Example 2.7: Bar with two sliders

In the mechanism of the figure, slider \(A\) has an instantaneous velocity and acceleration of \(10~m/s\) and \(5~m/s^2\) respectively, both pointing downwards. If the distance \(AB\) is \(15~m\), find:

1. The linear velocity of \(B\) and the angular velocity of bar \(AB\).

2. The linear acceleration of \(B\) and the angular acceleration of bar \(AB\).

Solution

Here we only need to study one rigid body, bar \(AB\) (Body 2), since it already contains both points of interest, \(A\) and \(B\). As there are no sliders moving along rotating bodies, we only need the rigid-body equations (there is no Coriolis acceleration).

From the statement, \(\vec{v}_A=-10\,\hat{j}\) and \(\vec{a}_A=-5\,\hat{j}\), and from the mechanism constraints, \(\vec{v}_B = v_B\,\hat{i}\) and \(\vec{a}_B = a_B\,\hat{i}\). Since the mechanism is planar, \(\vec{\omega}_2 = \omega_2\,\hat{k}\) and \(\vec{\alpha}_2 = \alpha_2\,\hat{k}\). The angle from \(A\) towards \(B\) is \(35^\circ\) clockwise (negative), so:

\[ \vec{r}_{AB} = 15 \cos(-35^\circ)\,\hat{i} + 15 \sin(-35^\circ)\,\hat{j} = 15 \cos 35^\circ\,\hat{i} - 15 \sin 35^\circ\,\hat{j} \]

1. Velocities

We write the velocity equation for points \(A\) and \(B\) of rigid body 2:

\[ \vec{v}_B = \vec{v}_A + \vec{\omega}_2 \times \vec{r}_{AB} \quad\rightarrow\quad v_B\,\hat{i} = -10\,\hat{j} + (\omega_2\,\hat{k}) \times \left(15 \cos 35^\circ\,\hat{i} - 15 \sin 35^\circ\,\hat{j}\right) \]

Expanding the cross products (\(\hat{k}\times\hat{i}=\hat{j}\), \(\hat{k}\times\hat{j}=-\hat{i}\)):

\[ v_B\,\hat{i} + 0\,\hat{j} = \left(15\,\omega_2 \sin 35^\circ\right)\hat{i} + \left(-10 + 15\,\omega_2 \cos 35^\circ\right)\hat{j} \]

Equating components gives a system of two scalar equations with two unknowns:

\[\begin{split} \left\{ \begin{aligned} v_B &= 15\,\omega_2 \sin 35^\circ \\ 0 &= -10 + 15\,\omega_2 \cos 35^\circ \end{aligned} \right. \quad\rightarrow\quad \omega_2 = 0.8138~\text{rad/s}~(\circlearrowleft), \qquad v_B = 7.002~\text{m/s}~(\rightarrow) \end{split}\]

2. Accelerations

With all velocities known, we write the rigid-body acceleration equation. Since \(\vec{\omega}_2 \perp \vec{r}_{AB}\), the normal term simplifies to \(\vec{\omega}_2 \times (\vec{\omega}_2 \times \vec{r}_{AB}) = -\omega_2^2\,\vec{r}_{AB}\):

\[ \vec{a}_B = \vec{a}_A + \vec{\alpha}_2 \times \vec{r}_{AB} - \omega_2^2\,\vec{r}_{AB} \]

\[ a_B\,\hat{i} = -5\,\hat{j} + (\alpha_2\,\hat{k}) \times \left(15 \cos 35^\circ\,\hat{i} - 15 \sin 35^\circ\,\hat{j}\right) - \omega_2^2 \left(15 \cos 35^\circ\,\hat{i} - 15 \sin 35^\circ\,\hat{j}\right) \]

\[ a_B\,\hat{i} + 0\,\hat{j} = \left(15\,\alpha_2 \sin 35^\circ - 15\,\omega_2^2 \cos 35^\circ\right)\hat{i} + \left(-5 + 15\,\alpha_2 \cos 35^\circ + 15\,\omega_2^2 \sin 35^\circ\right)\hat{j} \]

Equating components and solving for the two acceleration unknowns:

\[ \alpha_2 = -0.05686~\text{rad/s}^2~(\circlearrowright), \qquad a_B = -8.628~\text{m/s}^2~(\leftarrow) \]