By Anthony M. Bedford, Wallace Fowler
whereas masking the elemental ideas of mechanics in an example-driven structure, this cutting edge booklet emphasizes serious considering through proposing the reader with engineering occasions. Compelling photorealistic paintings, and a powerful photo application is helping readers to attach visually to the themes mentioned. gains robust assurance of FBDs and demanding ABET topics. For execs in mechanical, civil, aeronautical, or engineering mechanics fields.
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Additional info for Solutions Manual: Engineering Mechanics: Dynamics (5th Edition)
All rights reserved. This book is secure by way of Copyright and permission might be acquired from the writer ahead of any prohibited replica, garage in a retrieval procedure, or transmission in any shape or in anyway, digital, mechanical, photocopying, recording or likewise. eighty five Problem thirteen. 157 within the cam-follower mechanism, the slotted bar rotates with consistent angular pace ω = 10 rad/s and the radial place of the follower A relies on the profile of the desk bound cam. the trail of the follower is defined via the polar equation y A r r = 1 + zero. five cos(2θ ) m . make sure the speed of the cam follower whilst θ = 30◦ (a) by way of polar coordinates and (b) by way of cartesian coordinates. θ x answer: (a) θ = 30◦ , ω = dθ/dt = 10 rad/s, and α = zero. r = 1 + zero. five cos 2θ = 1. 25 m. dθ dr =− sin 2θ dt dt = −8. sixty six m/s. v= dθ dr er + r eθ dt dt = −8. 66er + 12. 5eθ (m/s). (b) vx = vr cos θ − vθ sin θ = −13. seventy five m/s, vy = vr sin θ + vθ cos θ = 6. 50 m /s. challenge thirteen. 158* In challenge thirteen. 157, make sure the acceleration of the cam follower whilst θ = 30◦ (a) by way of polar coordinates and (b) by way of cartesian coordinates. resolution: See the answer of challenge thirteen. 157. (a) d2r = −2θ 2 cos 2θ dt 2 = −100 m/s2 . a= d2r −r dt 2 dθ dt 2 er + r d2θ dr dθ eθ . +2 dt 2 dt dt = −225er − 173eθ (m/s 2). (b) ax = ar cos θ − aθ sin θ = −108 m/s 2, ay = ar sin θ + aθ cos θ = −263 m/s2 . 86 c 2008 Pearson schooling South Asia Pte Ltd. All rights reserved. This e-book is secure by way of Copyright and permission may be got from the writer ahead of any prohibited copy, garage in a retrieval method, or transmission in any shape or in anyway, digital, mechanical, photocopying, recording or likewise. challenge thirteen. 159 The cartesian coordinates of some degree P within the x−y aircraft are regarding its polar coordinates of the purpose by way of the equations x = r cos θ and y = r sin θ . express that the unit vectors i, j are relating to the unit vectors er , eθ via i = er cos θ − eθ sin θ and j = er sin θ + eθ cos θ . (b) starting with the expression for the placement vector of P when it comes to cartesian coordinates, r = xi + yj, derive Eq. (13. fifty two) for the location vector when it comes to polar coordinates. (c) through taking the time by-product of the placement vector of element P expressed by way of cartesian coordinates, derive Eq. (13. forty seven) for the speed when it comes to polar coordinates. y (a) eθ er r P θ x resolution: (a) From geometry (see Figure), the radial unit vector is er = i cos θ + j sin θ, and because the transverse unit vector is at correct angles: eθ = i cos θ + π 2 + j sin θ + π 2 = −i sin θ + j cos θ. remedy for i by means of multiplying er by means of cos θ, eθ through sin θ, and subtracting the ensuing equations: i = er cos θ − eθ sin θ . clear up for j by way of multiplying er by way of sin θ, and eθ via cos θ, and the implications: j = er sin θ + eθ cos θ (b) the placement vector is r = xi + yj = (r cos θ)i + (r sin θ)j = r(i cos θ + j sin θ). Use the result of half (a) expressing i, j by way of er , eθ : r = r(er cos2 θ − eθ cos θ sin θ + er sin2 θ + eθ sin θ cos θ) = rer (c) The time derivatives are: dr dr dθ =v=i cos θ − r sin θ dt dt dt +j dθ dr sin θ + r cos θ dt dt , from which v= dθ dr (i cos θ + j sin θ) + r (−i sin θ + j cos θ).