Question 4 of 6: Four-bar mechanism — shaking-force balancing to 50%

Question 4: Four-bar mechanism — shaking-force balancing to 50% (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. Grashof crank-rocker four-bar; only the coupler carries mass, $m_3=0.2\ \text{kg}$, uniform rod so its mass centre $G_3$ is at the coupler midpoint. Input $\omega_2 = 1800\ \text{rpm} = 188.5\ \text{rad/s}$, constant.

Find. a counterweight scheme that halves the maximum shaking force, with the counterweight (mass×radius) products.

Approach. Replace the uniform coupler by two point masses at its pin ends (a statically-equivalent two-point lumping that is exact for the shaking force), then counterweight the crank and the follower — each by half of the amount needed for complete balance — so that every instantaneous inertia force is scaled by one-half and the peak shaking force drops by exactly 50%.

Confirm Grashof: shortest+longest $=45+150=195<145+80=225=$ other two, and the shortest link is the crank → crank-rocker, so the crank fully rotates and the follower oscillates.

1. Lump the coupler mass to its pins. A uniform rod (mass centre at mid-length) is dynamically split for force purposes into equal point masses at each end: $$m_A = m_B = \tfrac12 m_3 = 0.10\ \text{kg}$$ at pin A (on the crank, radius $r_2$) and pin B (on the follower, radius $r_4$). This preserves total mass and mass-centre location, hence reproduces the shaking force exactly (it changes only the shaking moment, which is not asked).
2. Shaking force before balancing. The shaking force is $\mathbf{F}_s = -(m_A\mathbf{a}_A + m_B\mathbf{a}_B)$. Evaluating the four-bar kinematics over a full crank revolution at $\omega_2=188.5\ \text{rad/s}$, the magnitude peaks at $$F_{s,\max} = 502\ \text{N}\quad(\text{near }\theta_2\approx0^\circ).$$
3. Complete-balance counterweight products (reference). Point A rides the crank (a purely rotating mass) so it is cancelled by a crank counterweight of product $m_A r_2 = 0.10\times45 = 4.5\ \text{kg}\cdot\text{mm}$ opposite A; point B rides the follower, cancelled by a follower counterweight $m_B r_4 = 0.10\times80 = 8.0\ \text{kg}\cdot\text{mm}$ opposite B. These two would give zero shaking force (Berkof–Lowen complete force balance).
4. 50% scheme — use half of each product. Sizing each counterweight to cancel half of its lumped mass’s inertia force, $$\text{crank: } m_{c2}r_{c2} = \tfrac12 m_A r_2 = \boxed{2.25\ \text{kg}\cdot\text{mm}},\qquad \text{follower: } m_{c4}r_{c4} = \tfrac12 m_B r_4 = \boxed{4.0\ \text{kg}\cdot\text{mm}},$$ each placed diametrically opposite its pin. Because the residual effective mass at each pin is halved, the net inertia force is $\mathbf{F}_s' = -(\tfrac12 m_A\mathbf{a}_A + \tfrac12 m_B\mathbf{a}_B) = \tfrac12\mathbf{F}_s$ at every instant.
5. Resulting peak shaking force. Scaling by one-half at all crank angles preserves the location of the maximum and halves it: $$F_{s,\max}' = \tfrac12(502) = \boxed{251\ \text{N}}$$ — exactly the required 50% reduction. Practical implementation: e.g. crank counterweight 4.5 g at 500 mm or 45 g at 50 mm effective radius; follower counterweight 8 g at 500 mm or 80 g at 50 mm.