# 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 $AB$.

<img src="figs/problemas_cin_an/doble_deslizadera.png" alt="Bar with two sliders" width="450px">

:::{dropdown} 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:

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

**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:

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