Chapter 11.2, Problem 11.60P

(a)

To determine

The accelerations of A $\left(a_A\right)$ and B $\left(a_B\right)$ if $a_B>10{\text{mm/s}}^{\text{2}}$.

Answer to Problem 11.60P

The accelerations of A $\left(a_A\right)$ and B $\left(a_B\right)$ if $a_B>10{\text{mm/s}}^{\text{2}}$ are $\underset{\_}{20.0{\text{mm/s}}^2(\downarrow )}$ and $\underset{\_}{40.0{\text{mm/s}}^2(\downarrow )}$ respectively.

Explanation of Solution

Given information:

The relative change in position of block C with respect to block A $\left(y_{C/A}\right)$ is $280\text{mm}(\uparrow )$.

The relative velocity of collar B with respect to block A $\left(v_{B/A}\right)$ is $80\text{mm/s}(\downarrow )$.

The displacement of A $\left(y_A-{\left(y_A\right)}_0\right)$ is $160\text{mm}(\uparrow )$.

The displacement of B $\left(y_B-{\left(y_B\right)}_0\right)$ is $320\text{mm}(\downarrow )$.

Calculation:

Show the length and position of the cables as in Figure (1).

Choose the coordinate downward positive and right side positive.

Write the express for total lengths of cables 1:

$2y_A+2y_B+y_C=\text{constant}$

Differentiate the above equation with respective to time (t).

$2\frac{d{y}_A}{dt}+2\frac{d{y}_B}{dt}+\frac{d{y}_C}{dt}=\text{0}$

Denotes $\frac{d{y}_A}{dt}$ as $v_A$, $\frac{d{y}_B}{dt}$ as $v_B$ and $\frac{d{y}_C}{dt}$ as $v_C$.  Since, rate of change of any coordinate with respect to time is equal to the velocity.

$2v_A+2v_B+v_C=\text{0}$ (1)

Differentiate the equation (2) with respective to time (t).

$2\frac{d{v}_A}{dt}+2\frac{d{v}_B}{dt}+\frac{d{v}_C}{dt}=\text{0}$

Denotes $\frac{d{v}_A}{dt}$ as $a_A$, $\frac{d{v}_B}{dt}$ for $a_B$ and $\frac{d{v}_C}{dt}$ for $a_C$.

$2a_A+2a_B+a_C=\text{0}$ (2)

Write the express for total lengths of cables 2:

$\begin{array}{l}\left(y_D-y_A\right)+\left(y_D-y_B\right)=\text{constant}\\ y_D-y_A+y_D-y_B=\text{constant}\\ -y_A-y_B+2y_D=\text{constant}\end{array}$

Differentiate the above equation with respective to time (t).

$-\frac{d{y}_A}{dt}-\frac{d{y}_B}{dt}+\frac{2d{y}_D}{dt}=\text{0}$

Denotes $\frac{d{y}_A}{dt}$ as $v_A$, $\frac{d{y}_B}{dt}$ as $v_B$ and $\frac{d{y}_D}{dt}$ as $v_D$.  Since, rate of change of any coordinate with respect to time is equal to the velocity.

$-v_A-v_B+2v_D=\text{0}$ (3)

Differentiate the equation (3) with respective to time (t).

$-\frac{d{v}_A}{dt}-\frac{d{v}_B}{dt}+2\frac{d{v}_D}{dt}=\text{0}$

Denotes $\frac{d{v}_A}{dt}$ as $a_A$, $\frac{d{v}_B}{dt}$ for $a_B$ and $\frac{d{v}_D}{dt}$ for $a_D$.

$-a_A-a_B+2a_D=\text{0}$ (4)

At time (t) 0 sec the velocity (v) is zero.

Then, the initial position of cable A, cable B and cable C is equal.

${\left(y_A\right)}_0={\left(y_B\right)}_0={\left(y_C\right)}_0$

Calculate the position $\left(y_A\right)$ of block A using the relation:

$y_A={\left(y_A\right)}_0+v_{A}t+\frac{1}{2}{a}_A{t}^2$

Here, ${\left(y_A\right)}_0$ is initial position of block A.

Substitute 0 for $v_A$.

$\begin{array}{c}{y}_A={\left(y_A\right)}_0+0t+\frac{1}{2}{a}_A{t}^2\\ y_A={\left(y_A\right)}_0+\frac{1}{2}{a}_A{t}^2\end{array}$ . (5)

Calculate the position $\left(y_C\right)$ of block C using the relation:

$y_C={\left(y_C\right)}_0+v_{C}t+\frac{1}{2}{a}_C{t}^2$

Here, ${\left(y_C\right)}_0$ is initial position of block C.

Substitute 0 for $v_C$.

$\begin{array}{c}{y}_C={\left(y_C\right)}_0+0t+\frac{1}{2}{a}_C{t}^2\\ ={\left(y_C\right)}_0+\frac{1}{2}{a}_C{t}^2\end{array}$ . (6)

Calculate the position $\left(y_B\right)$ of block B using the relation:

$y_B={\left(y_B\right)}_0+v_{B}t+\frac{1}{2}{a}_B{t}^2$

Here, ${\left(y_B\right)}_0$ is initial position of block C.

Substitute 0 for $v_B$.

$\begin{array}{c}{y}_B={\left(y_B\right)}_0+0t+\frac{1}{2}{a}_B{t}^2\\ ={\left(y_B\right)}_0+\frac{1}{2}{a}_B{t}^2\end{array}$ . (7)

Write the expression for relative change in position of block C with respect to block A:

$y_{C/A}=y_C-y_A$

Substitute ${\left(y_C\right)}_0+\frac{1}{2}{a}_C{t}^2$ for $y_C$, ${\left(y_A\right)}_0+\frac{1}{2}{a}_A{t}^2$ for $y_A$ and ${\left(y_A\right)}_0={\left(y_C\right)}_0$.

$\begin{array}{c}{y}_{C/A}=\left({\left(y_C\right)}_0+\frac{1}{2}{a}_C{t}^2\right)-\left({\left(y_A\right)}_0+\frac{1}{2}{a}_A{t}^2\right)\\ =\frac{1}{2}\left(a_C-a_A\right)t^2\end{array}$ (8)

Write the equation for acceleration $a_C$ of block C:

Substitute $-280\text{mm}$ for $y_{C/A}$ and 2 sec for t in equation (5).

$\begin{array}{l}-280=\frac{1}{2}\left(a_C-a_A\right){\left(2\right)}^2\\ a_C=\left(a_A-140\right)\end{array}$ (9)

Calculate the acceleration of block A:

Substitute $\left(a_A-140\right)$ for $a_C$ in equation (2).

$\begin{array}{l}2a_A+2a_B+\left(a_A-140\right)=\text{0}\\ 3a_A+2a_B-140=0\\ a_A=\frac{1}{3}\left[140-2a_B\right]\end{array}$ (10)

Calculate the velocity $\left(v_B\right)$ of block B using the relation:

$v_B={\left(v_B\right)}_0+a_{B}t$

Substitute 0 for ${\left(v_B\right)}_0$.

$\begin{array}{c}{v}_B=0+a_{B}t\\ =a_{B}t\end{array}$

Calculate the velocity $\left(v_A\right)$ of block A using the relation:

$v_A={\left(v_A\right)}_0+a_{A}t$

Substitute 0 for ${\left(v_A\right)}_0$.

$\begin{array}{c}{v}_A=0+a_{A}t\\ =a_{A}t\end{array}$

Write the equation for relative velocity of block B with respect to block A

$v_{B/A}=v_B-v_A$

Substitute $a_{B}t$ for $v_B$ and $a_{A}t$ for $v_A$.

$\begin{array}{c}{v}_{B/A}=a_{B}t-a_{A}t\\ =\left(a_B-a_A\right)t\end{array}$

Substitute $80\text{mm/s}$ for $v_{B/A}$.

$80=\left(a_B-a_A\right)t$ (11)

Modify the equation (5).

$\begin{array}{l}{y}_A={\left(y_A\right)}_0+\frac{1}{2}{a}_A{t}^2\\ \left(y_A-{\left(y_A\right)}_0\right)=\frac{1}{2}{a}_A{t}^2\end{array}$

Substitute $160\text{mm}$ for $\left(y_A-{\left(y_A\right)}_0\right)$ in equation (

$160=\frac{1}{2}{a}_A{t}^2$ (12)

Modify the equation (7).

$\begin{array}{l}{y}_B={\left(y_B\right)}_0+\frac{1}{2}{a}_B{t}^2\\ \left(y_B-{\left(y_B\right)}_0\right)=\frac{1}{2}{a}_B{t}^2\end{array}$

Substitute $320\text{mm}$ for $\left(y_B-{\left(y_B\right)}_0\right)$.

$320=\frac{1}{2}{a}_B{t}^2$ (13)

Calculate the time (t):

Substrate equation (13) and (12).

$\begin{array}{l}320-160=\left(\frac{1}{2}{a}_B{t}^2-\frac{1}{2}{a}_A{t}^2\right)\\ 160=\frac{1}{2}\left(a_B-a_A\right)t^2\end{array}$

Substitute $80$ for $\left(a_B-a_A\right)t$.

$\begin{array}{l}160=\frac{1}{2}\left(80\times t\right)\\ t=\frac{\left(160\times 2\right)}{80}\\ t=4\sec \end{array}$

Calculate the acceleration $\left(a_A\right)$ of block A:

Substitute 4 sec for t in equation (12).

$\begin{array}{l}320=\frac{1}{2}{a}_B{\left(4\right)}^2\\ a_B=\frac{\left(320\times 2\right)}{4^2}\\ a_B=40{\text{mm/s}}^{\text{2}}(\downarrow )\end{array}$

Therefore, the accelerations of A $\left(a_A\right)$ and B $\left(a_B\right)$ if $a_B>10{\text{mm/s}}^{\text{2}}$ are $\underset{\_}{20.0{\text{mm/s}}^2(\downarrow )}$ and $\underset{\_}{40.0{\text{mm/s}}^2(\downarrow )}$ respectively.

(b)

To determine

The change in position $\left(y_D-{\left(y_D\right)}_0\right)$ block D when the velocity of block C is $600\text{mm/s}(\uparrow )$.

Answer to Problem 11.60P

The change in position $\left(y_D-{\left(y_D\right)}_0\right)$ block D when the velocity of block C is $600\text{mm/s}(\uparrow )$ is $\underset{\_}{375\text{mm}(\downarrow )}$.

Explanation of Solution

Given information:

The relative change in position of block C with respect to block A $\left(y_{C/A}\right)$ is $280\text{mm}(\uparrow )$.

The relative velocity of collar B with respect to block A $\left(v_{B/A}\right)$ is $80\text{mm/s}(\downarrow )$.

The displacement of A $\left(y_A-{\left(y_A\right)}_0\right)$ is $160\text{mm}(\uparrow )$.

The displacement of B $\left(y_B-{\left(y_B\right)}_0\right)$ is $320\text{mm}(\downarrow )$.

The velocity $\left(a_C\right)$ of block C is $600\text{mm/s}(\uparrow )$.

Calculation:

Calculate the acceleration $\left(a_C\right)$ of block C :

Substitute $20.0{\text{mm/s}}^2$ for $a_A$ in equation (9).

$\begin{array}{c}{a}_C=\left(20-140\right)\\ =-120{\text{mm/s}}^{\text{2}}\end{array}$

Calculate the acceleration $\left(a_D\right)$ of block D:

Substitute $20.0{\text{mm/s}}^2$ for $a_A$ and $40.0{\text{mm/s}}^2$ for $a_B$ in equation (4).

$\begin{array}{l}-20-40+2a_D=\text{0}\\ a_D\text{=}\frac{20+40}{2}\\ a_D=30{\text{mm/s}}^{\text{2}}\end{array}$

Calculate the time (t) using the relation below;

$v_C={\left(v_C\right)}_0+a_{C}t$

Here, ${\left(v_C\right)}_0$ is initial velocity of block C.

Substitute 0 for ${\left(v_C\right)}_0$, $-600\text{mm/s}$ for $v_C$ and $-120{\text{mm/s}}^{\text{2}}$ for $a_C$.

$\begin{array}{l}-600=0+\left(-120\times t\right)\\ t=\frac{-600}{-120}\\ t=5\sec \end{array}$

Calculate the change in positon $\left(y_D-{\left(y_D\right)}_0\right)$ block D after 5 sec using the relation:

$\begin{array}{l}{y}_D={\left(y_D\right)}_0+{\left(v_D\right)}_{0}t+\frac{1}{2}{a}_D{t}^2\\ \left(y_D-{\left(y_D\right)}_0\right)={\left(v_D\right)}_{0}t+\frac{1}{2}{a}_D{t}^2\end{array}$

Here, ${\left(y_D\right)}_0$ is initial position of block D.

Substitute 0 for ${\left(v_D\right)}_0$, 5 sec for t and $30{\text{mm/s}}^{\text{2}}$ for $a_D$.

$\begin{array}{c}\left(y_D-{\left(y_D\right)}_0\right)=\left(0\times 5\right)+\left(\frac{1}{2}\times \left(30\right)\times 5^2\right)\\ \text{=}375\text{mm}(\downarrow )\end{array}$

Therefore, the change in position $\left(y_D-{\left(y_D\right)}_0\right)$ block D when the velocity of block C is $600\text{mm/s}(\uparrow )$ is $\underset{\_}{375\text{mm}(\downarrow )}$.