Question 6 of 6: Question 6 (paper Question VI): Velocity analysis of a sliding-link mechanism
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3 (Statics and Dynamics) — 2016-May. 3 hours, closed book (one 8.5″×11″ formula sheet permitted). Part A (Statics) offers 3 questions, answer any 2; Part B (Dynamics) offers 3 questions, answer any 2. Every printed question is solved below (all 6), even though the exam instructions require only 4.
Given. Rigid link AB, length 800 mm, at 28° to the horizontal; slider A moves in a horizontal guide with $v_A=4\text{ m/s}$ to the left (constant); slider B moves in a vertical guide. Point "a" lies on the link 600 mm from A (200 mm from B).
Given data
Quantity
Value
Link length, AB
800 mm
Point a from A
600 mm
Link angle to horizontal
28°
$v_A$
4 m/s (horizontal, left)
Find. (A) angular velocity $\omega_{AB}$; (B) velocity of slider B; (C) magnitude, direction and sense of the velocity of point "a".
Fig. 6 – slider A (horizontal guide), slider B (vertical guide), link AB at 28°, with the solved velocity vectors.
Approach. Apply the rigid-body relative-velocity equation $\mathbf{v}_B=\mathbf{v}_A+\boldsymbol{\omega}\times\mathbf{r}_{B/A}$. Slider B's velocity is constrained to be vertical (its guide), which supplies one scalar equation to solve directly for $\omega_{AB}$; the same relation, applied to point "a", then gives its velocity. The result is cross-checked using the instantaneous-centre-of-rotation method.
Position vector A to B. With B up and to the left of A at 28°:
$$\mathbf{r}_{B/A}=0.8(-\cos28^{\circ},\ \sin28^{\circ})=(-0.7065,\ 0.3756)\text{ m}$$
Relative-velocity equation, x-component. $v_{Bx}=v_{Ax}-\omega r_{B/A,y}$. Since B's guide is vertical, $v_{Bx}=0$, and $v_{Ax}=-4\text{ m/s}$:
$$0=-4-\omega(0.3756)\ \Rightarrow\ \boxed{\omega_{AB}=-10.65\text{ rad/s}}$$
(the negative sign indicates clockwise rotation of the link).
Velocity of B (y-component). $v_{By}=v_{Ay}+\omega r_{B/A,x}=0+(-10.65)(-0.7065)$:
$$\boxed{v_B=7.52\text{ m/s, vertically upward}}$$
Velocity of point a. With $\mathbf{r}_{a/A}=0.6(-\cos28^{\circ},\sin28^{\circ})=(-0.5297,\ 0.2817)\text{ m}$:
$$v_{ax}=-4-\omega(0.2817)=-1.00\text{ m/s},\qquad v_{ay}=\omega(-0.5297)=5.64\text{ m/s}$$
$$\boxed{v_a=\sqrt{v_{ax}^2+v_{ay}^2}=5.73\text{ m/s},\ \text{directed }100.1^{\circ}\text{ from the +x axis (up and slightly left)}}$$
Cross-check via instantaneous centre (IC). The IC lies at the intersection of the vertical through A (perpendicular to the horizontal $v_A$) and the horizontal through B (perpendicular to the vertical $v_B$), giving $IC=(0,\,0.3756)$. Then $\omega=v_A/|IC\!-\!A|=4/0.3756=10.65$ rad/s, and $v_B=\omega\,|IC\!-\!B|=10.65(0.7065)=7.52$ m/s, and $v_a=\omega\,|IC\!-\!a|=10.65(0.5380)=5.73$ m/s – all three match Steps 2–4 exactly.
