# Mechanical Linkage Question - Velocity Diagram

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?

• Welcome to Engineering! This looks like a homework question. Do you know if your approach is incorrect? Do you know the correct answer? Please edit your question showing us your work, your answer and what it should be. – Wasabi Apr 17 '16 at 15:04
• – John Alexiou Apr 17 '16 at 17:26
• @ja72 - I think your proposition is overkill. – Karlo Apr 18 '16 at 11:15
• It is a quick tool for checking static and kinematic problems. – John Alexiou Apr 18 '16 at 12:36

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}

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.