Question 6 (paper Question VI) — Angular Velocities of a Slider-Linkage Mechanism (Part B · Dynamics, equal value): 04-BS-3, May 2018, Q6

Question 6 (paper Question VI) — Angular Velocities of a Slider-Linkage Mechanism (Part B · Dynamics, equal value)

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. Link AB is pinned to a fixed wall at A, horizontal, length 2 m, to B. Link BC is pinned at B, vertical at this instant, length 2 m, to slider C. Slider C is constrained to move along a fixed groove inclined 45° from the vertical, with speed $v_C=10$ m/s down the groove.

Given data
Quantity	Value
Length AB	2 m (horizontal)
Length BC	2 m (vertical, at this instant)
Groove angle from vertical	45°
Slider speed, $v_C$	10 m/s (down the groove)

Find. The angular velocities $\omega_{AB}$ and $\omega_{BC}$.

Figure 6 (reconstructed). Fixed pin A, link AB to B, link BC to slider C on the 45° inclined groove.

Approach. Place A at the origin with x horizontal, y vertical. Write the rigid-body velocity relation $\mathbf{v}_C=\mathbf{v}_B+\boldsymbol\omega_{BC}\times\mathbf{r}_{C/B}$, with $\mathbf{v}_B=\boldsymbol\omega_{AB}\times\mathbf{r}_{B/A}$, and match components against the known direction of $\mathbf{v}_C$ along the groove.

Velocity of B. With $\mathbf{r}_{B/A}=(2,0,0)$ m and $\boldsymbol\omega_{AB}=\omega_{AB}\hat{k}$: $\mathbf{v}_B=\omega_{AB}\hat{k}\times(2,0,0)=(0,2\omega_{AB},0)$.
Velocity of C in terms of both unknowns. With $\mathbf{r}_{C/B}=(0,-2,0)$ m and $\boldsymbol\omega_{BC}=\omega_{BC}\hat{k}$: $\mathbf{v}_C=\mathbf{v}_B+\omega_{BC}\hat{k}\times(0,-2,0)=(2\omega_{BC},\,2\omega_{AB},\,0)$.
Known velocity of C. Down the groove (45° from vertical): $\mathbf{v}_C=10(\sin45^\circ,-\cos45^\circ)=(7.071,-7.071)$ m/s.
Match components and solve. $2\omega_{BC}=7.071\Rightarrow\boxed{\omega_{BC}=3.54\text{ rad/s (counter-clockwise)}}$. $2\omega_{AB}=-7.071\Rightarrow\boxed{\omega_{AB}=3.54\text{ rad/s (clockwise)}}$.

Final results
Quantity	Value
$\omega_{AB}$	3.54 rad/s, clockwise
$\omega_{BC}$	3.54 rad/s, counter-clockwise

Topic: Rigid-body planar kinematics — relative-velocity equation for a linkage with a sliding constraint.
Key relations: $\mathbf{v}_Q=\mathbf{v}_P+\boldsymbol\omega\times\mathbf{r}_{Q/P}$ applied link-by-link; a slider's velocity is constrained to lie along its fixed guide.
Why this works: writing $\mathbf{v}_C$ two ways — built up through both links from the fixed pin, and independently as a known-direction slide along the groove — gives two scalar equations in the two unknown angular velocities, solved directly without needing an instantaneous-centre construction.
Common pitfall: forgetting that the slider's velocity is fixed in DIRECTION (along the groove) but not decomposed arbitrarily — the two components of $\mathbf{v}_C$ must be resolved consistently from the single known groove angle before matching them to the linkage's velocity expression.
Source: Hibbeler, Engineering Mechanics: Dynamics, Ch. 16 (Relative-motion analysis: velocity; instantaneous centre of zero velocity).

Question 6 of 6: Question 6 (paper Question VI) — Angular Velocities of a Slider-Linkage Mechanism (Part B · Dynamics, equal value)

Question 6 (paper Question VI) — Angular Velocities of a Slider-Linkage Mechanism (Part B · Dynamics, equal value)