Chapter 15.6, Problem 15.218P

To determine

Find the acceleration of collar A when $c=80\text{mm}$.

Answer to Problem 15.218P

The acceleration of collar A is $\underset{\_}{a_A=-\left(9.51\text{mm}/{\text{s}}^2\right)\text{j}}$.

Explanation of Solution

Given information:

The collar B moves towards D at a constant speed of $v_B=50\text{mm}/\text{s}$.

The length of rod AB is $l_{AB}=300\text{mm}$

The distance $c=80\text{mm}$

Calculation:

Calculate the position vector $\left(r\right)$ as shown below.

The position of A as follows:

$r_A=y\text{j}$

The position of D as follows:

$r_D=\left(90\text{mm}\right)\text{i}$

The position of C as follows:

$r_C=\left(180\text{mm}\right)\text{k}$

The position of D with respect to C.

$r_{D/C}=r_D-r_C$

Substitute $\left(90\text{mm}\right)\text{i}$ for $r_D$ and $\left(180\text{mm}\right)\text{k}$ for $r_C$.

$r_{D/C}=\left(90\text{mm}\right)\text{i}-\left(180\text{mm}\right)\text{k}$

Calculate the length of CD $\left(l_{CD}\right)$ as shown below.

$\begin{array}{c}{l}_{CD}=\sqrt{{90}^2+{\left(-180\right)}^2}\\ =\sqrt{40,500}\\ =201.246\text{mm}\end{array}$

The position of B with respect to C.

$r_{B/C}=\frac{c\left(r_{D/C}\right)}{180}$

Substitute $80\text{mm}$ for c and $\left(90\text{mm}\right)\text{i}-\left(180\text{mm}\right)\text{k}$ for $r_{D/C}$.

$\begin{array}{c}{r}_{B/C}=\frac{80\left(90\text{i}-\text{180k}\right)}{180}\\ =\left(40\text{mm}\right)\text{i}-\left(80\text{mm}\right)\text{k}\end{array}$

The position of B.

$r_B=r_C+r_{B/C}$

Substitute $\left(40\text{mm}\right)\text{i}-\left(80\text{mm}\right)\text{k}$ for $r_{B/C}$ and $\left(180\text{mm}\right)\text{k}$ for $r_C$.

$\begin{array}{c}{r}_B=\left(180\text{mm}\right)\text{k}+\left(\left(40\text{mm}\right)\text{i}-\left(80\text{mm}\right)\text{k}\right)\\ =\left(40\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{k}\end{array}$

The position of A with respect to B.

$r_{A/B}=r_A-r_B$

Substitute $y\text{j}$ for $r_A$ and $\left(40\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{k}$ for $r_B$.

$\begin{array}{c}{r}_{A/B}=y\text{j}-\left(\left(40\text{mm}\right)\text{i}+\left(100\text{mm}\right)\text{k}\right)\\ =-\left(40\text{mm}\right)\text{i}+y\text{j}-\left(100\text{mm}\right)\text{k}\end{array}$ (1)

Calculate the length value of y of AB using the relation as shown below.

$l_{AB}^2={\left(-40\right)}^2+y^2+{\left(-100\right)}^2$

Substitute $300\text{mm}$ for $l_{AB}$.

$\begin{array}{l}{300}^2=11,600+y^2\\ y^2=78,400\\ y=280\text{mm}\end{array}$

Calculate the position of A with respect to B $\left(r_{A/B}\right)$ as shown below.

Substitute $280\text{mm}$ for $y$ in Equation (1).

$r_{A/B}=-\left(40\text{mm}\right)\text{i}+\left(280\text{mm}\right)\text{j}-\left(100\text{mm}\right)\text{k}$

Calculate the velocity at B $\left(v_B\right)$ as shown below.

${\text{v}}_B=v_B\times \frac{r_{D/C}}{l_{CD}}$

Substitute $201.246\text{mm}$ for $l_{CD}$, $50\text{mm}/\text{s}$ for $v_B$,and $\left(90\text{mm}\right)\text{i}-\left(180\text{mm}\right)\text{k}$ for $r_{D/C}$.

$\begin{array}{c}{\text{v}}_B=\frac{50\left(90\text{i}-180\text{k}\right)}{201.246}\\ =22.3607\text{i}-44.7214\text{k}\\ =\left(22.3607\text{mm}/\text{s}\right)\text{i}-\left(44.7214\text{mm}/\text{s}\right)\text{k}\end{array}$

Calculate the velocity of collar A with respect to B $\left(v_{A/B}\right)$ as shown below.

${\text{v}}_{A/B}={\omega }_{AB}\times r_{A/B}$

The velocity of collar A with respect to B ${\text{v}}_{A/B}$ is perpendicular to $r_{A/B}$. Hence, $r_{A/B}\cdot {\text{v}}_{A/B}=0$.

Calculate the velocity of collar A $\left(v_A\right)$ as shown below.

${\text{v}}_A={\text{v}}_B+{\text{v}}_{A/B}$

Apply the dot product of $r_{A/B}$ on both sides of the Equation.

$\begin{array}{c}{r}_{A/B}\cdot {\text{v}}_A=r_{A/B}\cdot \left({\text{v}}_B+{\text{v}}_{A/B}\right)\\ =r_{A/B}\cdot {\text{v}}_B+r_{A/B}\cdot {\text{v}}_{A/B}\end{array}$

Substitute $0$ for $r_{A/B}\cdot {\text{v}}_{A/B}$.

$r_{A/B}\cdot {\text{v}}_A=r_{A/B}\cdot {\text{v}}_B$

Substitute $-\left(40\text{mm}\right)\text{i}+\left(280\text{mm}\right)\text{j}-\left(100\text{mm}\right)\text{k}$ for $r_{A/B}$, $\left(v_A\right)\text{j}$ for $v_A$, and $\left(22.3607\text{mm}/\text{s}\right)\text{i}-\left(44.7214\text{mm}/\text{s}\right)\text{k}$ for ${\text{v}}_B$.

$\begin{array}{l}\left(-40\text{i}+280\text{j}-100\text{k}\right)\cdot \left(v_A\text{j}\right)=\left(-40\text{i}+280\text{j}-100\text{k}\right)\cdot \left(\text{22}\text{.3607i}-44.7214\text{k}\right)\\ 280v_A=-894.428+\mathrm{4.472.14}\\ 280v_A=3,577.712\\ v_A=12.7775\text{mm}/\text{s}\end{array}$

Calculate the relative velocity of collar A with respect to B $\left({\text{v}}_{A/B}\right)$ as shown below.

${\text{v}}_{A/B}={\text{v}}_A-{\text{v}}_B$

Substitute $\left(12.7775\text{mm}/\text{s}\right)\text{j}$ for ${\text{v}}_A$ and $\left(22.3607\text{mm}/\text{s}\right)\text{i}-\left(44.7214\text{mm}/\text{s}\right)\text{k}$ for ${\text{v}}_B$.

$\begin{array}{c}{\text{v}}_{A/B}=\left(12.7775\text{mm}/\text{s}\right)\text{j}-\left(\left(22.3607\text{mm}/\text{s}\right)\text{i}-\left(44.7214\text{mm}/\text{s}\right)\text{k}\right)\\ =-\left(22.3607\text{mm}/\text{s}\right)\text{i}+\left(12.7775\text{mm}/\text{s}\right)\text{j}-\left(44.7214\text{mm}/\text{s}\right)\text{k}\end{array}$

$\begin{array}{c}{\left({\text{v}}_{A/B}\right)}^2={\left(-22.3607\right)}^2+{\left(12.7775\right)}^2+{\left(44.7214\right)}^2\\ =2,663.3{\left(\text{mm}/\text{s}\right)}^2\end{array}$

The acceleration of collar B is ${\text{a}}_B=0$.

The acceleration of collar A is ${\text{a}}_A=a_A\text{j}$.

Calculate the acceleration of collar A with respect to B $\left({\text{a}}_{A/B}\right)$ as shown below.

${\text{a}}_{A/B}={\alpha }_{A/B}\times r_{A/B}+{\omega }_{AB}\times {\text{v}}_{A/B}$

The acceleration ${\alpha }_{A/B}\times r_{A/B}$ is perpendicular to $r_{A/B}$. Hence, $r_{A/B}\cdot \left({\alpha }_{A/B}\times r_{A/B}\right)=0$.

Calculate the acceleration of collar A $\left({\text{a}}_A\right)$ as shown below.

${\text{a}}_A={\text{a}}_B+{\text{a}}_{A/B}$

Substitute ${\alpha }_{A/B}\times r_{A/B}+{\omega }_{AB}\times {\text{v}}_{A/B}$ for ${\text{a}}_{A/B}$ and $0$ for ${\text{a}}_B$.

$\begin{array}{c}{\text{a}}_A=0+{\alpha }_{A/B}\times r_{A/B}+{\omega }_{AB}\times {\text{v}}_{A/B}\\ ={\alpha }_{A/B}\times r_{A/B}+{\omega }_{AB}\times {\text{v}}_{A/B}\end{array}$

Apply the dot product of $r_{A/B}$ on both sides of the Equation.

$\begin{array}{c}{r}_{A/B}\cdot {\text{a}}_A=r_{A/B}\cdot {\alpha }_{A/B}\times r_{A/B}+r_{A/B}\cdot {\omega }_{A/B}\times {\text{v}}_{A/B}\\ =r_{A/B}\cdot {\alpha }_{A/B}\times r_{A/B}+{\text{v}}_{A/B}\cdot r_{A/B}\times {\omega }_{A/B}\end{array}$

Substitute $0$ for $r_{A/B}\cdot {\alpha }_{A/B}\times r_{A/B}$ and $-{\text{v}}_{A/B}$ for $r_{A/B}\times {\omega }_{A/B}$.

$\begin{array}{c}{r}_{A/B}\cdot {\text{a}}_A=0+{\text{v}}_{A/B}\cdot \left(-{\text{v}}_{A/B}\right)\\ =-{\left({\text{v}}_{A/B}\right)}^2\end{array}$

Substitute $-\left(40\text{mm}\right)\text{i}+\left(280\text{mm}\right)\text{j}-\left(100\text{mm}\right)\text{k}$ for $r_{A/B}$, $a_A\text{j}$ for ${\text{a}}_A$, and $2,663.3{\left(\text{mm}/\text{s}\right)}^2$ for ${\left({\text{v}}_{A/B}\right)}^2$.

$\begin{array}{l}\left(-40\text{i}+280\text{j}-100\text{k}\right)\cdot a_A\text{j}=-2,663.3\\ 280a_A=-2,663.3\\ a_A=-9.51\text{mm}/{\text{s}}^2\end{array}$

Therefore, the acceleration of collar A is $\underset{\_}{a_A=-\left(9.51\text{mm}/{\text{s}}^2\right)\text{j}}$.