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Oct 30, 2023

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California State University, Long Beach College of Engineering Department of Mechanical Engineering MAE 375-Kinematics Dynamics Mechanisms Instructor: Lab #3 Group #2 Student Name Collaboration Annika Lambach Shakir Shihabudeen Thomas Phan Benny Matecny

CALIFORNIA STATE UNIVERSITY, LONG BEACH MAE 375: Kinematics and Dynamics of Mechanisms Lab3: Position Analysis II A. Rezaei Show your work clearly. Illegible work will not be given full credit. Follow the submission instruction for the files carefully. Include your name as the first comment line in all program files. Staple your work (extra page, printouts, etc.) to this handout for submission. NAME: Total score: 10 pts 1. [5 pt] Write a script called, Group#Grashof , to determine the Grashof condition of a fourbar linkage. The program must ask the user to input a row vector containing the four link lengths in the order of ⇥ L 1 L 2 L 3 L 4 ⇤ and return the respective Grashof condition as a text output as ”Grashof”, ”Special Grashof”, or ”Non-Grashof”. In the case of Grashof, the program needs to compute if the mechanism is “Crank-Rocker”, “Double-Crank” or, “Double-Rocker”. The program should output an error for an invalid link length i.e. zero and negative and if the user input less than or more than 4 link lengths. You can use link lengths from Problem 2.15 (Pg 84) to check your program. Submit the m-file in the Dropbox , called Lab 3, on Beachboard. Hint: Check out the MATLAB built-in function, sort() , to help sort the link lengths. 2. [5pt] Solve Problem 4-9 and 4-10 using the link parameters found in Row (a) of Table P4-2 (Pg 216). Write your analysis in the space below. Attach extra sheets if needed. 1 Annika Lambach, Thomas Phan, Shakir Shihabudeen, Shakir Shihabudeen

[Y - 9] i Ay 1.4sin(4s) 4 Ay = 0.9899 2 1- Ay sinc= 1-At= sin= p a = sin () x = 0.143 Oz = 180 + x = 180 + 0.143 0 = 180. 143 Slider Position d = L cos0z +Lycosx d = 1.4cos(4S) +4cos (0.145) d = 4.9899

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[n-10] R2-Ps- Ry-R, = 0 Rz = aio--> allosztjsinGz) M Gijte Rz = beits--blcoss+jsinEs) · Gre ! I L ! Ry = ceit--c(cost+jsintn) I d I jO, R = de- - dIcoS8, tjsint.) Real Parts acOSO2-bLosO,-(Costy-d1050, = 0 ⑦ , = 0 acostz-bcustz-<costy - d = 0 Imaginary Parts jasintz - jbsints-jesinty-josino, = 0 ⑦ , = 0 asinz-bsinGz-CSinGy = 0 Os= sin ( ) -> sint( - 1) = -0.14 I crossed) d = acosOz-bcosts, ->4coslYS)-410S)-0) = -3.01 Osz= sint(-) + 180 = 180. (opens d = ACOSOz-boss, d = 1.4cosLYs)-4c0S (18.1) d = 4.98 = 5.0