo></mover>}+a_B=0,\\ or\\ \text{<mover accent="’true'"><mi>r</mi><mo>¨</mo></mover>}=-a_B\mathrm{...............}\left(1\right)\end{array}$

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:

$\begin{array}{l}\text{Applying force balance in y-direction:}\\ +\downarrow \sum F_y=m_B{a}_B;\\ m_{B}g-T=m_B{a}_B\mathrm{..........}\left(3\right)\\ AddEq.\left(2\right)\text{ and }Eq.\left(3\right)toe\lim \text{i}\text{n}ateT,\\ m_{B}g=m_A{a}_A\sec \theta +m_B{a}_B\mathrm{....................}\left(4\right)\end{array}$

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

$\begin{array}{l}\text{The radial component of acceleration of block A:}\\ a_r=\ddot{r}-r{\dot{\theta }}^2\\ also,\\ a_r=a_A\cdot e_r=-a_A\cos \theta \\ \therefore \ddot{r}-r{\dot{\theta }}^2=-a_A\cos \theta \mathrm{.................}\left(5\right)\end{array}$

$\begin{array}{l}Noth\text{i}\text{n}gthat\text{i}\text{n}itially\dot{\theta }=0,u\sin gEq.\left(1\right)toe\lim \text{i}\text{n}ate\ddot{r},andchang\text{i}\text{n}gsignsgives\\ a_B=a_A\cos \theta \mathrm{.......................}\left(6\right)\\ Substitut\text{i}\text{n}g\text{ Eq}\text{. }\left(6\right)\text{ into}\left(4\right)\text{and solving for }{a}_A,\\ a_A=\frac{m_{B}g}{m_A\sec \theta +m_B\cos \theta }=\frac{\left(25\right)\left(9.81\right)}{20\sec {30}^o+25\cos {30}^o}\\ \Rightarrow a_A=5.48m\text{/}{s}^2\end{array}$

Conclusion:

To determine

(c)

The initial acceleration of block B.

Answer to Problem 12.71P

The initial acceleration of block B: $a_B=4.75m/s^2$

Explanation of Solution

Given information:

Blocks are released from rest when $r=0.8m\text{ and }\theta {\text{=30}}^{\circ }$.

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

From part (b), The initial acceleration of block A: $a_A=5.48m/s^2$

Calculations:

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

$\begin{array}{l}{a}_B=a_A\cos \theta \\ substitut\text{i}\text{n}gvalues,\\ a_B=5.48\cos {30}^{\circ }\\ \Rightarrow a_B=4.75m/s^2\end{array}$

Conclusion:The initial acceleration of block B is $a_B=4.75m/s^2$