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Mechanics of Machines MET 305
- 1. Example 3 In the toggle mechanism shown in Fig. 8.16, the slider D is constrained to move on a horizontal path. The crank OA is rotating in the counter-clockwise direction at a speed of 180 r.p.m. increasing at the rate of 50 rad/s . The dimensions of the various links are as follows: OA = 180 mm ; CB = 240 mm ; AB = 360 mm ; and BD = 540 mm. For the given configuration, find 1.Velocity of slider D and angular velocity of BD, and 2.Acceleration of slider D and angular acceleration of BD. Solution. Given : N = 180 r.p.m. or = 2 p × 180/60 = 18.85 rad/s ; OA = 180 mm= 0.18 m ; CB = 240 mm = 0.24 m ; AB = 360 mm = 0.36 m ; BD = 540 mm = 0.54 m We know that velocity of A with respect to O or velocity of A , v AO = vA = wOA × OA = 18.85 × 0.18 = 3.4 m/s ...(Perpendicular to OA ) 1. Velocity of slider D and angular velocity of BD First of all, draw the space diagram to some suitable scale, as shown in Fig. ( a ). Now the velocity diagram, as shown in Fig. ( b ), is drawn as discussed below: 1. Since O and C are fixed points, therefore these points lie at one place in the velocity diagram. Draw vector oa perpendicular to OA , to some suitable scale, to represent the velocity of A with respect to O or velocity of A i.e. v or v , such that AO A vector v AO = v AO = v A = 3.4 m/s figure 2. Since B moves with respect to A and also with respect to C , therefore draw vector ab perpendicular to AB to represent the velocity of B with respect to A i.e. v , and draw vector cb BA perpendicular to CB to represent the velocity of B with respect to C ie. v . The vectors ab and cd intersect at b . 3. From point b , draw vector bd perpendicular to BD to represent the velocity of D with respect to B i.e. vDB , and from point c draw vector cd parallel to CD ( i.e., in the direction of motion DB of the slider D ) to represent the velocity of D i.e vD . D By measurement, we find that velocity of B with respect to A , v BA = vector ab = 0.9 m/s Velocity of B with respect to C , v BC = vector cb = 2.8 m/s Velocity of D with respect to B , v DB = vector bd = 2.4 m/s and velocity of slider D , v D = vector cd = 2.05 m/s Ans. Angular velocity of BD We know that the angular velocity of BD , WBD = v DB /BD = 2.4 /.54 Ans 4.5 rad/s 2. Acceleration of slider D and angular acceleration of BD Since the angular acceleration of OA increases at the rate of 50 rad/s , i.e. a = 50 rad/s2 , α AO = 50 rad/s2 therefore Tangential component of the acceleration of A with respect to O, αt AO = α AO x OA = 50 x 0.18 = 50 0.18 9 m/ s2
- 2. Radial component of the acceleration of A with respect to O , (v A O )2 (3.4)2 a r A O = --------- ----------- = 63.9 m/s2 OA 0.18 Radial component of the acceleration of B with respect to A, (v B A)2 (0.9)2 a r BA = ------------- ----------------- = 2.25 m/s2 BA 0.36 Radial component of the acceleration of B with respect to C , (v BC )2 (2.8)2 ar BC = ----------- ---------- = 32.5 m/s2 CB 0.24 and radial component of the acceleration of D with respect to B , (v DB )2 (2.4)2 ar DB = ----------- ---------- = 10.8 m/s2 BD 0.54 Now the acceleration diagram, as shown in Fig. ( c ), is drawn as discussed below: 1. Since O and C are fixed points, therefore these points lie at one place in the acceleration diagram. Draw vector o'x parallel to OA , to some suitable scale, to represent the radial component of the acceleration of A with respect to O i.e. such that arAO vector o'x = arAO = 63.9 m/s2 2. From point x , draw vector xa' perpendicular to vector o'x or OA to represent the tangential component of the acceleration of A with respect to O i.e. a tAO , such that vector o'a = a tAO 9 m/s2 3. Join o'a' . The vector o'a' represents the total acceleration of A with respect to O or acceleration of A i.e. aAO or aA. 4. Now from point a' , draw vector a'y parallel to AB to represent the radial component of the acceleration of B with respect to A i.e a rBA. such that , vector a'y = a rBA 2.25 m/s2 5. From point y , draw vector yb' perpendicular to vector a'y or AB to represent the tangential component of the acceleration of B with respect to A i.e. atBA whose magnitude is yet unknown. 6. Now from point c' , draw vector c'z parallel to CB to represent the radial component of the acceleration of B with respect to C i.e ar BC.such that vector c'z =atBC = 32.5 m/s2 7. From point z , draw vector zb' perpendicular to vector c'z or CB to represent the tangential component of the acceleration of B with respect to C i.e. The vectors yb' and zb' intersect at b' . Join c' b' . The vector c' b' represents the acceleration of B with respect to C i.e. aBC . 8. Now from point b' , draw vector b's parallel to BD to represent the radial component of the acceleration of D with respect to B i.e a rDB. such that, vector b's = arDB = 10.8 m/s2
- 3. 9. From point s , draw vector sd' perpendicular to vector b's or BD to represent the tangential component of the acceleration of D with respect to B i.e. at DB whose magnitude is yet unknown. 10. From point c' , draw vector c'd' parallel to the path of motion of D (which is along CD )to represent the acceleration of D i.e. aD . The vectors sd' and c'd' intersect at d' . By measurement, we find that acceleration of slider D , aD = vector c'd' = Ans.13.3 m/s2 Angular acceleration of BD By measurement, we find that tangential component of the acceleration of D with respect to B , at DB = vector sd’ = vector 38.5 m/s We know that angular acceleration of BD , αBD = αBD / BD = 38.5 71.3 rad/s2 (Clockwise)