Relative-Velocity Analysis of a Rigid Linkage (Part B — Dynamics): 04-BS-3, May 2013, Q4

Question IV: Relative-Velocity Analysis of a Rigid Linkage (Part B — Dynamics) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Straight rigid rod ADB (A and B each 0.6 m from D, rod inclined 45°); separate rigid rod CD (0.6 m, inclined 30°), pinned to rod ADB at D. Slider A moves only horizontally, $v_A=2.4$ m/s to the right. Slider B moves only vertically. Slider C moves only vertically.

Figure 4 — rod ADB is straight through D at 45°; rod CD is pinned to it at D and inclined 30°.

Find. The velocities of sliders B and C at the instant shown.

Approach. Rod ADB is one rigid body: use $\mathbf v_B=\mathbf v_A+\boldsymbol\omega_1\times\mathbf r_{B/A}$, with $v_A$ known (horizontal) and $v_B$'s direction known (vertical), to solve for $\omega_1$ and hence $v_B$ and $v_D$. Then treat rod CD as a second rigid body pinned at D (velocity now known) and C (direction known, vertical) to solve for $\omega_2$ and $v_C$.

Set up coordinates and solve rod ADB. With D at the origin, $A=(-0.424,-0.424)$ m, $B=(0.424,0.424)$ m (both 0.6 m from D along the 45° line). $\mathbf r_{B/A}=(0.849,0.849)$ m. Writing $\mathbf v_B=\mathbf v_A+\omega_1\hat k\times\mathbf r_{B/A}=(2.4-0.849\omega_1,\,0.849\omega_1)$ and requiring the x-component to vanish (B moves only vertically), $$\omega_1=\frac{2.4}{0.849}=2.828\text{ rad/s (CCW)}$$
Velocity of B. $$v_B = 0.849(2.828) = \boxed{2.40\text{ m/s (upward)}}$$
Velocity of D (midpoint of the rod). $\mathbf r_{D/A}=(0.424,0.424)$ m, so $$\mathbf v_D=\mathbf v_A+\omega_1\hat k\times \mathbf r_{D/A}=(2.4-1.2,\,1.2)=(1.2,\,1.2)\text{ m/s}$$
Rod CD. With $C=(-0.520,0.300)$ m relative to D, $\mathbf r_{D/C}=(0.520,-0.300)$ m. Writing $\mathbf v_D=\mathbf v_C+\omega_2\hat k\times\mathbf r_{D/C}$ with $\mathbf v_C=(0,v_{Cy})$ and matching both components, $$0.300\,\omega_2 = 1.2 \;\Rightarrow\; \omega_2=4.00\text{ rad/s}, \qquad v_{Cy}=1.2-0.520(4.00)=-0.879\text{ m/s}$$

Quantity	Value
$\omega_1$ (rod ADB)	2.83 rad/s CCW
$v_B$	2.40 m/s upward
$\omega_2$ (rod CD)	4.00 rad/s
$v_C$	0.879 m/s downward

Concept & Theory

Topic. Rigid-body relative-velocity equation applied sequentially through a two-link mechanism.

Key relations. $\mathbf v_B=\mathbf v_A+\boldsymbol\omega\times\mathbf r_{B/A}$; slider constraint: velocity component along the fixed track direction is unconstrained, the perpendicular component is zero.

Why this works. Each rigid link has one scalar unknown ($\omega$) once one point's full velocity and a second point's direction are known; chaining links through their shared pin (D) propagates the solution from the driven input (A) through to the far output (C).

Common pitfall. Treating the bent-looking sketch of "ADB" as two separately-rotating links instead of recognizing A, D, B are collinear points on one straight rigid rod; and forgetting that D, being the midpoint of rod ADB, has its own (generally non-zero, non-axis-aligned) velocity that must be computed before it can drive rod CD.

Source. Hibbeler, Engineering Mechanics: Dynamics (14th ed.), Ch. 16 — Planar Kinematics of a Rigid Body, §16.5 Relative-Motion Analysis: Velocity.

Question 4 of 6: Relative-Velocity Analysis of a Rigid Linkage (Part B — Dynamics)

Question IV: Relative-Velocity Analysis of a Rigid Linkage (Part B — Dynamics) (20 marks)