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Problem 15.248

D. 4 in. 8 in. C. B. 8 in. A. Problem 15.248. Knowing that at the instant shown crank BC has a constant angular velocity of 45 rpm clockwise, determine the acceleration (a) of point A , (b) of point D. Problem 15.248. D. 4 in. 8 in. C. B. 8 in. A. v = r w. y. r. x.

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Problem 15.248

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  1. D 4 in. 8 in. C B 8 in. A Problem 15.248 Knowing that at the instant shown crank BC has a constant angular velocity of 45 rpm clockwise, determine the acceleration (a) of point A, (b) of point D.

  2. Problem 15.248 D 4 in. 8 in. C B 8 in. A v = rw y r x w Solving Problems on Your Own Knowing that at the instant shown crank BC has a constant angular velocity of 45 rpm clockwise, determine the acceleration (a) of point A, (b) of point D. 1. Determine velocities in a body rotating about a fixed axis: In vector form the velocity in the body is given by: v = wxr Where v , w , and r are the velocity of the point, the angular velocity of the body, and the position vector from the axis to the point. The magnitude of the velocity is given by: v = rw where v,r, and w are the magnitudes of the corresponding vectors.

  3. Problem 15.248 D 4 in. 8 in. C B 8 in. A w A rB/A + vB/A B Solving Problems on Your Own Knowing that at the instant shown crank BC has a constant angular velocity of 45 rpm clockwise, determine the acceleration (a) of point A, (b) of point D. 2. Determine velocities in a body under general plane motion: Velocities can be determined either by method of instantaneous center of rotation, or by considering the motion of the body as the sum of a translation and a rotation. C vA vA vB A A vA = vB B B B A vA vB = vA + vB/AvB/A = wxrB/A

  4. Problem 15.248 D 4 in. 8 in. C B 8 in. A Solving Problems on Your Own Knowing that at the instant shown crank BC has a constant angular velocity of 45 rpm clockwise, determine the acceleration (a) of point A, (b) of point D. 3. Determine the acceleration in a body rotating about a fixed axis: In vector form the acceleration is given by: a = axr + wx ( wxr ) The normal and tangential components are given by: an = rw2at = ra where w and a are the angular velocity and angular acceleration, respectively. at = ra y w an = rw2 x a

  5. Problem 15.248 D 4 in. 8 in. C B 8 in. A y’ wk A x’ ak aB/A (aB/A)n (aB/A)t B (c) Solving Problems on Your Own Knowing that at the instant shown crank BC has a constant angular velocity of 45 rpm clockwise, determine the acceleration (a) of point A, (b) of point D. 4. Determine accelerations in a body under general plane motion: Draw (a) a plane motion diagram, (b) translation diagram, and (c) rotation diagram. A A aA aA + = B B aA aB (a) (b) aB = aA + aB/AoraB = aA + (aB/A )n + (aB/A )t

  6. Problem 15.248 Solution D 4 in. 8 in. C B rad min 2 p 8 in. wBC = (45 rpm)( ) = 1.5p rad/s A s rev 60 D vB B vA A Determine velocities in a body rotating about a fixed axis. Calculating the velocity of point B in crank BC. vB = wBC rB/C = (1.5p rad/s)(4 in) = 18.85 in/s vB = 18.85 in/s vB wBC C Determine velocities in a body under general plane motion. B Since point A is forced to move in the vertical direction, and the direction of the velocity of point B is up, the angular velocity of bar AD is zero: wAD = 0

  7. Problem 15.248 Solution D 4 in. 8 in. C B 8 in. A (aB)n = (wBC )2 rB/C = (1.5p rad/s)2 (4 in) (aB)n = 88.83 rad/s 2 Determine accelerations in a body rotating about a fixed axis. Calculating the acceleration of point B in crank BC. Since the angular velocity of crank BC is constant, aBC= 0 and (aB)t = 0. The normal component of the acceleration at point B is: wBC (aB)n C B aBC = 0

  8. Problem 15.248 Solution D 4 in. 8 in. C B 8 in. A aADis assumed CCW D 30o (aD/B)t B + aAD (aA/B)t 30o A Determine accelerations in a body under general plane motion. (aD)y (aD)x aB D D = aB aB B B aA aB A A The acceleration aA must be vertical

  9. Acceleration of point A: Problem 15.248 Solution + xcomponent: + y component: D D D aB aB aAD B + = B B (aA/B)t aA 30o aB A A A aA = aB + (aA/B)t [aA ] = [88.83 ] + [ 8 aAD 30o ] Equating the x and y components: 0 = 88.82 + 8 aAD cos 30o aAD = - 12.82 rad/s2 (aADis actually CW) aA = 0 + 8 (- 12.82)sin 30o aA = - 51.3 in/s2aA = 51.3 in/s2

  10. Acceleration of point D: Problem 15.248 Solution 30o (aD/B)t aD = aB + (aD/B)t [(aD)x ] + [(aD)y ] = [88.83 ] + [ 8 aDA 30o ] + x component: (aD)y = 0 - ( 8 )(- 12.82 ) sin 30o (aD)y = 51.3 in/s2aD = 184.9 in/s2 16.1o + y component: (aD)y aB D (aD)x D D aB aB aAD B + = B B A A A Equating the x and y components: (aD)x = 88.82 - ( 8 )(- 12.82 ) cos 30o (aD)x = 177.7 in/s2

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