The system shown in the Figure is guided by two blocks in A and B, that move in the fixed grooves. If the speed of A is 2m/s down, determine the speed of B at time θ = 45º

enter image description here

In the textbook they said w = 14.1rad/s (counterclockwise) and v(B)=2m/s

i consider counterclockwise moment positive

if i use the vector BA

$$v_B=v_A+w_{BA} x \vec{r}_{BA} = -2\vec{j} + w_{BA}\vec{k} \times (-0.2cos(45)\vec{i}+0.2sin(45)\vec{j})$$

$$= -0.141w_{BA}\vec{i}+ (-0.1412w_{BA}-2)\vec{j}$$

and since $-0.1412w_{BA}-2=0\Leftrightarrow w_{BA}=-14.1rad/s$ (since is negative, the direction is clockwise) and $v(B)=2i$.

But if i use the vector AB:

$$v_B = v_A+w_{BA}\times r_{AB} $$

$$= -2\vec{j} + w_{BA}\vec{k} \times (+0.2cos(45)\vec{i}-0.2sin(45)\vec{j}) $$

$$= 0.141w(BA)\vec{i}+(-2+0.141w(BA))\vec{j}$$

and since $$-2+0.1412w_{BA}=0 <=> w_{BA}=14.1rad/s$$ (since is positive, the direction is counterclockwise) and $v(B)=2i$.

So I got two different results for w(BA) when using vector AB or BA. Which vector should I use when I apply the equation v(B)=v(A)+v(B/A) and why?

Similar case

in this case i use the vectors AB and BD, and i got the opposite signals for the angular speed of the connecting rod BD and for the piston speed P (although the value in module is correct) Should i had used these vectors?

enter image description here

we know from the exercice that w(AB)=209.44rad/s, β=13.95 and we want to discover v(P) and w(BD)


v(B)=v(A)+w(AB) * r(AB)

=0+ w(AB)k * (0.075cos(40)i+0.075sen(40)j)




=v(B)+w(BD)k * (0.2cos(13.95)i-0.2sen(13.95)j)

=(-10.1i+12.04j) + 0.194w(BD)j+0.0482w(BD)i

=(-10.1i+0.0482w(BD))i + (12.04+0.194w(BD))j

12.04+0.194w(BD)=0 <=> w(BD)=-62.06rad/s


enter image description here

  • $\begingroup$ This is an interesting question. I started by working out the height of A as $ h_A = 0.2\ cos\Theta $. Then I was going to differentiate both sides so I could set $ \frac {dh}{dt} = 2 $ (m/s). But now I'm stuck. What do we do with $ 0.2 \frac {d\ cos\Theta}{dt} $? $\endgroup$
    – Transistor
    Feb 13 '21 at 12:04
  • $\begingroup$ @Transistor Perhaps use the chain rule to write it in terms of the rate of change of $\theta$? Then do the same thing (i.e. differentiate with respect to time and use the chain rule) for the equivalent trigonometric equation for the position of pin joint $\mathsf{B}$. You should find yourself with two linear simultaneous equations in two unknowns, one of those unknowns being the rate of change of $\theta$, the other being the velocity of pin joint $\mathsf{B}$. $\endgroup$ Feb 13 '21 at 12:40
  • $\begingroup$ @sputnik I strongly suggest you write out in words your definition of $w_{\mathsf{B}\mathsf{A}}$. You may find this makes it very clear that one of your two alternative starting equations is true and the other isn't. $\endgroup$ Feb 13 '21 at 12:53
  • $\begingroup$ @Transistor I wrote the following steps as a comment, but it didn't render very nicely, so I added another answer. $\endgroup$
    – NMech
    Feb 13 '21 at 13:06
  • 2
    $\begingroup$ I’m voting to close this question because OP has removed the question by editing it & has closed their account. $\endgroup$
    – Fred
    May 20 '21 at 10:51

I think the main thing that is confusing you is the following, is that you think that there is an option for the positive rotation. While in some cases this is true, since you deciced on employing the cross product $\omega\times r$ for the calculation, (and since you already have a x-y coordinate system, and defined $\vec{i}, \vec{j}, \vec{k}$) the positive rotation is determined by the right hand rule. So the positive direction for rotation is counterclockwise.

The next point is that correct formula is: $$v_B = v_A + \omega\times r_{B|A}$$


  • $v_B = 2\vec{i} = \begin{bmatrix}2\\0\\0\end{bmatrix}$, (note this only happens because the angle is 45 degrees and there is symmetry in the problem, and in the general case this is an unknown)
  • $v_A = -2\vec{j} = \begin{bmatrix}0\\-2\\0\end{bmatrix}$
  • $\omega = \omega\vec{k} = \begin{bmatrix}0\\0\\ \omega\end{bmatrix}$: the angular velocity is the same irrespective of the point. You can see in this problem that whichever point you select (A or B), the body rotates CCW with respect to that point.
  • $r_{B|A} = 0.2(\frac{\sqrt{2}}{2}\vec{i} -\frac{\sqrt{2}}{2}\vec{j})= 0.2 \frac{\sqrt{2}}{2}\begin{bmatrix}1\\-1\\0\end{bmatrix}$: is the vector connecting to B starting from A

if you use the column notation you get:

$$\begin{bmatrix}2\\0\\0\end{bmatrix} = \begin{bmatrix}0\\-2\\0\end{bmatrix} + \begin{bmatrix}0\\0\\ \omega\end{bmatrix}\times \left( 0.2 \frac{\sqrt{2}}{2}\begin{bmatrix}1\\-1\\0\end{bmatrix}\right) $$

$$\begin{bmatrix}2\\0\\0\end{bmatrix} = \begin{bmatrix}0\\-2\\0\end{bmatrix} + 0.2 \omega\frac{\sqrt{2}}{2} \left(\begin{bmatrix}0\\0\\1 \end{bmatrix}\times \begin{bmatrix}1\\-1\\0\end{bmatrix}\right) $$

$$\begin{bmatrix}2\\0\\0\end{bmatrix} = \begin{bmatrix}0\\-2\\0\end{bmatrix} + 0.2 \omega\frac{\sqrt{2}}{2} \begin{bmatrix}1\\1\\0\end{bmatrix} $$

Both equations for x and y are identical :

$$2= 0.2 \omega\frac{\sqrt{2}}{2} \rightarrow$$ $$ \omega = \frac{4}{0.2\sqrt{2}}=\frac{20}{\sqrt{2}}=14.1[rad/s]$$

  • $\begingroup$ you'll get symmetrical angular velocity because $r_{B|A} = - r_{A|B}$ therefore, $\omega\times r_{B|A} = -\omega\times r_{A|B}$. Hope that makes sense. $\endgroup$
    – NMech
    Feb 13 '21 at 15:30
  • $\begingroup$ Additionally you should note that $r_{AB} = r_{B|A}$. I read the latter, position vector of B with respect to A. Similarly $r_{BA} = r_{A|B}$. That also can introduce confusion. $\endgroup$
    – NMech
    Feb 13 '21 at 15:34
  • $\begingroup$ in your latest example you can use either $$v_D = v_B+\omega_{BD}\times r_{D|B}=v_B+\omega_{BD}\times r_{BD}$$ OR $$v_B = v_D+\omega_{BD}\times r_{DB}=v_D+\omega_{BD}\times r_{B|D}$$ $\endgroup$
    – NMech
    Feb 13 '21 at 15:53
  • $\begingroup$ From what I see in your results you calculate $\omega_{DB}=-62$ which makes sense (BD will move clockwise). Did you not calculate negative value? $\endgroup$
    – NMech
    Feb 13 '21 at 15:58
  • $\begingroup$ So they have +62? $\endgroup$
    – NMech
    Feb 13 '21 at 16:01

@Transistor although this is not a response (because the problem was IMHO the positive direction), the following is rendered much better here, and its a different approach to the same problem.

$$h_A = 0.2\ cos\Theta \rightarrow $$ $$\frac{d h_A}{dt} = 0.2\frac{d \cos\theta}{dt}$$ then $$v_A = -0.2\sin\theta \frac{d \theta}{dt}$$, where

  • $\frac{d \theta}{dt}=\omega$.

Therefore: $$v_A =- 0.2\sin\theta \omega$$.

$$v_B =0.2\cos\theta \omega$$.

Since $\theta= 45^o$, you can use $v_A=2[m/s]$ and $\theta=\pi/4$ to calculate $\omega$ and then afterwards to calculate $v_B$.


doing it the hard way:

Let A and B be distance from the corner

A = r*cosϑ                 B = r*sinϑ
dA/dt = (-r*sinϑ)(dϑ/dt)   dB/dt = (r*cosϑ)(dϑ/dt)

          (dB/dt) / (dA/dt) = -cotϑ 


dB/dt = (dA/dt)(-cotϑ)

The scale dimension (r=0.2m) and the angluar velocity (dϑ/dt) both cancel, and the relationship of velocities is determined purely by the angle.

The negative sign indicates that when A is increasing, B is decreasing and vice versa.

per comment: for the second problem you can still use trigonometry to solve it. Brief attempt below, can't promise that it is correct. index-card

the negative sign indicates that when alpha is increasing, X (i.e. AD) is reducing, i.e. D moves to the left.

  • $\begingroup$ @sputnik - I am a little unclear about what you are asking, but it sounds like just a matter of sign convention. if we define AB = B-A then B = A + AB = A - BA $\endgroup$
    – Pete W
    Feb 13 '21 at 16:06
  • $\begingroup$ @sputnik - i don't know, looks complicated. Try to be really explicit about how you define the sign and direction of each variable. Keep trying until you get it right, it is really important. $\endgroup$
    – Pete W
    Feb 13 '21 at 16:40

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