Chapter 15.4, Problem 15.123P

The right leg of an athlete on a rowing machine can be modeled as a linkage as shown, where A represents the ankle (which is stationary), K the knee, and H the hip. At the instant when θ = 75°, the shank AK has an angular velocity of 1 rad/s and an angular acceleration of 1.5 rad/s², both counterclockwise. Determine the velocity and acceleration of the hip H at this instant in time.

Fig. P15.123

To determine

Find the velocity and acceleration of the hip H at the instant.

Answer to Problem 15.123P

The velocity and acceleration of the hip H are $\underset{\_}{\left(0.374\text{m/s}\right)\text{i}}$ and $\underset{\_}{\left(-0.485{\text{m/s}}^2\right)\text{i}}$.

Explanation of Solution

Given information:

The ankle, knee and the hip are denoted by A, K, and H.

The ankle A is stationary.

The value of the angle is $\theta =75°$.

The angular velocity of the shank AK is $1\text{rad/s}\left(\text{Clockwise}\right)$.

The angular acceleration of the shank AK is $1.5{\text{rad/s}}^2$$\left(\text{Clockwise}\right)$.

Calculation:

Show the modelled linkage as shown in Figure 1.

Refer to Figure 1.

Consider the distance KH and KA are denoted by $L_T=442\text{mm}$ and $L_S=375\text{mm}$.

Show the relation between the distances KH and KA as follows:

$L_s\sin \theta =d+h$ (1)

$L_T\sin \varphi =d$ (2)

Modify Equation (1) using Equation (2).

$\begin{array}{l}{L}_s\sin \theta =L_T\sin \varphi +h\\ L_T\sin \varphi =L_s\sin \theta -h\\ \sin \varphi =\frac{L_s\sin \theta -h}{L_T}\\ \varphi ={\sin }^{-1}\left(\frac{L_s\sin \theta -h}{L_T}\right)\end{array}$

Substitute $442\text{mm}$ for $L_T$, $375\text{mm}$ for $L_S$, and $310\text{mm}$ for $h$ and $75°$ for $\theta$.

$\begin{array}{c}\varphi ={\sin }^{-1}\left(\frac{375\sin \left(75°\right)-310}{442}\right)\\ ={\sin }^{-1}\left(\frac{362.222-310}{442}\right)\\ ={\sin }^{-1}\left(0.118149\right)\\ =6.7853°\end{array}$

Show the relative velocity for AK as follows:

$\begin{array}{l}{\text{v}}_K={\text{v}}_A+{\text{ω}}_{AK}\times {\text{r}}_{K/A}\\ {\text{v}}_K={\text{v}}_A+{\omega }_{AK}\text{k}\times {\text{r}}_{K/A}\end{array}$

Show the relative velocity for KH as follows:

$\begin{array}{l}{\text{v}}_H={\text{v}}_K+\left({\text{ω}}_{KH}\right)\times {\text{r}}_{H/K}\\ v_H\text{i}={\text{v}}_K+\left({\omega }_{KH}\text{k}\right)\times {\text{r}}_{H/K}\\ v_H\text{i}={\text{v}}_A+{\omega }_{AK}\text{k}\times {\text{r}}_{K/A}+{\text{ω}}_{KH}\times {\text{r}}_{H/K}\end{array}$ (3)

Consider the position of the point K with respect to A is denoted by ${\text{r}}_{K/A}=-L_S\cos \theta \text{i}+L_S\sin \theta \text{j}$.

Consider the position of the point H with respect to K is denoted by ${\text{r}}_{H/K}=-L_T\cos \varphi \text{i}-L_T\sin \varphi \text{j}$.

Substitute $-L_T\cos \varphi \text{i}-L_T\sin \varphi \text{j}$ for ${\text{r}}_{H/K}$, $-L_S\cos \theta \text{i}+L_S\sin \theta \text{j}$ for ${\text{r}}_{K/A}$, and 0 for ${\text{v}}_A$.

$\begin{array}{l}{v}_H\text{i}=\left[\begin{array}{l}0+{\omega }_{AK}\text{k}\times \left(-L_S\cos \theta \text{i}+L_S\sin \theta \text{j}\right)\\ +{\omega }_{KH}\text{k}\times \left(-L_T\cos \varphi \text{i}-L_T\sin \varphi \text{j}\right)\end{array}\right]\\ v_H\text{i}=\left[\begin{array}{l}-L_S{\omega }_{AK}\cos \theta \text{j}-L_S{\omega }_{AK}\sin \theta \text{i}\\ -L_T{\omega }_{KH}\cos \varphi \text{j}+L_T{\omega }_{KH}\sin \varphi \text{i}\end{array}\right]\\ v_H\text{i}=\left[\begin{array}{l}\left(-L_S{\omega }_{AK}\cos \theta -L_T{\omega }_{KH}\cos \varphi \right)\text{j}\\ \text{+}\left(-L_S{\omega }_{AK}\sin \theta +L_T{\omega }_{KH}\sin \varphi \right)\text{i}\end{array}\right]\end{array}$ (4)

Equate the j component of the Equation (4).

$\begin{array}{l}\left(-L_S{\omega }_{AK}\cos \theta -L_T{\omega }_{KH}\cos \varphi \right)=0\\ L_T{\omega }_{KH}\cos \varphi =-L_S{\omega }_{AK}\cos \theta \\ {\omega }_{KH}=\frac{-L_S\cos \theta }{L_T\cos \varphi }\times {\omega }_{AK}\end{array}$ (5)

Substitute $442\text{mm}$ for $L_T$, $375\text{mm}$ for $L_S$, $6.7853°$ for $\varphi$, $1\text{rad/s}$ for ${\omega }_{AK}$, and $75°$ for $\theta$.

$\begin{array}{c}{\omega }_{KH}=\frac{-375\cos \left(75°\right)}{442\cos \left(6.7853°\right)}\times 1\\ =-0.22114\text{rad/s}\end{array}$

Equate i component of the Equation (4).

$v_H=\left(-L_S{\omega }_{AK}\sin \theta +L_T{\omega }_{KH}\sin \varphi \right)$ (6)

Substitute $442\text{mm}$ for $L_T$, $375\text{mm}$ for $L_S$, $6.7853°$ for $\varphi$, $1\text{rad/s}$ for ${\omega }_{AK}$, $-0.22114\text{rad/s}$ for ${\omega }_{KH}$ and $75°$ for $\theta$.

$\begin{array}{c}{v}_H=-375\times 1\times \sin \left(75°\right)+442\times \left(-0.22114\right)\sin \left(6.7853°\right)\\ =0.37377\text{m/s}\\ \text{=}\left(0.37377\text{m/s}\right)\text{i}\end{array}$

Thus, the velocity at H is $\underset{\_}{\left(0.374\text{m/s}\right)\text{i}}$.

Consider AK.

Show the relation between the acceleration of A and K as follows:

$\begin{array}{l}{\text{a}}_K={\text{a}}_A-{\omega }_{KA}^2{\text{r}}_{K/A}+\text{α}\times {\text{r}}_{K/A}\\ {\text{a}}_K={\text{a}}_A-{\omega }_{KA}^2{\text{r}}_{K/A}+\left({\alpha }_{KA}\text{k}\right)\times {\text{r}}_{K/A}\end{array}$

Substitute $-L_S\cos \theta \text{i}+L_S\sin \theta \text{j}$ for ${\text{r}}_{K/A}$, $1\text{rad/s}$ for ${\omega }_{KA}$, and 0 for ${\text{a}}_A$.

$\begin{array}{l}{\text{a}}_K=0-1^2\times \left(-L_S\cos \theta \text{i}+L_S\sin \theta \text{j}\right)+{\alpha }_{KA}\text{k}\times \left(-L_S\cos \theta \text{i}+L_S\sin \theta \text{j}\right)\\ {\text{a}}_K=L_S\cos \theta \text{i}-L_S\sin \theta \text{j}-L_S{\alpha }_{KA}\cos \theta \text{j}-L_S{\alpha }_{KA}\sin \theta \text{i}\\ {\text{a}}_K=\left(L_S\cos \theta -L_S{\alpha }_{KA}\sin \theta \right)\text{i}+\left(-L_S\sin \theta -L_S{\alpha }_{KA}\cos \theta \right)\text{j}\end{array}$

Consider HK.

Show the relation between the acceleration of H and K as follows:

$\begin{array}{l}{\text{a}}_H={\text{a}}_K-{\omega }_{HK}^2{\text{r}}_{H/K}+\text{α}\times {\text{r}}_{H/K}\\ {\text{a}}_H={\text{a}}_K-{\omega }_{HK}^2{\text{r}}_{H/K}+\left({\alpha }_{HK}\text{k}\right)\times {\text{r}}_{H/K}\end{array}$

Substitute $-L_T\cos \varphi \text{i}-L_T\sin \varphi \text{j}$ for ${\text{r}}_{H/K}$, $-0.22114\text{rad/s}$ for ${\omega }_{HK}$, and $\left(L_S\cos \theta -L_S{\alpha }_{KA}\sin \theta \right)\text{i}+\left(-L_S\sin \theta -L_S{\alpha }_{KA}\cos \theta \right)\text{j}$ for ${\text{a}}_K$.

$\begin{array}{c}{\text{a}}_H=\left[\begin{array}{l}\left(L_S\cos \theta -L_S{\alpha }_{KA}\sin \theta \right)\text{i}+\left(-L_S\sin \theta -L_S{\alpha }_{KA}\cos \theta \right)\text{j}\\ -{\left(-0.22114\right)}^2\left(-L_T\cos \varphi \text{i}-L_T\sin \varphi \text{j}\right)+\left({\alpha }_{HK}\text{k}\right)\times \left(-L_T\cos \varphi \text{i}-L_T\sin \varphi \text{j}\right)\end{array}\right]\\ a_H\text{i}=\left[\begin{array}{l}\left(L_S\cos \theta -L_S{\alpha }_{KA}\sin \theta \right)\text{i}+\left(-L_S\sin \theta -L_S{\alpha }_{KA}\cos \theta \right)\text{j}\\ -\left(-0.0489L_T\cos \varphi \text{i}-0.0489L_T\sin \varphi \text{j}\right)+\left(-L_T{\alpha }_{HK}\cos \varphi \text{j}+L_T{\alpha }_{HK}\sin \varphi \text{i}\right)\end{array}\right]\\ a_H\text{i}=\left[\begin{array}{l}\left(L_S\cos \theta -L_S{\alpha }_{KA}\sin \theta +0.0489L_T\cos \varphi +L_T{\alpha }_{HK}\sin \varphi \right)\text{i}\\ +\left(-L_S\sin \theta -L_S{\alpha }_{KA}\cos \theta +0.0489L_T\sin \varphi -L_T{\alpha }_{HK}\cos \varphi \right)\text{j}\end{array}\right]\end{array}$ (7)

Equate j component of the Equation (7).

$\begin{array}{l}0=\left(-L_S\sin \theta -L_S{\alpha }_{KA}\cos \theta +0.0489L_T\sin \varphi -L_T{\alpha }_{HK}\cos \varphi \right)\\ L_T{\alpha }_{HK}\cos \varphi =-L_S\sin \theta -L_S{\alpha }_{KA}\cos \theta +0.0489L_T\sin \varphi \\ {\alpha }_{HK}=\frac{-L_S\sin \theta -L_S{\alpha }_{KA}\cos \theta +0.0489L_T\sin \varphi }{L_T\cos \varphi }\end{array}$

Substitute $442\text{mm}$ for $L_T$, $375\text{mm}$ for $L_S$, $6.7853°$ for $\varphi$, $1.5{\text{rad/s}}^2$ for ${\alpha }_{AK}$, and $75°$ for $\theta$.

$\begin{array}{l}{\alpha }_{HK}=\frac{-375\sin \left(75°\right)-375\times 1.5\cos \left(75°\right)+0.0489\times 442\sin \left(6.7853°\right)}{442\cos \left(6.7853°\right)}\\ {\alpha }_{HK}=-1.1512{\text{rad/s}}^2\end{array}$

Equate i component of the Equation (7).

$a_H=\left(L_S\cos \theta -L_S{\alpha }_{KA}\sin \theta +0.0489L_T\cos \varphi +L_T{\alpha }_{HK}\sin \varphi \right)$

Substitute $442\text{mm}$ for $L_T$, $375\text{mm}$ for $L_S$, $6.7853°$ for $\varphi$, $1.5{\text{rad/s}}^2$ for ${\alpha }_{AK}$, and $75°$ for $\theta$, and $-1.1512{\text{rad/s}}^2$ for ${\alpha }_{HK}$.

$\begin{array}{c}{a}_H=\left(\begin{array}{l}375\cos \left(75°\right)-375\times 1.5\times \sin \left(75°\right)+\\ 0.0489\times 442\cos \left(6.7853°\right)+442\times \left(-1.1512\right)\sin \left(6.7853°\right)\end{array}\right)\\ =-0.48492{\text{m/s}}^2\\ =\left(-0.485{\text{m/s}}^2\right)\text{i}\end{array}$

Thus, the acceleration at H is $\underset{\_}{\left(-0.485{\text{m/s}}^2\right)\text{i}}$.