Question

Movement of the solenoid plunger shown in the figure below gives pin A of the mechanism a velocity of 75 mm/s horizontally to the right for some short interval of its motion. For this instant, determine angular velocity was using the method of instant centres. Horizontal distance AD = 320 mm, vertical distance BD = 105 mm and CD = 105 mm. CD - AD Answer: Check (Continued from the previous question) For the mechanism above, determine angular velocity WCD- Answer: Check (Continued from the previous question) For the mechanism above, determine velocity Vg of pin B. Answer:

Answer 1

From the configuration given at the instant VA = 75 mm/s ! 105 mm 320 mm tano=105 320 o = 18.16599 618-1659 320 from traingle ACD using casine rule co²= AC² + AD² – 2x (AC) CAD) Cas 18 1659 1052 = 320² + AC² - 2x (AC) x ( 320) Cas 18-1659 so Ac² – 608 1 AC +91375 =0 Solving . AC = 336 78 or 271-31 So, we get 2 values of AC So using AC = 336.78 mm we get <COA = 90° which is not admissible acces ding to the limon configuration

AC= 271-31 mm 105 271-31 18-160 18-1659 xde 320 Sine sule 105 271:31 Sina Svi (18.1659) we get : x = 53-6665° to feel ... wing Instant center Method Angular velocity of bar AC A 5 .. risc i А/ AC of the 18.1659 53-66 we see that we see that (IC AD) Forsu a right angled triangle w 36.33 € L 53.660 : LA = AD tau 53.66 LAG = 320 tan 83.66 = 435.0937 mm

using Instant center WAC = WAC al rape repic 435.09 AR 320 mm (Ic)(O) = 200 AD Sin 36-33 (Ic) (0) = 320 Sin 36.33 = 540 mm ... (Ic) (c) = (Ic) (O) - (CD) or repo = (S40 - 105) reno = 435.09 дио и NA = 75mm RAAD = re, 10 = 435.09 mm / i oo WACE OVA = ve RALD reID : | WAC = (son) ont radje

Augalas velocity of AC = 0.17 radse WAC = WAB - rigid body Motion Augular velocity of AB = 0.17 soud/s 7 & from Ô VA - ve Зраар RAAD = 26/10 .: { VA = ve = 75 mols Angulas velocity of Color ) ) - To find velocity of fin B From configuration (BF=AD ↑ Icg 2B/IC - TVR F -- 7- 435.09 : BF = 320 mm BD = 105 mm 7 18-1659 320

(BD = AF) (IC) (F) = (f/2) - BD (IC) (F) = (435.09 - 105) (1) (F) = 330.09 mm From A (IC) (F) (B) (2 94 ) = [(T ( X ) ² + (FB) = 330.094 + 3202 RC = √ 330.09h +320 & B / C = 459.73 mm D : velocity of point (B) WAB = VA = VB IC WAB RA/IC Blic BL RAIC ..Va = a +с ху RA/IC 459.73 x 75 435.09 t up = 79 • 24 mul