To determine Determine the required torque. Explanation Show the diagram of the part of the mechanism as in Figure (1). Express the angular velocity of the link O A . ω O A = 2 π N 60 (I) Here, the angular velocity of the link O A is ω O A and rotational velocity of the link O A is N . Express the velocity of point A . V → A = V o + ω → O A O A → (II) Here, the velocity of point A is V → A and velocity of point O is V o . Express the velocity of point B . V → B = V → B + ( ω B / A k ^ ) A B → (III) Here, the velocity of the point B is V → B , the angular velocity of the point B with respect to the point A is ω B / A . Express the change in kinetic energy of the link A B . K = 1 2 m ( V B 2 − V A 2 ) (IV) Here, mass of the link is m and change in kinetic energy is K . Express the change in potential energy of the coupler. P = m g h (V) Here, potential energy change is P , gravitational acceleration is g and height is h . Express the moment of inertia of the link A B . I = m ( A B ) 2 12 (VI) Express the change in kinetic energy of the link A B . T = 1 2 I ( ω B / A ) 2 (VII) Here, change in kinetic energy is T and mass moment inertia of link is I . Express the change in total energy of the link. U = M θ (VIII) Here, change in the total energy is U , moment applied is M and the angle turned by crank is θ Apply the conversation of the energy. U = K + T + P (IX) Conclusion: Substitute 60 rev / min for N in Equation (I). ω O A = 2 π ( 60 rev / min ) 60 = 2 π rad / s From the diagram of the mechanism calculate the distance D A . D A = 190 mm − 80 cos 45 ° Apply Pythagoras theorem in triangle A B D . x = ( 190 mm − 80 cos 45 ° ) 2 − ( 240 mm ) 2 = 3196.046 = 56.53 mm Express the angle ϕ in the triangle A B D using rules of trigonometry. sin ϕ = 240 mm 190 mm + 80 mm ( cos 45 ° ) = 0.973 ϕ = 76.65 ° Express the unit position vector along O A . O A → = { ( 80 mm ) [ 10 − 3 m mm ] cos 45 ° i ^ + ( 80 mm ) [ 10 − 3 m mm ] cos 45 ° j ^ } Substitute { ( 80 mm ) [ 10 − 3 m mm ] cos 45 ° i ^ + ( 80 mm ) [ 10 − 3 m mm ] cos 45 ° j ^ } for O A → , 0 for V , and ( 2 π rad / s ) k ^ for ω O A in Equation (II). V → A = 0 + ( 2 π rad / s ) k ^ { ( 80 mm ) [ 10 − 3 m mm ] cos 45 ° i ^ + ( 80 mm ) [ 10 − 3 m mm ] cos 45 ° j ^ } = ( 2 π rad / s ) k ^ [ ( 0.707 ) i ^ + ( 0.707 ) j ^ ] Express the vector product. V → A = | i j k 0 0 2 π 0.707 0.707 0 | = [ 0 − 2 π ( 0.707 ) ] i ^ − [ 0 − 2 π ( 0.707 ) ] j ^ = − 4

8th Edition

ISBN: 9781118885840

Author: James L. Meriam, L. G. Kraige, J. N. Bolton

Publisher: WILEY

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Chapter 6.5, Problem 108P

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