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Problem 2 For the quick-return mechanism shown below, the rod DC rotates with a constant angular velocity of 1 rad/s CCW. a) For the position shown find the angular velocity of member AB and the velocity of sliding of block C within the member AB. b) If a force P 1 kN is applied vertically at B, determine the couple moment M that must be applied to member DC. Also find the force transmitted through the slider joint at C Neglect the masses of all members. Dimensions are in cm. 30 D M 10 B b) Energy method (No masses) with ΣΣ Tk => P.V+M.y=0 The velocity of B: VB AB. (-sin 30° + cos 30°) VB 0.4x0.5x(-sin 30° + cos 30°) The force at P: P = -1000 j and P.V-1000 x (0.4 x 0.5 x cos 30°) = -173.2 then Eq. 4-173.2+M x 10 => M = 173.2 N.m F.B.D. of DC 30° 30° 30° M N is perpendiculare to the axes of slip along the axis of transmission ΣM/D = 0 (a = 0, constant angular velocity) M-N x 0.1 x cos 30° = 0 N = 173.2 0.1 x cos 30°-2000N

Problem 2 For the quick-return mechanism shown below, the rod DC rotates with a constant angular velocity of 1 rad/s CCW. a) For the position shown find the angular velocity of member AB and the velocity of sliding of block C within the member AB. b) If a force P 1 kN is applied vertically at B, determine the couple moment M that must be applied to member DC. Also find the force transmitted through the slider joint at C Neglect the masses of all members. Dimensions are in cm. 30 D M 10 B b) Energy method (No masses) with ΣΣ Tk => P.V+M.y=0 The velocity of B: VB AB. (-sin 30° + cos 30°) VB 0.4x0.5x(-sin 30° + cos 30°) The force at P: P = -1000 j and P.V-1000 x (0.4 x 0.5 x cos 30°) = -173.2 then Eq. 4-173.2+M x 10 => M = 173.2 N.m F.B.D. of DC 30° 30° 30° M N is perpendiculare to the axes of slip along the axis of transmission ΣM/D = 0 (a = 0, constant angular velocity) M-N x 0.1 x cos 30° = 0 N = 173.2 0.1 x cos 30°-2000N

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Please explain the energy method rule in the first line, with a description of how to put the numbers and where we got them from, please, in detail.
Problem 2 For the quick-return mechanism shown below, the rod DC rotates with a constant angular velocity of 1 rad/s CCW. a) For the position shown find the angular velocity of member AB and the velocity of sliding of block C within the member AB. b) If a force P 1 kN is applied vertically at B, determine the couple moment M that must be applied to member DC. Also find the force transmitted through the slider joint at C Neglect the masses of all members. Dimensions are in cm. 30 D M 10 B
Transcribed Image Text:Problem 2 For the quick-return mechanism shown below, the rod DC rotates with a constant angular velocity of 1 rad/s CCW. a) For the position shown find the angular velocity of member AB and the velocity of sliding of block C within the member AB. b) If a force P 1 kN is applied vertically at B, determine the couple moment M that must be applied to member DC. Also find the force transmitted through the slider joint at C Neglect the masses of all members. Dimensions are in cm. 30 D M 10 B
b) Energy method (No masses) with ΣΣ Tk => P.V+M.y=0 The velocity of B: VB AB. (-sin 30° + cos 30°) VB 0.4x0.5x(-sin 30° + cos 30°) The force at P: P = -1000 j and P.V-1000 x (0.4 x 0.5 x cos 30°) = -173.2 then Eq. 4-173.2+M x 10 => M = 173.2 N.m F.B.D. of DC 30° 30° 30° M N is perpendiculare to the axes of slip along the axis of transmission ΣM/D = 0 (a = 0, constant angular velocity) M-N x 0.1 x cos 30° = 0 N = 173.2 0.1 x cos 30°-2000N
Transcribed Image Text:b) Energy method (No masses) with ΣΣ Tk => P.V+M.y=0 The velocity of B: VB AB. (-sin 30° + cos 30°) VB 0.4x0.5x(-sin 30° + cos 30°) The force at P: P = -1000 j and P.V-1000 x (0.4 x 0.5 x cos 30°) = -173.2 then Eq. 4-173.2+M x 10 => M = 173.2 N.m F.B.D. of DC 30° 30° 30° M N is perpendiculare to the axes of slip along the axis of transmission ΣM/D = 0 (a = 0, constant angular velocity) M-N x 0.1 x cos 30° = 0 N = 173.2 0.1 x cos 30°-2000N
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