Chapter 3 HW Solution

 Merilyn Kennedy
 4 months ago
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1 Chapter 3 HW Solution Problem 3.6: I placed an xy coordinate system at a convenient point (origin doesn t really matter). y 173 x The positions of both planes are given by r B = v B ti + 173j mi (1) r A = 1i + v A t(.77i +.77j) mi (2) The distance between A and B is the length (magnitude) of vector r AB, which is the vector from B to A (or the reverse; distance is the same). So we have r BA = r B r A (3) However, we re really interested in the magnitude of the vector r BA, denote that magnitude by length l: l(t) = r BA = (74t 1)i + ( 276t + 173)j mi (4) The magnitude (Euclidean norm) of a vector is simply the square root of the sum of the components squared, so l(t) = (74t 1) 2 + ( 276t + 173) 2 mi (5) Parts (a) and (b) of the problem should have been reversed: first you have to find the time, then you can find the distance. This is just a standard function minimization problem; e.g. take the first derivative and set it equal to zero. (b) The differentiation is simpler if you realize that when l(t) is at a minimum, so is [l(t)] 2. This removes the square root, and d [l(t)] 2 = d [ 81, 652t 2 11, 296t + 39, 929 ] = 163, 34t 11, 296 = (6) dt dt Solving (13) yields the time when the minimum separation occurs, which is If the planes leave at 6: p.m., then the time of minimum separation is t = 11, 296 =.675 hr = 4.52 min (7) 163, 34 t min = 6 : 4 : 31 p.m. (8) (a) To find the actual separation distance, substitute t min (expressed in fractional hours of (14)) into equation (12). The result I found was l min = mi (9) Although not required, I couldn t resist plotting separation distance vs time (MATLAB plot on next page); it seems to agree with my result. 1
2 Separation distance (mi) Time (min) (c) Although not required, you can also do this problem with ADAMS, you have to set up two bodies with the velocities of the planes, and a PointtoPoint measure for the distance between the two planes. You get the following plot: The minimum of the ADAMS plot occurs at t = 4.6 minutes, which is pretty close to the previous result. ADAMS separation distance is miles; again pretty close. The Problem 3.8: Do this analytically. Velocity of A along this line Velocity of B is known Points A and B are both on link 3, so they re related by the 2 pts on a body equation: v A = v B + ω 3 r BA (1) Velocity v B is known (along lower plane), the direction of velocity v A is known (along upper plane), and the angular velocity ω 3 is in the k direction (perpendicular to the plane). From the angles given, the upper plane is at angle of 2
3 15 from the horizontal. From inspection, block A is moving to the left, so we have v A ( cos 15 i sin 15 j) = 4i + ω 3 k.4( cos 3 i + sin 3 j) (11) Separating the i and j equations, there are i :.9659v a = 4.2ω 3 (12) j :.2588v a =.3463ω 3 (13) In matrix form, equations (12) (13) are Solving, we get [ ] [ ] va = ω 3 [ ] [ ] va m/s = ω rad/s [ ] 4 (14) (15) In terms of vectors, we have v A = v A ( cos 15 i sin 15 j), so v A = i j m/s ω 3 = k rad/s (16) (17) So link 3 is rotating CCW, which I think agrees with the sketch. And block A is sliding a little faster than block B (49 m/s compared with 4 m/s). Problem 3.9 In the 4bar mechanism shown below, link 2 is driven at a constant angular velocity of ω 2 = 45 rad/s CCW. We want to find the angular velocities ω 3 and ω You will need the angles I found above in the analysis. You are to do this problem both analytically and using ADAMS. (a) Analytical Solution. This can be done using only 2 point on a body throughout. Start by finding the velocity of A: Next relate the velocities of A and B: v A = v O2 +ω 2 r O2A = 45k ( 2i j) = 155.9i 9j in/s (18) }{{} = v B = v A + ω 3 r AB (19) 3
4 where ω 3 = ω 3 k rad/s and r AB = 6.78i j in. Substituting for v A and evaluating, we get Now relate the velocities of B and O 4 : v B = ( ω 3 )i + ( ω 3 )j in/s (2) where ω 4 = ω 4 k rad/s and r BA = 5.22i j in. Evaluting this, we get Equate (2) and (22) to obtain v B = v O4 +ω 4 r BO4 (21) }{{} = v B = 1.81ω 4 i 5.22ω 4 j in/s (22) i : ω 3 = 1.81ω 4 (23) j : ω 3 = 5.22ω 4 (24) I like to express these in matrix form: I solved these with MATLAB to yield [ ] [ ] ω3 = ω 4 ω 3 = 1.42k rad/s ω 4 = 15.39k rad/s [ ] (25) (26) (27) So both angular velocities are CCW, and link 4 is much faster than link 3. I guess that looks okay. (b) ADAMSSolution. An ADAMS screenshot of the mechanism (at θ 2 = 12 ) is shown below. The velocity plot is shown on the next page. 4
5 Here s the velocity plot, with lines drawn at 12. The results at the angle agree with the analytical. 2. ADAMS Analysis of Problem 3.9 Angular Velocity of Links 3 & Angular Velocity (rad/sec) Link 3 Angular Velocity (rad/s) Link 4 Angular Velocity (rad/s) Angle (deg) Problem 3.11 (ADAMS Only). A screenshot of my linkage in the initial position is shown below: 5
6 The velocity plot for this problem is shown below. Note that the velocity of point C is quite large near the limits of motion (typical). 8 Velocity of Point C and Angular Velocity of Link Velocity (ft/sec) Velocity of C (Xcomponent) Velocity of C (Y component) Omega 3 (rad/sec) Angular Velocity (rad/sec) Angle (deg) Problem 3.15: A position analysis using the loop closure equation shows that r AO4 = mm (28) Angle of AB with horizontal = , (29) and both these values will be needed. The figure is shown below, with those numerical values mm n n (a) Analytical Velocity: For the analytical velocity analysis, you ll need to use both the 2 points on a body and the one point moving on a body equations. Find the velocity of A using points O 2 (stationary) and A and the two points on a body relationship: So the velocity of A is known. v A = ω 2 r O2A = 225i 3897j mm/s (3) 6
7 Next find the velocity of A again, but now you relate links 3 and 4. You know the path of A relative to body 4. For this situation use the one point moving on a body equation, with A as the point, and 4 as the body, therefore written as follows: v A = v A4 + 4 v A (31) Consider equation (31) very carefully!! Point A 4 is point A in the figure. However...point A 4 is a point that is coincident with A, but FIXED TO BODY 4. You may think of it as a hypothetical extension of body 4 up to point A. The path of A 4 is a circular arc centered at O 4. Therefore, the velocity v A4 is tangential to that circle, and hence perpendicular to AB. So we know the direction of v A4, but not its magnitude. Finally, term 4 v A is the velocity of A relative to body 4. I visualize this by mentally fixing body 4, then examining the motion of A. All right, let s solve the problem. Referring to equation (31), we know velocity v A, it s given in equation (3). Next express v A4 as an unknown magnitude in a known direction. This can either be done using v A4 multiplied by the direction of the velocity, or using ω 4 and the cross product. Since the problem statement asks for the angular velocities of 3 and 4 (they re equal), I ll do that: v A4 = ω 4 r O4A = ω 4 k ( i j) = 37.51ω 4 i ω 4 j (32) Now express 4 v A as an unknown magnitude in a known direction: 4 v A = 4 v A (cos(11.17 i sin(11.17 j) = 4 v A (.9811i.1937j) (33) }{{} along AB Substituting into (31) and separating the i and j components, we get Angular velocities: Solving (34) and (35), we get results So the angular velocity vector of links 3 and 4 is i : 37.5ω v A3/4 = 225 (34) j : ω v A3/4 = 3897 (35) ω 4 = ω 3 = 22 rad/s (36) v A3/4 = 1453 mm/s (37) ω 3 = ω 4 = 22 k rad/s (CCW) (38) Velocity of point B: Knowing ω 3 we can relate the velocity of B to the velocity of A: v B = v A + ω 3 r AB = v A + 22k (392.42i 77.49j) = 545.3i j mm/s =.5453i j m/s = m/s (39) (4) (41) I expressed the last result for v B in polar form; this may be easier to visualize. (b) ADAMS Analysis. The path of Point B is shown at right. The yellow bar across the center is simply the initial position of link 3. What is NOT shown in this plot is the speed (magnitude of velocity) of point B as it moves along the path. In particular, the y velocity is quite large as θ 2 is near zero. Hopefully this will be shown in the velocity plots on the next page. 7
8 The plot of the velocity of point B appears below; the y velocity is large near θ 2 =. 15 Velocity of Point B 1 5 Velocity (m/sec) X Velocity of B (m/s) Y Velocity of B (m/s) Link 2 Angle (deg) At θ 2 = 15 the values for the velocity components are (v B ) x =.546 m/s (v B ) y = m/s (42) (43) which agree quite well with the analytical solution. The plot of ω 3 (same as ω 4 ) is: 25. Angular Velocity of Links 3 & 4 Same for both links. Angular Velocity (rad/sec) Link 2 Angle (deg) At θ 2 = 15 the value of the angular velocity is which also agrees well. ω 3 = ω 4 = rad/s (44) 8
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