Question

Question + 130 marks An inverted low. The -10 in totales verted crank-slider mechanism is shown below. The ground link makes angle of 25° with the izontal line as shown. The crank rotates at a constant angular velocity of 300 rad/sec (CCW). Mechanism dimensions are 10 inches, F. 40 inches a) Choose a proper angle for the angular orientation of the follower (link 4) and the coupler (link) 0, 0, and conduct the displacement analysis by means of the loop.closure equation. Determine the general expressions for the two dependent variables and s. Here is the distance between the node B and pin A. b) Compute the values and s for 0, 60'. c) Conduct the velocity analysis, and determine the general expressions for the two velocny parameters, and s. d) Compute the values of , and for 0,= 60. e) Determine the Coriolis acceleration agen for e, = 60 Not to scale B3, B4 OM Ground

Answer 1

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6 for O₂ = 60° ton (04-25) = 105n (60-25- 10S /60-28) - 408/335) 10ros 160-25- 4oras (935) = 22.6405 -0.80684 - 28.0608 - 04 - 25 & ton(-0.80684) = 25+ 141. 1020 - 166.1020 5- 10 cos(60-25) – 40 105 835) = 36.056 inches. cos (166.1020-25) and s= 36.056 inks. = 60 Oy = / 66: 102 O4 = 166.102° Velocity analysis Differentiating equation (i) and (ii) w. Not time we get -s sin(04-25). 64 + cos(04-25). S =-8 Siv (02-25). Cli) sos/04-25). 6y + sin/ u-25).5 = 105/02-25). 6, Peat Moltiplying (all) with sing04-25) and (iv) with cos(84-25) we get and substrective liieien in all.get + Sn19-25) Sn04-28) - - soby (1082/94-25) tsin? (864 - 25)) = 0; 07 (10510,-Joyou-3) S. - 8. 6 cos/0, -25 - 04+25) on = 0., 105/02 - 04) volonit, two for we can find Ś 2 58 (04-02) the general expossion og ands. sim. lordy s = 8 Those are parometer

for 64 - O-60° szi öz cos (02-04) = 10 x 300 x 105/60-16611020) 36.056 = – 23.076 red moons 23.076 rod/sec. (cw) s = 8.6 5.6(04-02) = 10x 300 x 5.r (166. 02-60) = 2882:308 roks soon

a) The general expression for two dependent variables are

$\tan (\theta _{4 }-25)= (r_{2}*\sin (\theta _{2}-25)-r_{1}*\sin (335))/(r_{2}*\cos (\theta _{2}-25)-r_{1}*\cos (335))$

  

s = (r2 cos(02 – 25) - r1 cos(335))/ cos(04 – 25)

$= (r_{2}\sin (\theta _{2}-25)- r_{1}\sin (335))/\sin (\theta _{4}-25)$

b) For $\theta _{2}=60 , \theta _{4}= 166.102 , s = 36.056 inches$

c) The general expression for two velocity parameters are

  $\theta ^{\cdot }_{4}= (r_{2}*\theta ^{\cdot }_{2}*\cos (\theta _{2}-\theta _{4}))/ s$

$s^{\cdot }= r_{2}*\theta ^{\cdot }_{2}*\sin (\theta _{4}-\theta _{2})$

d) For   $\theta _{2}= 60, \theta ^{\cdot }_{4}= 23.076 rad/sec (cw) , s^{\cdot }= 2882.308 inches/sec.$