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Homework3asol.spring11

# Homework3asol.spring11 - ME 352 Machine Design I Spring...

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1 ME 352 - Machine Design I Name of Student ________________________________ Spring Semester 2011 Lab Section Number ____________________________ Homework No. 3 (30 points). Due at the beginning of lecture on Friday, February 4th. Consider Problem 3.15, see Figure P3.15, page 160. For this homework, only solve for the angular velocity of link 3 and the sliding velocity of link 3 relative to link 4. You do not need to determine the velocity of point B. To solve this problem use the following two methods: (i) The method of kinematic coefficients. Begin by clearly documenting your solution to the position analysis problem. Then clearly show all the steps for the method of kinematic coefficients beginning with the vector loop equation. List the unknown variables, and any constraint equations. Use Cramers rule to determine the first-order kinematic coefficients. Specify the determinant of the coefficient matrix. Also, determine when the mechanism is in a singular configuration. (ii) The method of instantaneous centers of velocity. Draw the mechanism to a good scale and accurately measure the locations of the instantaneous centers of velocity. Then use the appropriate measurements to determine the first-order kinematic coefficients of the mechanism. (iii) Compare the answers for the angular velocity of link 3 and the sliding velocity of link 3 relative to link 4 that you obtained from Part (i) with the answers that you obtained from Part (ii).

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2 Solution to Homework Set 3 (30 Points). Position Analysis. The known link lengths and angles for the inverted slider-crank mechanism 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 shown in Figure 1. Figure 1. Vectors for the inverted slider-crank mechanism. (i) 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 = ( 2 b ) (ii) 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) that is, the angle O AO 18.83 ° ( 3 c ) Therefore, the angular position of link 3 is 33 4 (30 18.83 ) 11.17 θ=θ =− °− °=− ° (4a) or 4 348.83 θ =θ = ° (4b)
3 Velocity Analysis. (i) The method of kinematic coefficients.

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Homework3asol.spring11 - ME 352 Machine Design I Spring...

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