2D - Kinematics - Linkage System using Vector Algebra

Question

I have this question that I dont know how to solve correctly :

My question is, how do I find $V_B$ ? I will find the angular velocities myself, but I want to know the method to get $V_B$ ?

I know I can start by using $V_C=V_B+V_{C/B}$, but then I'm not sure what to do next...is the direction of $V_B$ and $V_{C/B}$ the same ? I'm using vector algebra (cross product with i and j etc).

More specifically :

$$\vec{V}_C=\vec{V}_B+\vec{V}_{C/B}$$

$$\vec{V}_C = -1 \hat{j}$$

$$\vec{V}_B= ?$$

$$\vec{V}_{C/B}= \omega_{CB} \hat{k} \times \vec{r}_{CB}$$

Shouldn't $\vec{V}_B = \omega_{CB} \hat{k} \times \vec{r}_CB$ ??? If not, then what should it be ?

If you need more info please let me know.

Answer 1

The first thing I would do is get the current angles.

Link BC is the hypotenuse of a right triangle with base 100 and height 100 (120-20). So it is at 45 degrees, so its instantaneous velocity and acceleration along its length is sqrt(1/2) times Vc and Ac. Its instantaneous tangential velocity is the same, because it's 45 degrees.

Now you need to know the angle ABC. AB is the hypotenuse of a right triangle with base 200 and height 120, so the tangent of its angle above the horizontal is arctan(120/200) = 30.964 degrees.

So angle ABC is 135-30.964 = 104.036 degrees. So the velocity and acceleration of B perpendicular to AB (i.e. tangential) is the lengthwise velocity and acceleration of BC times cos(104.036 - 90) = cos(14.036) = .9701.

The radial velocity and acceleration of vector AB is the tangential velocity and acceleration of B divided by the length of AB, which you know because it's the hypotenuse of a right triangle.

To get the radial velocity and acceleration of BC, just take its tangential velocity and acceleration and divide by its length.

Just push the numbers through. Then you can just bring in the vector math at each point instead of working with degrees and trig.

Answer 2

What do you know. You know the positions of the pins

$$ \vec{r}_A = (0,0,0)\,{\rm m} $$ $$ \vec{r}_B = (0.200,0.120,0)\,{\rm m} $$ $$ \vec{r}_C = (0.300,0.020,0)\,{\rm m} $$

You know the linear velocities of $A$ and $C$

$$ \vec{v}_A = (0,0,0)\,{\rm m\cdot s^{-1}} $$ $$ \vec{v}_C = (0,-1,0)\,{\rm m\cdot s^{-1}} $$

and you have two unknown rotational accelerations for links 1 and 2.

$$ \vec{\omega}_1 = (0,0,\omega_1) \,{\rm rad\cdot s^{-1}}$$ $$ \vec{\omega}_2 = (0,0,\omega_2) \,{\rm rad\cdot s^{-1}}$$

and the kinematic relationships

$$ \vec{v}_B = \vec{v}_A + \vec{\omega}_1 \times (\vec{r}_B-\vec{r}_A ) $$ $$ \vec{v}_C = \vec{v}_B + \vec{\omega}_2 \times (\vec{r}_C-\vec{r}_B ) $$

where $\times$ is the cross product operator. The above are solved for the two unknown rotational velocities, and two linear velocities $\vec{v}_B=((\vec{v}_B)_x,(\vec{v}_B)_y,0))$. Expand as:

$$ ((\vec{v}_B)_x,(\vec{v}_B)_y,0) = (-0.120 \omega_1, 0.200 \omega_1,0) $$ $$ (0,-1,0) = (0.100 \omega_2, 0.100 \omega_2,0) + ((\vec{v}_B)_x,(\vec{v}_B)_y,0)$$

and solve for the 4 unknowns. I get the following system

$$ \begin{array}{cc} (\vec{v}_B)_x = -0.120\,\omega_1 & 0 = 0.100\,\omega_2 + (\vec{v}_B)_x \\ (\vec{v}_B)_y = 0.200\,\omega_1 & -1 = 0.100\,\omega_2 + (\vec{v}_B)_y \end{array} $$

The solution here is $(\vec{v}_B)_x=0.375\,{\rm m\cdot s^{-1}}$, $(\vec{v}_B)_y=\text{-}0.625\,{\rm m\cdot s^{-1}}$, $\omega_1=\text{-}3.125\,{\rm rad\cdot s^{-1}}$, $\omega_2=\text{-}3.75\,{\rm rad\cdot s^{-1}}$.

---

Similarly for accelerations you use these vector equations

$$ \vec{a}_B = \vec{a}_A + \vec{\omega}_1 \times (\vec{v}_B-\vec{v}_A )+ \vec{\alpha}_1 \times (\vec{r}_B-\vec{r}_A ) $$ $$ \vec{a}_C = \vec{a}_B + \vec{\omega}_2 \times (\vec{v}_C-\vec{v}_B ) + \vec{\alpha}_2 \times (\vec{r}_C-\vec{r}_B ) $$

with $\vec{a}_B = (0,-3,0)\,{\rm m\cdot s^{-2}}$.

to get the rotational accelerations $\alpha_1$ and $\alpha_2$ and linear acceleration $\vec{a}_B$ of pin $B$.