Question 4 of 6: Question 4 (paper Question IV) — Instantaneous Velocity of a Slider-Link Mechanism (Part B · Dynamics, 20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination May 2015 — 04-BS-3, Statics and Dynamics, 3 hours, closed book (one 8½″×11″ self-prepared note sheet permitted; Casio or Sharp approved calculator only). Candidates complete 2 of 3 questions from Part A (Statics) and 2 of 3 from Part B (Dynamics); all 6 are solved below.
Check — the source figure marks the AD segment at 45° and the DC segment at 30° but prints no angle for segment DB; in the drawing DB lies at approximately the same 45° as AD (43–45°), i.e. A, D and B lie on ONE STRAIGHT rigid rod (consistent with the name "member ADB", as opposed to a bent link). Solved on that reading: a straight 4 m rod at 45°, with D at its 2 m midpoint, pin-connected there to the separate 2 m link DC at 30°.
Given. Slider A moves horizontally at vA = 8 m/s. Rigid straight rod ADB (4 m total, D at the 2 m midpoint) is inclined at 45° to the horizontal; B slides in a vertical guide. A separate 2 m link DC, inclined at 30° to the horizontal, is pin-connected to the rod at D; C slides in a second vertical guide.
Given data
Quantity
Value
vA (horizontal)
8 m/s
AD = DB
2 m each (rod ADB straight, 45°)
DC
2 m (30° to horizontal)
Find. The velocity of blocks B and C at the instant shown.
Figure 4 — slider A (horizontal guide), straight rod ADB at 45°, pin-connected link DC at 30° to slider C (vertical guide).
Approach. Rod ADB is a single rigid body: use its instantaneous centre of zero velocity (IC), located where the perpendiculars to vA (vertical, through A) and vB (horizontal, through B) meet, to get its angular velocity and the velocity of D. Link DC is a second rigid body sharing point D (pin connection allows a different angular velocity); repeat the rigid-body velocity equation for DC using the known vD and the constraint that C moves vertically.
Rod ADB — instantaneous centre. With A at the local origin, $B=(2.828,2.828)$ m (4 m at 45°). The IC lies on the vertical line through A and the horizontal line through B: $IC_1=(0,2.828)$.
$$\omega_{ADB} = \dfrac{v_A}{|IC_1A|} = \dfrac{8}{2.828} = \boxed{2.83\ \text{rad/s}}$$
Velocity of B. $|IC_1B| = 2.828$ m (horizontal distance):
$$v_B = \omega_{ADB}\,|IC_1B| = 2.828(2.828) = \boxed{8.00\ \text{m/s}\ \uparrow}$$
Velocity of D. $D=(1.414,1.414)$ m, $\vec{r}_{D/IC_1}=(1.414,-1.414)$:
$$\vec{v}_D = \omega_{ADB}\,\hat{k}\times\vec{r}_{D/IC_1} = (4.00,\ 4.00)\ \text{m/s}$$
Link DC — rigid-body equation. With C at $(-0.318, 2.414)$ m (D minus 2 m at 30° above horizontal, up-left), $\vec{r}_{D/C}=(1.732,-1.00)$ m, and $\vec{v}_C=(0,v_{Cy})$ (vertical guide):
$$\vec{v}_D = \vec{v}_C + \omega_{DC}\,\hat{k}\times\vec{r}_{D/C}$$
$$(4.00,4.00) = (0,v_{Cy}) + \omega_{DC}(1.00,\,1.732)$$
Solve for ωDC and vC. Matching the x-component: $\omega_{DC}=4.00/1.00=\boxed{4.00\ \text{rad/s}}$. Matching the y-component:
$$v_{Cy} = 4.00 - 4.00(1.732) = 4.00-6.928 = \boxed{-2.93\ \text{m/s}\ (\text{i.e. 2.93 m/s}\downarrow)}$$
