Question 15.SP.15: Crank AB of the engine system of Sample Prob. 15.7 has a con...

MODELING and ANALYSIS:
Motion of Crank AB. Since the crank rotates about A with constant  \omega_{A B} = 2000 rpm = 209.4 rad/s, you have  \alpha_{A B} = 0. The acceleration of B is therefore directed toward A (Fig. 1) and has the magnitude of

\begin{aligned}&a_B=r \omega_{A B}^2=\left(\frac{3}{12}  \mathrm{ft}\right)(209.4  \mathrm{rad} / \mathrm{s})^2=10,962  \mathrm{ft} / \mathrm{s}^2 \\&\mathbf{a}_B=10,962  \mathrm{ft} / \mathrm{s}^2 \text{⦫} 40^{\circ}\end{aligned}

Motion of Connecting Rod BD. The angular velocity  \omega_{B D}  and the value of β were obtained in Sample Prob. 15.7 using relative velocity equations:

\boldsymbol{\omega}_{B D}=62.0  \mathrm{rad} / \mathrm{s} \circlearrowleft               \quad \beta=13.95^{\circ}

Resolve the motion of BD into a translation with B and a rotation about B (Fig. 2). Resolve the relative acceleration  \alpha_{D / B}  into normal and tangential components:

\begin{aligned}\left(a_{D / B}\right)_n=(B D) \omega_{B D}^2=\left(\frac{8}{12}  \mathrm{ft}\right)(62.0  \mathrm{rad} / \mathrm{s})^2 &=2563  \mathrm{ft} / \mathrm{s}^2 \\\left(\mathrm{a}_{D / B}\right)_n &=2563  \mathrm{ft} / \mathrm{s}^2 \text{⦩} 13.95^{\circ} \\\left(a_{D / B}\right)_t=(B D) \alpha_{B D} &=\left(\frac{8}{12}\right) \alpha_{B D}=0.6667 \alpha_{B D} \\\left(\mathrm{a}_{D / B}\right)_t &=0.6667 \alpha_{B D} \text{⦨} 76.05^{\circ}\end{aligned}

Although  \left(\mathbf{a}_{D / B}\right)_t  must be perpendicular to BD, its sense is not known.

Noting that the acceleration  a_D  must be horizontal, you have

\begin{aligned}\mathbf{a}_D &=\mathbf{a}_B+\mathbf{a}_{D / B}=\mathbf{a}_B+\left(\mathbf{a}_{D / B}\right)_n+\left(\mathbf{a}_{D / B}\right)_t \\{\left[a_D \leftrightarrow\right] } &=\left[10,962 ⦫  40^{\circ}\right]+\left[2563  ⦩  13.95^{\circ}\right]+\left[0.6667 \alpha_{B D} \text{ ⦨ } 76.05^{\circ}\right]\end{aligned}              (1)

Equating x and y components, you obtain the following scalar equations, as

-a_D=-10,962 \cos 40^{\circ}-2563 \cos 13.95^{\circ}+0.6667 \alpha_{B D} \sin 13.95^{\circ}

0=-10,962 \sin 40^{\circ}+2563 \sin 13.95^{\circ}+0.6667 \alpha_{B D} \cos 13.95^{\circ}

Solving the equations simultaneously gives  \alpha_{B D}=+9940  \mathrm{rad} / \mathrm{s}^2  and  a_D=+9290  \mathrm{ft} / \mathrm{s}^2.  The positive signs indicate that the senses shown on the vector polygon (Fig. 3) are correct.

\begin{aligned}\alpha_{B D} &=9940  \mathrm{rad} / \mathrm{s}^2 \circlearrowleft \\\mathbf{a}_D &=9290  \mathrm{ft} / \mathrm{s}^2 \leftarrow\end{aligned}

REFLECT and THINK: In this solution, you looked at the magnitude and direction of each term in Eq. (1) and then found the x and y components. Alternatively, you could have assumed that  a_D  was to the left,  \alpha_{B D}  was positive, and then substituted in the vector quantities to get

\begin{aligned}\mathbf{a}_D=& \mathbf{a}_B  +  \alpha \mathbf{k} \times \mathbf{r}_{D / B}-\omega^2 \mathbf{r}_{D / B} \\-a_D \mathbf{i}=&-a_B \cos 40^{\circ} \mathbf{i}  –  a_B \sin 40^{\circ} \mathbf{j}  +  \alpha_{B D} \mathbf{k} \times(l \cos \beta \mathbf{i}  –  l \sin \beta \mathbf{j}) \\& \quad-\omega_{B D}^2(l \cos \beta \mathbf{i}-l \sin \beta \mathbf{j}) \\=&-a_B \cos 40^{\circ} \mathbf{i}  –  a_B \sin 40^{\circ} \mathbf{j}  +  \alpha_{B D} l \cos \beta \mathbf{j}  +  \alpha_{B D} l \sin \beta \mathbf{i} \\&-\omega_{B D}^2 l \cos \beta \mathbf{i}  +  \omega_{B D}^2 l \sin \beta \mathbf{j}\end{aligned}

Equating components gives

These are identical to the previous equations if you substitute in the numbers.