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# ch05 - Solutions to Chapter 5 Exercise Problems Problem 5.1...

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- 178 - Solutions to Chapter 5 Exercise Problems Problem 5.1 For the mechanism shown, do the following: a) Write the vector equation of the above linkage. b) Write the x and y displacement equations. c) Find the velocity component equations. d) Find the acceleration component equations. A B θ C a b c φ Solution Position Analysis a + b = c In component form acos ± + bcos ² = ccos0 asin ± + bsin ² = csin0 Substituting in the constant numbers acos ± + bcos ² = c asin ± + bsin ² = 0 Velocity Analysis ± a ˙ ² sin ²± b ˙ ³ sin ³ = ˙ c a ˙ ² cos ² + b ˙ ³ cos ³ = 0 Acceleration Analysis ± a( ˙ ˙ ² sin ² + ˙ ² 2 cos ² ) ± b( ˙ ˙ ³ sin ³ + ˙ ³ 2 cos ³ ) = ˙ ˙ c a( ˙ ˙ ² cos ²± ˙ ² 2 sin ² ) + b( ˙ ˙ ³ cos ³± ˙ ³ 2 sin ³ ) = 0

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- 179 - Problem 5.2 In the mechanism in Problem 5.1, determine ˙ ± analytically for the following values: a = 1 cm , b = 4 cm , ± = 60 ˚ , ˙ ± = 10 rad / sec Solution A B C 2 3 4 φ θ r 1 r 2 r 3 The analysis can be conducted using the equations in Table 5.4 with M=2, J=3 The known input information is: ± 1 = 0 ° ; ± 2 = ± = 60 ° ; ˙ ± 2 = ˙ ± = 10 rad/sec; r 2 = AB = 1 cm; r 3 = BC = 4 cm; r 4 = 0 cm; Start with the position analysis, and first compute constants A and B: A = 2r 4 (cos ± 1 cos ± 4 + sin ± 1 sin ± 4 ) ² 2r 2 (cos ± 1 cos ± 2 + sin ± 1 sin ± 2 ) = ± 2*1*(cos0 ° cos60 ° + sin0 ° sin60 ° ) = ± 1 B = r 2 2 + r 4 2 ± r 3 2 ± 2r 2 r 4 (cos ² 2 cos ² 4 + sin ² 2 sin ² 4 ) = 1 2 ± 4 2 = ± 15 The desired configuration of the linkage corresponds to the position of the slider with the larger x coordinates. Therefore ± = + 1 . Then r 1 = ± A + ² A 2 ± 4B 2 = ± ( ± 1) + ( ± 1) 2 ± 4( ± 15) 2 = 4.405 Then ± 3 is given by ± 3 = tan ² 1 [ r 1 sin ± 1 + r 4 sin ± 4 ² r 2 sin ± 2 r 1 cos ± 1 + r 4 cos ± 4 ² r 2 cos ± 2 ] = tan ² 1 [ ² 1 ³ sin60 ° 4.405 ² 1 ³ cos60 ° ] = ² 12.504 ° For the velocity ˙ ± , solve the linear set of velocity equations,
- 180 - cos ± 1 r 3 sin ± 3 sin ± 1 ² r 3 cos ± 3 ³ ´ µ ˙ r 1 ˙ ± 3 · ¸ ¹ º » ¼ = ² r 2 ˙ ± 2 sin ± 2 r 2 ± 2 cos ± 2 · ¸ ¹ º » ¼ or cos0 ° 4sin( ± 12.504 ° ) sin0 ° ± 4cos( ± 12.504 ° ) ² ³ ´ µ ˙ r 1 ˙ 3 · ¸ ¹ º » ¼ = ± 1 ½ 10sin60 ° 1 ½ 10cos60 ° · ¸ ¹ º » ¼ then 1 ± 0.866 0 ± 3.905 ² ³ ´ µ ˙ r 1 ˙ 3 · ¸ ¹ º » ¼ = ± 8.660 5 · ¸ ¹ º » ¼ or ˙ r 1 ˙ ± 3 ² ³ ´ µ · = ¸ 9.768 ¸ 1.280 ² ³ ´ µ · Therefore, ˙ ± = ² ˙ ³ 3 = 1.280 rad/sec, CW. Problem 5.3 In the mechanism shown, ˙ s = ± 10 in/ s and ˙ ˙ s = 0 for the position corresponding to ± = 60 ˚ . Find ˙ ± and ˙ ˙ ± for that position using the loop equation approach. 3 4 φ 2 s 10 inches Solution The vector equation is r 3 = r 2 + r 1 In component form, this equation becomes: r 3 cos ± 3 = r 2 cos ± 2 + r 1 cos ± 1 r 3 sin ± 3 = r 2 sin ± 2 + r 1 sin ± 1

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- 181 - 3 4 φ 2 s r 1 r 2 r 3 Substituting the constant values ± 1 = 0 ° and ± 2 = 90 ° gives r 3 cos ± = r 1 r 3 sin ± = r 2 The component equations for velocity are: ˙ r 3 cos ± ² r 3 ˙ ± sin ± = ˙ r 1 ˙ r 3 sin ± + r 3 ˙ ± cos ± = 0 The component equations for acceleration are: ˙ ˙ r 3 cos ± ² r 3 ˙ ± sin ±² r 3 ˙ ˙ ± sin ± ² r 3 ˙ ± 2 cos ± = ˙ ˙ r 1 ˙
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ch05 - Solutions to Chapter 5 Exercise Problems Problem 5.1...

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