Homework3sol.fall10

Homework3sol.fall10 - ME 560 Kinematics Fall Semester 2010...

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1 ME 560 Kinematics Fall Semester 2010 Homework No. 3 Wednesday, September 8th Consider Problem 3.15, see Figure P3.15, page 160. For this homework, only solve for the angular velocity of link 3 and the angular velocity of link 4. To solve the problem use the following two methods: (i) The method of kinematic coefficients. (ii) The method of instantaneous centers of velocity. Draw the Kennedy circle and list the primary and the secondary instant centers. Compare the answers obtained from Part (i) with the answers obtained from Part (ii).
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2 Solution to Homework Set 3. Position Analysis. The known link lengths and angles are 1 R 125 mm, = 2 R7 5 m m , = o 1 0, θ= and o 2 150 . The unknown position variables can be calculated from trigonometry. Consider the triangle 24 O AO , see Figure 1 below. Using the law of cosines gives 22 2 34 1 2 1 2 2 RR R 2 R R c o s = +− θ (1) where the input angle o 2 150 . Substituting the given numerical data into this equation gives ( ) () ( ) () 2 34 R 125 75 2 125 75 cos150 =+ ° (2a) Therefore, the distance from the ground pin 4 O to pin A (connecting links 2 and 3) is 34 R 193.62 mm = (2b) Using the law of sines gives 34 1 4 2 R R sin O AO sin O O A = ∠∠ (3a) Substituting the numerical data into this equation gives 14 2 34 Rs i n OOA 125 sin150 sin O AO R 193.62 ° ∠= = (3b) Therefore, the angle is O AO 18.83 (3c) and 33 4 (30 18.83 ) 11.17 θ=θ =− °− °=− ° (3d) Velocity Analysis. (i) The method of kinematic coefficients. A suitable choice of vectors for the inverted slider-crank mechanism are shown in Figure 1. Figure 1. The Vectors for a Kinematic Analysis of the Mechanism.
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3 The vector loop equation (VLE) for the mechanism, see Figure 1, can be written as 1 I? ? RR R 0 23 4 √√ + −= (4) It is important to note that the vector R 34 is from a point on link 3 (that is, pin A) to a point on link 4 (that is, the ground pin 4 O ). Therefore, the magnitude and the direction of this vector will change with a change in the input position of the mechanism. Since the angle 34 3 , θ= θ see Eq.
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Homework3sol.fall10 - ME 560 Kinematics Fall Semester 2010...

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