--- title: "Mechanism AOB driven by gear D, tangential force" subject: "Applied Mechanics (Dynamics)" chapter: 5 key: "semester_3/applied_mechanics_dynamics:mechanism-aob-gear-d-tangential-force" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/mechanism-aob-gear-d-tangential-force site: BCT Engineering --- # Mechanism AOB driven by gear D, tangential force Applied Mechanics (Dynamics) · Chapter 5: Plane Motion of Rigid Bodies: Forces, Moments, and Accelerations - Asked in: 2078 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/mechanism-aob-gear-d-tangential-force - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-5 ## Question The portion AOB of the mechanism is actuated by gear D and at the instant shown has a clockwise angular velocity of 8 rad/s and a counter clockwise angular acceleration of 40 rad/s$^2$. Determine tangential force exerted by gear D. Take $m_E = 4$ kg, $\bar k_E = 85$ mm and $m_{OB} = 3$ kg. [Figure: gear D meshing with gear E (radius 120 mm) pinned at O; arm OB attached to gear E, with a 40 mm dimension shown on the arm.] ## Short answer The gear and arm rotate together about O. Compute their total moment of inertia about O, then apply F_D r_E = I_O α. With given masses and dimensions, I_O = 0.1889 kg·m²; the tangential force exerted by gear D is F_D = I_O α / r_E = 62.97 N (≈63.0 N). ## Answer Gear $E$ and the arm $OB$ rotate together about $O$, so the tangential force $F_D$ of gear $D$ at the pitch radius $r_E = 0.120$ m must supply the torque needed to give both bodies the angular acceleration $\alpha = 40$ rad/s². The angular velocity (8 rad/s) does not enter the moment equation: both bodies rotate about the fixed axis $O$, and the centripetal terms produce no moment about $O$. **Assumptions.** Arm $OB$ is a uniform slender rod of length $OB = 0.400$ m pivoted at its end $O$ (the "40 mm" in the figure is read as 400 mm). At this instant the rod is vertical, or the mechanism turns in a horizontal plane, so the weights have no moment about $O$. If the rod length differs, use $I_{OB} = \tfrac13 m L^2$ with the actual $L$. #### Moments of inertia about O $$ \bar I_E = m_E\bar k_E^2 = 4(0.085)^2 = 0.0289\ \text{kg m}^2 $$ $$ I_{OB} = \tfrac13 m_{OB}L^2 = \tfrac13(3)(0.400)^2 = 0.160\ \text{kg m}^2 $$ $$ I_O = I_E + I_{OB} = 0.1889\ \text{kg m}^2 $$ #### Equation of motion about O $$ F_D\,r_E = I_O\,\alpha $$ $$ F_D = \frac{I_O\alpha}{r_E} = \frac{0.1889\times40}{0.120} = 62.97\ \text{N} $$ The moment is $M_O = I_O\alpha = 7.56$ N m. **Answer: the tangential force of gear $D$ on gear $E$ is about $62.97$ N ($\approx 63.0$ N)**, acting in the sense of the counter-clockwise angular acceleration. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the 8 rad/s angular velocity not enter the moment equation about O? A. It is too small B. Both bodies rotate about fixed O and the centripetal terms give no moment about O C. It cancels with alpha D. Gear D removes it Answer: B. Normal (omega^2) terms act through O, so only alpha matters in sum M_O = I_O alpha. (Key point: Role of angular velocity) 2. Which equation gives the tangential force F_D? A. F_D r_E = I_O alpha B. F_D = I_O alpha C. F_D r_E = I_O omega D. F_D = m alpha r_E Answer: A. Moment of F_D about O, F_D r_E, equals I_O alpha for the combined gear and arm. (Key point: Equation of motion) 3. Gear E has I = 0.0289 kg m^2 and arm OB (3 kg, 0.4 m, pivoted at end) has I = (1/3) m L^2. I_O total is: A. 0.0289 kg m^2 B. 0.1889 kg m^2 C. 0.160 kg m^2 D. 0.0489 kg m^2 Answer: B. I_OB = (1/3)(3)(0.16) = 0.160; add 0.0289 to get 0.1889 kg m^2. (Key point: Total inertia) 4. F_D = I_O alpha / r_E = 0.1889 x 40 / 0.120 is about: A. 7.56 N B. 629.7 N C. 62.97 N D. 15.1 N Answer: C. 7.556 N m / 0.120 m = 62.97 N. 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