Mechanical Linkage Question - Velocity Diagram

Question

Can someone confirm the velocity of B, given point C has velocity of 3 units/s down and point A has a velocity of 5 units/s upwards?

My approach has been to resolve for the tangent velocity component of B with respect to A, and tangent component of velocity B with respect to C, then found the resultant B using the two aforementioned tangents?

Any thoughts?

Answer 1

This is how to approach this problem:

The rule is that the relative rotation center between two bodies (point B) lies on the line connecting the instant centers of rotation of each body (points D and E).

Point D is located to the left of A because a positive rotation makes A go upwards with $$\omega_{AB} = \frac{v_A}{a}$$

Similarly point E is located to the right of C causing a positive rotation for the point to move down and $$\omega_{CB} = \frac{v_C}{c}$$

I am going to color code the motion in order to make it more clear

Since B belongs to both AB and CB it means that

$$ v_B = \omega_{AB} (g+h) = \omega_{CB} h $$

Those are four equations to be used with the geometry of a, c, g and h to be used for the solution. I had to consider the following angles to solve this

I got $\theta_C = 0.403696$ and $\theta_A = 0.11990$. By projecting the various lengths along the x and y axes I get the following relationships

$$\begin{aligned} g \cos(\theta_B) + (200-a) & = c \\ g \sin(\theta_B) & = 150 \\ (g+h) \cos(\theta_B) + (200-a) &= 250 \cos(\theta_C) \\ (g+h) \sin(\theta_B) & = 250 \cos(\theta_A) \end{aligned} $$

From the kinematics (first set of equations) we substitute $a=\frac{5}{\omega_{AB}}$, $c=\frac{3}{\omega_{CB}}$, $h=\frac{v_B}{\omega_{CB}}$ and $g=v_B \left( \frac{1}{\omega_{AB}} - \frac{1}{\omega_{CB}} \right)$ above to get four equations for four unknowns.

My solution is

$$\begin{aligned} \theta_B &= 0.585446 \\ \omega_{AB} & = 0.014515 \\ \omega_{CB} & = 0.036685 \\ v_B &= 6.51982 \end{aligned} $$

Answer 2

Assume that the link AB makes an angle $a[t]$ with the horizontal and the link CB makes an angle $c[t]$ with the horizontal.

Equating the position of B (x and y coordinates) computed w.r.t both A and C, we get two equations.

$$ 250 \cos (a(t))+200=250 \cos (c(t)) $$ $$ 250 \sin (c(t))+150=250 \sin (a(t)) $$

These equations can be solved for $a[t]$ and $c[t]$.

$$ a(t)=\tan ^{-1}\left(\frac{3+4 \sqrt{3}}{3 \sqrt{3}-4}\right),\ c(t)=\tan ^{-1}\left(\frac{4 \sqrt{3}-3}{4+3 \sqrt{3}}\right)$$

The velocity of B (x and y coordinates) can also be computed in two ways. So we get two more equations.

$$ -250 c'(t) \sin (c(t))=-250 a'(t) \sin (a(t))$$ $$ 250 c'(t) \cos (c(t))+\text{vc}=250 a'(t) \cos (a(t))+\text{va}$$

These equations can be solved for $a'[t]$ and $c'[t]$.

$$ a'(t)=-\frac{\left(\sqrt{3}-4\right) (\text{va}-\text{vc})}{1250},\ c'(t)=\frac{\left(4+\sqrt{3}\right) (\text{va}-\text{vc})}{1250}$$

The velocity of B can now be computed as $$ \left\{-250 c'(t) \sin (c(t)),250 c'(t) \cos (c(t))+\text{vc}\right\}$$

or

$$ \left\{-250 a'(t) \sin (a(t)),250 a'(t) \cos (a(t))+\text{va}\right\} $$

Substituting the values of $a[t]$, $c[t]$, $a'[t]$, $c'[t]$ and $va=5$, $vc=-3$ gives the same result in both cases.

$$ \left\{-3.60, 5.43\right\} $$

The point B is moving upwards and to the left.