Question

The Geneva mechanism shown is used to provide an intermittent rotary motion of disk S. Disk D weighs 2 lb and has a radius of gyration of 0.9 in. and disk S weighs 6 lb and has a radius of gyration of 1.5 in. The motion of the system is controlled by a couple M applied to disk D. A pin Pis attached to disk D and can slide in one of the six equally spaced slots cut in disk S. It is desirable that the angular velocity of disk S be zero as the pin enters and leaves each of the six slots; this will occur if the distance between the centers of the disks and the radii of the disks are related as shown. Knowing disk D rotates with a constant counterclockwise angular velocity of 7.6 rad/s and the friction between the slot and pin P is negligible, determine when φ = 150° the couple M and the magnitude of the force pin P applied to disk S. (Round the final answer to three decimal places.) 

The couple M is _______  lb·in. 

The magnitude of the force pin that is applied to the disk is _______  lb.

Answer 1

(2881 JB O B. sino 1.25 ap- apt apps + 2 us un Solution : Law of cosines 3 = 1,25 +2.50 -2 (-25) (2.5) corso r= 1.54 qiuin. 2.50 in law of sines. sing > ß = sin /125 sin 30g 1.54 914 het disk s be a rotating point P. on the disk Up' = r Ws = 15491 ws DB ap = -46 KXV Plows uplo. [1.54912S BR] + [15491 was 7B] motion relative to trame Upls = uzß Coriolis Acceleration up = Vp + Up/s. = {115491 Ws Pyp] + [u DF (1915 - 23794° aph=u re

9p= (isuaiss ne] +losti wes zf]+[upf) + [2 wg. UNE mob'on about desk D (koration about B) up = (BP) wo = 1.25 x 76 = 9.5 infec 17 830 ap:= {@p_o D60] +[(BP ) w os 30) 0 +1.25 (1.65 40 ap = 72:2 in /see a 30 faluate the two expressions for up and resowe into component Puß = 1.5ugiu ws = 9.5 cos (30 + 8) 5.61118 radtsee Ws = bol the two I Bou = 9.5 sin (30 +P) = 9.5 sins3794 u= 7.6655 inlee Equate Expresions ap and into components. AB: 1.5491425-2 ws n = 72.2 72-2 sin ($+30) + 2 x 5.61181,6655) resolve sin (30+R) Isa 1.54914 93.14366 rad lecê

Is as= ( 9 2015 alebo J) 6+)(23 1426. Lad kB) 0.27 118 loft 1.20 =0 since xo=0 Diskas 2 Em = Emest Pr = 2 Ls. P (usually) 0:27 118 12 P= 2.1000 lb Disk Do- r=90°-38-8 = 36-206" +5 Emo = {(mBlett: m-(p sint) (1.25 ) ft m 1.25 X (2.1000) sin 36,206 12 m = 0. 1292 lb ft M 1.5504 en in G Renet couple m= 1.5soy loin magurense of the pin that applied to dick es 2.1000 lb