Question!1 For the four bar mechanism shown below and 02-240°, compute (i)The an

Question

Question!1 For the four bar mechanism shown below and 02-240°, compute (i)The angular positions of the coupler and the follower (ii) The angular velocities of the coupler and the follower (iii) The angular accelerations of the coupler and the follower, (iv) The x and y components of the acceleration of the mass center of the follower (v)The x and y components of the shaking force (the crank and coupler are assumed massless) an (vi) Torque, T2, required to overcome the inertia of the follower Not to scale 04 4 T2 Geometry and material property parameters ??.-240 mm; Ad-45 mm; '4B-175 mm; BB-155 mm; B.G4-90 mm; m4-0.5 kg; Ic-0.045 kgem2 The input link and coupler are massless Operational parameters OF50 rad/s(ccw), ?2-0 rad/s*

Explanation / Answer

(i) By drawing the mechanism with proper scale we get, ?3 = 42.42° (coupler position) and ?4 = 149.32° (follower position).

(ii) Given, ?2 = 50 rad/s

Hence, linear velocity of point A with respect to A0, VA = ?2* A0 A = 50*45 = 2250 mm/s

By drawing velocity diagram and taking measurements we get linear velocity of the point B with respect to the point A, VBA = 2351 mm/s

Hence, angular velocity of AB = 2351/175 = 13.43 rad/s (angular velocity of the coupler)

From velocity diagram, linear velocity of the point B with respect to the point B0 = = 710 mm/s

Hence, angular velocity of B B0 = 710/155 = 4.58 rad/s (angular velocity of the follower)

(iii) Drawing acceleration diagram and taking measurements we get, tangential acceleration of the point B with respect to A, atBA = 14370 mm/s2

and tangential acceleration of the point B with respect to B0, atBB0 = 78120 mm/s2

So, angular acceleration of the coupler = 14370/175 = 82.1 rad/s2

and angular acceleration of the follower = 78120/155 = 504 rad/s2

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.