1ME 560 Kinematics Fall Semester 2010Homework No. 4 Wednesday, September 15th Consider the sliding-block linkage shown in Example 2.5, see Figure 2.24, page 77. The position solution for the input angle o2135θ=is presented on pages 78 and 79. The answers, which can be obtained from the law of cosines and the law of sines, are 344ROA12.59 inches==and .o3434104.64θ= θ= θ=Assume that the input link 2 is rotating with an angular velocity 25 rad sω=/clockwise and an angular acceleration 2210 rad sα=/counterclockwise. The length of link 4; i.e., 4OC20 inches=and the origin of the XY reference frame is chosen to be coincident with the ground pin O2as shown in Figure 1. Use the method of kinematic coefficients to determine the following: (i) the relative velocity and acceleration between links 3 and 4 (denoted as 34R±and 34R±±). (ii) the magnitude and direction of the angular velocity and angular acceleration of links 3 and 4. (iii) the velocity and acceleration of point C. Give the magnitude and direction of each vector. (iv) the unit tangent and normal vectors of point C. Show the vectors on Figure 1. (v) the radius of curvature of the path of point C. Show on Figure 1. (vi) the X and Y coordinates of the center of curvature of the path of point C. Show on Figure 1. Figure 1. The Sliding-Block Linkage.
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2Solution to Homework Set 4. (A). The vectors for the sliding-block linkage is shown in Figure 2. Figure 2. The Vector Loop for the Sliding-Block Linkage. The vector loop Equation (VLE) for the sliding-block linkage is ??01234IRRR√√√+−=(1) The X and Y components of Equation (1) are 11223434coscoscos0RRRθθθ+−=(2a) 11223434sinsinsin0RRRθθθ+−=(2b) Differentiating Equations (2) with respect to the input position θ2gives 223434343434sinsincos0RRRθθ θθ′′−+−=(3a) 223434343434coscossin0RRRθθ θθ′′−−=(3b)