Question 6 of 6: Question 6 (paper Question VI) — Four-Bar Linkage Velocity and Acceleration Analysis (Part B · Dynamics, equal value)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 04-BS-3, Statics and Dynamics — 2019-Dec. Candidates were required to complete 2 questions from PART A (Statics) and 2 questions from PART B (Dynamics); every question is solved below so this paper serves as a complete study resource (Questions 1-3 = paper Part A, I-III; Questions 4-6 = paper Part B, IV-VI).
Check — source reconstruction notes. (2) Question 1's isometric figure gives explicit 1 m / 1.5 m dimensions along the rod, but plotting them on a single straight line makes the six rigid-body equilibrium equations inconsistent (no solution exists). Reading the printed figure, the horizontal run bends twice: A→C is 2 m along the first horizontal direction (with the vertical 1.5 m branch C→D), C→E is 1 m along the perpendicular horizontal direction (this is where the smooth journal bearing sits), and E→F is a further 1.5 m continuing the first horizontal direction to the point that carries the 250 N cylinder. This reading is the only one of the plausible alternatives that returns a statically-determinate, non-degenerate system, and it lands on the clean values reported below — strong corroborating evidence for the reconstruction. (3) Question 4's dimensions place the pin O exactly 90 mm above the floor, equal to the sphere radius (90 mm R); the sphere therefore contacts the floor exactly when the rod is horizontal, which is used as the impact configuration. (4) Question 6 numbers the four-bar linkage O-A-B-C from the source's own 100 mm / 250 mm / 75 mm / 50 mm dimensions; O and C are the fixed pivots, OA and CB are the driven/driving cranks and AB is the coupler.
Question 6 (paper Question VI) — Four-Bar Linkage Velocity and Acceleration Analysis (Part B · Dynamics, equal value)
Given. Fixed pivots O(0,0) and C(250,50) mm. Link OA vertical, $A=(0,100)$ mm. Crank CB horizontal at this instant, length 75 mm, so $B=(175,50)$ mm. Coupler AB connects A and B, length $\sqrt{175^2+50^2}=182.0$ mm. $\omega_{CB}=2$ rad/s CCW, constant ($\alpha_{CB}=0$).
Find. (a) $\omega_{OA}$ and $\omega_{AB}$; (b) $\alpha_{OA}$ and $\alpha_{AB}$.
Figure 6. Four-bar linkage O-A-B-C; OA and CB are the driven/driving cranks, AB is the coupler.
Approach. Apply the rigid-body relative-velocity equation $\vec v_B=\vec v_A+\vec\omega_{AB}\times\vec r_{AB}$, with $\vec v_A=\vec\omega_{OA}\times\vec r_{OA}$ and $\vec v_B=\vec\omega_{CB}\times\vec r_{CB}$ known from the fixed pivots; solve the resulting 2×2 system for $\omega_{OA},\omega_{AB}$. Repeat with the relative-acceleration equation (including centripetal terms) for the accelerations.
Velocity of B (rotation about fixed C). $\vec r_{CB}=B-C=(-75,0)$ mm:
$$\vec v_B=\omega_{CB}\hat k\times\vec r_{CB}=2(0,-(-75))=(0,-150)\text{ mm/s}$$
Velocity compatibility, $\vec v_B=\vec v_A+\vec\omega_{AB}\times\vec r_{AB}$. With $\vec r_{OA}=(0,100)$, $\vec r_{AB}=B-A=(175,-50)$ mm:
$$(0,-150)=\omega_{OA}(-100,0)+\omega_{AB}(50,175)$$
Solving the two scalar equations simultaneously:
$$\boxed{\omega_{OA}=-0.429\text{ rad/s}}\ (\text{i.e. }0.429\text{ rad/s CW}),\qquad \boxed{\omega_{AB}=-0.857\text{ rad/s}}\ (\text{CW})$$
Acceleration of B (rotation about fixed C, $\alpha_{CB}=0$).
$$\vec a_B=-\omega_{CB}^2\vec r_{CB}=-4(-75,0)=(300,0)\text{ mm/s}^2$$
Acceleration compatibility, $\vec a_B=\vec a_A+\vec\alpha_{AB}\times\vec r_{AB}-\omega_{AB}^2\vec r_{AB}$, with $\vec a_A=\alpha_{OA}(-100,0)-\omega_{OA}^2(0,100)$. Substituting the known $\omega_{OA},\omega_{AB}$ and collecting terms gives two scalar equations in $\alpha_{OA},\alpha_{AB}$:
$$300=-100\alpha_{OA}+50\alpha_{AB}-128.6,\qquad 0=-18.37+175\alpha_{AB}+36.73$$
Solving: $\boxed{\alpha_{AB}=-0.105\text{ rad/s}^2}$, then $\boxed{\alpha_{OA}=-4.34\text{ rad/s}^2}$ (both clockwise, i.e. CB's constant rotation is currently decelerating the swing-back of OA and AB).
