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

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Unformatted text preview: 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. B φ b θ c Solution Position Analysis a+b=c In component form a cos + bcos = ccos0 asin + bsin = csin 0 Substituting in the constant numbers a cos + bcos = c asin + bsin = 0 Velocity Analysis a ˙ sin b ˙ sin = c ˙ ˙ cos + b ˙ cos = 0 a Acceleration Analysis a(˙˙sin + ˙ 2 cos ) b(˙˙sin + ˙ 2 cos ) = c ˙˙ ˙˙ cos ˙ 2 sin ) + b(˙˙ cos ˙ 2 sin ) = 0 a( C a A - 178 - Problem 5.2 In the mechanism in Problem 5.1, determine ˙ analytically for the following values: a = 1 cm, Solution B 2 A r2 θ r1 r3 b = 4cm, = 60˚, ˙ = 10 rad / sec 3 φ C 4 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; r4 = 0 cm; r2 = AB = 1 cm; r3 = BC = 4 cm; Start with the position analysis, and first compute constants A and B: A = 2r4(cos 1 cos 4 + sin 1 sin 4) 2r2 (cos 1 cos 2 + sin 1 sin 2 ) = 2 *1*(cos0° cos60° + sin0°sin60°) = 1 2 2 2 B = r2 + r4 r3 2r2 r4(cos 2 cos 4 + sin 2 sin 4 ) = 12 42 = 15 The desired configuration of the linkage corresponds to the position of the slider with the larger x coordinates. Therefore = + 1 . Then r1 = A + Then 3 A2 4B = ( 1) + ( 1)2 4( 15) = 4.405 2 2 is given by 3= tan 1[ r1sin 1 + r4 sin r1 cos 1 + r4 cos 4 4 r2 sin r2 cos 2 ]= 2 tan 1[ 1 sin60° ] = 12.504° 4.405 1 cos60° For the velocity ˙ , solve the linear set of velocity equations, - 179 - cos sin or 1 1 r3 sin 3 ˙1 r r2 ˙ 2 sin = r3 cos 3 ˙ 3 r2 2 cos 2 2 cos0° sin0° then 1 0 or r ˙1 ˙3 = 1 10sin60° 4sin( 12.504°) ˙1 r = 1 10cos60° 4 cos( 12.504°) ˙3 0.866 ˙1 r = 3.905 ˙ 3 9.768 1.280 8.660 5 Therefore, ˙ = ˙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 ˙ ˙ ˙ and ˙˙ for that position using the loop equation approach. s 3 4 = 60˚. Find 2 10 inches φ Solution The vector equation is r3 = r2 + r1 In component form, this equation becomes: r3 cos 3 = r2 cos 2 + r1 cos 1 r3 sin 3 = r2 sin 2 + r1 sin 1 - 180 - s 4 r2 r1 3 2 φ r3 Substituting the constant values r3 cos = r1 r3 sin = r2 1 = 0° and 2 = 90° gives The component equations for velocity are: r r3 cos r3 ˙ sin = ˙1 ˙ ˙ cos = 0 r3 sin + r3 ˙ The component equations for acceleration are: r r3 ˙˙ cos 2˙3 ˙ sin r3˙˙sin r r˙ sin + 2˙3 ˙ cos + r3˙˙ cos ˙3 The known input information is: = 60° so r3 = r2 = 10 = 11.547 sin60° sin r1 = r3 cos = 11.547cos60° = 5.774 Solve the velocity equations: cos sin or cos60° sin60° or 0.5 10 ˙3 r = 0.866 5.774 ˙ then 5 r ˙3 ˙ = 0.75 11.547sin60° ˙3 r = 11.547cos60° ˙ 10 0 r3 sin r3 cos r r3 ˙1 ˙ = ˙ 0 10 0 r2 = 10 in rs ˙1 = ˙ = 10 in / s ˙˙ = ˙˙ = 0 r1 s r3 ˙ 2 cos = ˙˙ r1 r3 ˙ 2 sin = 0 Therefore ˙ = 0.75 rad/sec, CCW. - 181 - Solve the acceleration equations: cos sin or cos60° sin60° or then 11.547sin60° ˙˙ r3 2( 5)(0.75)sin60° + 11.547 0.752 cos60° = 11.547cos60° ˙˙ 2( 5)(0.75)cos60° + 11.547 0.752 sin60° r3 sin r3 cos r3 r1 r ˙˙ ˙˙ + 2˙3 ˙ sin + r3 ˙ 2 cos = ˙˙ 2˙3 ˙ cos + r3 ˙ 2 sin r 3.248 0.5 10 ˙˙ r3 = 9.375 0.866 5.774 ˙˙ 6.495 r ˙˙3 = ˙˙ 0.650 Therefore ˙˙ = 0.650 rad / sec2 , CCW. Problem 5.4 In the mechanism in Problem 5.3 assume that ˙ is 10 rad/s CCW. Use the loop equation approach to determine the velocity of point B4 for the position defined by = 60˚. Solution B r1 3 r3 φ The vector equation is: r3 = r2 + r1 In component form, this equation becomes: r3 cos 3 = r2 cos 2 + r1 cos 1 r3 sin 3 = r2 sin 2 + r1 sin 1 Substituting the constant values r3 cos = r1 r3 sin = r2 The component equations for velocity are: 1 4 r2 2 = 0 and 2 = 90° gives - 182 - r r3 cos r3 ˙ sin = ˙1 ˙ ˙ cos = 0 r3 sin + r3 ˙ The known input information is: = 60° ; so ˙ = 10 ; r2 = 10 inches; r3 = r2 = 10 = 11.547 sin60° sin r1 = r3 cos = 11.547cos60° = 5.774 Solve for the velocities: r3 ˙ cos = 11.547 10cos60° = 66.667 sin60° sin ˙ sin = 66.667 cos60° 11.547 10sin60° = 133.333 r3 r1 = ˙3 cos ˙r r3 = ˙ Therefore vB4 = 133.333 cm/sec. Problem 5.5 In the mechanism given, point A is moving to the right with a velocity of 10 cm/s. Use the loop equation approach to determine the angular velocity of link 3. Link 3 is 10 cm long, and is 120˚ in the position shown. 4 B 3 φ A 2 Solution The vector loop equation is: r2 = r1 + r3 In component form, this equation becomes: r2 cos r2 sin 2 = r cos 1 + r3 cos 3 1 2 = r sin 1 + r3 sin 3 1 1 Substituting the constant values = 0, 3 = , and 2 = 90° gives - 183 - 4 B r3 r2 3 A φ r1 2 0 = r1 + r3 cos r2 = r3 sin The component equations for velocity are: 0 = ˙1 r3 ˙ sin r r2 = r3 ˙ cos ˙ The known input information is: = 120° ; so r2 = r3 sin = 10sin120 = 8.66 r1 = r3 cos = 10cos120° = 5.00 Solve for ˙ : ˙= r 10 ˙1 = = 1.155 10sin120° r3 sin r3 = 10 cm; r1 = vA = 10 cm/sec ˙ Therefore ˙ = 1.155 rad/sec, CCW. Problem 5.6 Resolve Problem 5.5 if is 150˚. Solution - 184 - 4 B r2 r3 3 φ r1 A 2 Position Analysis The basic loop equation is: r1 + r3 = r2 In component form r1 cos 1 + r3 cos = r2 cos 2 r1 sin 1 + r3 sin = r2 sin 2 Substituting in the constant numbers r1 cos(0 ) + r3 cos = r2 cos( 90°) r1 sin(0 ) + r3 sin = r2 sin(90°) or r1 + r3 cos = 0 r3 sin = r2 Then, r1 = r3 cos = 10 cos(150°) = 8.66 r2 = r3 sin = 10 sin(150°) = 5.0 Velocity Analysis r1 = r3 ˙3 sin ˙ ˙3 = ˙ r= 10 1 = 2.0 rad / sec r3 sin 10 sin(150°) r2 = r3 ˙3 cos = 10( 2.0)cos(150°) = 17.32 in / sec ˙ Acceleration Analysis ˙1 = r3 ˙˙3 sin + r3 ( ˙3 ) cos r˙ 2 2 ˙˙ r3 ( ˙3 ) cos r 1 0 10(2 )2 cos(150°) ˙˙3 = = = 6.928 rad / sec2 r3 sin 10 sin(150°) - 185 - ˙˙ = r3 ˙˙3 cos r2 Then r3 ( ˙3 ) sin = 10( 6.928)cos(150°) 10( 2)2 sin(150°) = 80 in / sec 2 2 vA 2 = 17.32 in / sec aA 2 = 80 in / sec2 Problem 5.7 The mechanism shown is a marine steering gear called Raphson’s slide. AB is the tiller, and CD is the actuating rod. If the velocity of rod CD is a constant 10 inches per minute to the right, use the loop-equation approach to determine the angular acceleration of the tiller. ψ = 300˚ A 2 6' C 3 4 B 1 D Solution ψ = 300˚ A r1 r2 2 r3 C 3 4 B 1 D The vector equation is: r2 = r1 + r3 In component form this equation becomes: - 186 - r2 cos r2 sin 2 = r cos 1 + r3 cos 3 1 2 = r sin 1 + r3 sin 3 1 3= Substituting the constant values q1 = 0, r2 cos = r1 r2 sin = r3 The component equations for velocity is: r2 cos r2 ˙ sin ˙ r2 sin + r2 ˙ cos ˙ = ˙1 r =0 90° gives The component equations for acceleration is: 2˙2 ˙ sin r r2 ˙˙ sin r2 ˙˙ cos r r˙ sin + 2˙2 ˙ cos + r2 ˙˙ cos ˙2 The known input information is: = 300° ; so r2 = r3 = 6 = 6.928 sin300° sin r1 = r2 cos = 6.928cos300° = 3.464 r3 = 6' ; r1 = 10 in/min; ˙ ˙˙ = 0 ; r1 r2 ˙ 2 cos r2 ˙ 2 sin = ˙˙ r1 =0 To solve the velocities: cos sin or cos300° sin300° or 0.5 6 r ˙2 = 0.866 3.464 ˙ then 6 r ˙2 = 1.25 ˙ 6.928sin300° ˙2 r = 6.928cos300° ˙ 10 0 r2 sin r2 cos r1 r ˙ ˙2 = ˙ 0 10 0 To solve the accelerations: cos sin or cos300° sin300° or 6.928sin300° ˙r2 2(6)(1.25)sin300° + 9.628 1.252 cos300° ˙ = 6.928cos300° ˙˙ 2(6)(1.25)cos300° + 9.628 1.252 sin300° r2 sin r2 cos r ˙˙ + 2˙2 ˙ sin + r2 ˙ 2 cos r1 r ˙˙2 = ˙˙ 2 ˙2 ˙ cos + r2 ˙ 2 sin r - 187 - 0.5 6 r2 ˙˙ = 0.866 3.464 ˙˙ then 10.826 r2 ˙˙ = 2.165 ˙˙ 7.578 16.875 Therefore ˙˙ = 2.165 rad / min2 CCW. Problem 5.8 Use loop equations to determine the velocity and acceleration of point B on link 2 when Make point A the origin of your reference coordinate system. 10 in. 3 = 30˚. y 1ω 4 = 1 rad sec 1α4 = 0 2 3 B θ3 4 A x Solution y 2B r3 3 θ3 r2 r1 A x The vector equation is: r3 = r1 + r2 In component form, this equation becomes: r3 cos 3 = r1 cos 1 + r2 cos 2 r3 sin 3 = r1 sin 1 + r2 sin 2 Substituting the constant values 1 = 0 and 2 = 90° gives - 188 - r3 cos 3 = r1 r3 sin 3 = r2 The component equations for velocity are: r3 cos 3 r3 ˙ 3 sin ˙ r3 sin 3 + r3 ˙3 cos ˙ 0 r 3 = ˙2 3= The component equations for acceleration are: r3 r ˙˙ cos 3 2˙3 ˙ 3 sin 3 r3˙˙3 sin r r3 ˙˙ sin 3 + 2˙3 ˙ 3 cos ˙ 3 + r3˙˙3 cos The known input information is: so 3 = 30° ; 3 3 r3 ˙ 32 cos r3 ˙ 32 sin 3=0 r2 3 = ˙˙ ˙3 = 4 =1 rad/sec; ˙˙3 = 4 = 0; r1 = 10 inches; r3 = r1 = 10 = 11.547 cos 3 cos30° r2 = r3 sin 3 = 11.547sin30° = 5.774 To solve the velocities: ˙ r3 = r3 3 sin 3 = 11.547 1 sin30° = 6.667 ˙ cos 3 cos30° ˙ 3 cos 3 = 6.667sin30° + 11.547 1 cos30° = 13.333 r2 = ˙3 sin 3 + r3 ˙r Therefore vB2 = ˙2 = 13.333 in/sec. r Solve for the accelerations: 2r3 ˙ 3 sin ˙ r3 ˙˙ = 3 + r ˙˙3 sin 3 + r ˙2 cos 3 3 33 cos 3 2 6.667 1 sin30° + 11.547 12 cos30° = 19.245 = cos30° r r r˙ = ˙˙3 sin 3 + 2˙3 ˙ 3 cos 3 + r3˙˙3 cos 3 r3 ˙ 32 sin 3 ˙2 = 19.245 sin30° + 2 6.667 1 cos30° 11.547 12 sin30° = 15.397 Therefore aB2 = ˙˙2 = 15.397 in / sec2 . r - 189 - Problem 5.9 In the mechanism shown, = 30˚ , 2 = 1 rad / s CCW, and determine the velocity and acceleration of point B on link 4. Y 3 in 2 = 0 . Use loop equations to A 2 3 θ X B Solution Y 4 A r2 3 B 4 2 r3 30˚ r1 X Position Analysis r1 + r3 = r2 In component form, r1 cos0 + r3 cos90 = r2 cos r1sin 0 + r3 sin90 = r2 sin or r1 = r2 cos r3 = r2 sin Solving for r2 and r3 , r2 = r1 / cos = 3 / cos30 = 3.464 in. r3 = r2 sin = 3.464sin30 = 1.732 in Velocity Analysis r1 + r3 = r2 ˙˙˙ In component form, - 190 - or r1 = ˙2 cos r2 ˙ sin ˙r 0 = ˙2 sin + r2 ˙ cos r r2 = r2 ˙ cos / sin ˙ r1 = ˙2 cos r2 ˙ sin ˙r r2 = 3.464(1)[cos(30) / sin(30)] = 6.00 in / sec ˙ r1 = 6.00cos30 3.464(1)sin30 = 6.928 in / sec ˙ or Acceleration Analysis r1 r r ˙˙ + ˙˙3 = ˙˙2 In component form, r r1 r ˙˙ = ˙˙2 cos 2˙2 ˙ sin r2˙˙sin 0 = ˙˙2 sin + 2˙2 ˙ cos + r2˙˙ cos r r or r2 ˙˙ = c [ 2˙r ˙ cos + r s˙˙inos 2 2 r2 ˙ 2 cos r2 ˙ 2 sin r2 ˙ 2 sin ˙˙ = [˙˙2 r2 ˙ 2 ]cos r1 r or r [2˙2 ˙ + r2˙˙]sin 3.464(0)cos30 3.464(1)2 sin30 = 24.249 in / sec 2 sin30 ˙˙ = [24.249 3.464(1)2] cos30 [2( 6.00)(1) + 3.464(0)]sin30 r1 = 18.000 + 6.000 = 24.000 in / s2 r2 ˙˙ = [ 2(6.00)(1)cos30 Therefore, vB4 = r1 = 6.928 in / s ˙ aB4 = ˙˙ = 24.000 in / s2 r1 - 191 - Problem 5.10 In the mechanism for Problem 5.9, assume that vB4 is a constant 10 in/s to the left and Use loop equations to determine the angular velocity and acceleration of link 3. Solution Y A r2 vB4 B 4 3 r1 2 1θ 3 is 45˚. . r3 X The basic loop equations is: r2 = r1 + r3 In component form this equation becomes: r2 cos r2 sin 2 = r cos 1 + r3 cos 3 1 2 = r sin 1 + r3 sin 3 1 1 Substituting the constant values (r1 = 3, r2 cos r2 sin When and 2 2= r 1 2= r 3 = 0, 3 = 90) gives = 30, 2 r2 = 3 / cos r3 = r2 sin 2 = 3 / cos30° = 3.464 = 3.464sin30° = 1.732 For the velocities, r2 = ˙1 + ˙3 ˙rr or in component form, r2 cos ˙ r2 sin ˙ r2 ˙ 2 sin 2 + r2 ˙ 2 cos 2 = ˙1 r 2 =0 2 Substituting the know values gives 0.866 (3.464)(0.5) ˙2 r ˙2 = 0.5 (3.464)(0.866) or 10 0 - 192 - 8.661 r ˙2 ˙ 2 = 1.443 Therefore, ˙ 2 = 1.443 rad/sec, CCW. For the accelerations, r r1 r3 ˙˙2 = ˙˙ + ˙˙ or in component form, ˙˙ cos r2 ˙˙ sin r2 2˙2 ˙2 sin r r2 2 + 2˙ ˙ 2 cos 2 r2˙˙2 sin 2 + r2˙˙2 cos 2 2 2 r2 ˙ 2 cos 2 ˙ 2 sin r2 2 = ˙˙ r1 2 =0 2 Substituting the know values gives 0.866 (3.464)(0.5) ˙˙2 r 2( 8.661)(1.443)sin30° + (3.464)(1.4432 )cos30° = 0.5 (3.464)(0.866) ˙˙2 2( 8.661)(1.443)cos30° + (3.464)(1.4432 )sin30° or 7.213 r2 ˙˙ = ˙˙2 7.216 Therefore, ˙˙2 = 7.216 rad / sec 2 , CCW. Problem 5.11 For the mechanism in the position shown, link 2 is the driver and rotates with a constant angular velocity of 100 rad/s CCW. Write vector loop equations for position, velocity, and acceleration, and solve for the velocity and acceleration of point C on link 4. AB = 0.9" AD = 1.7", BC = 2.6", h = 0.8", C 3 B 2 φ A θ1 h D 4 1 = 6˚, = 120˚ Solution - 193 - Y 3 B 2 r' 3 r3 r1 r4 r' 4 C 4 r2 A θ'4 D θ4 X The basic loop equation for the mechanism is: r1 + r4 = r2 + r3 In component form of this equation becomes: r1 cos 1 + r4 cos 4 = r2 cos 2 + r3 cos 3 r1sin 1 + r4 sin 4 = r2 sin 2 + r3 sin 3 r1, r2 , r4, and 1 are constants, and 2 =+ 1 and 4 = 90° + 3. The component equations for the velocities are: r4 ˙ 4 sin 4 = r2 ˙ 2 sin 2 + ˙3 cos 3 r3 ˙ 3 sin r ˙ 4 cos 4 = r2 ˙ 2 cos 2 + ˙3 sin 3 + r3 ˙ 3 cos 3 r4 r ˙3 = ˙4 , ˙2 = ˙ . And the component equations for the accelerations are: r4˙˙4 sin 4 r4 ˙ 2 cos 4 4 = r2˙˙2 sin 2 r2 ˙ 2 cos 2 + ˙˙ cos 3 2˙3 ˙ 3 sin 3 r3˙˙3 sin 3 r3 ˙2 cos 3 r3 r 2 3 ˙˙4 cos 4 r4 ˙ 2 sin 4 r4 4 = r2˙˙2 cos 2 r2 ˙2 sin 2 + ˙˙3 sin 3 + 2˙3 ˙ 3 cos 3 + r3˙˙3 cos 3 r3 ˙ 2 sin 3 r r 2 3 The known input information is: r1 = 1.7 , r2 = 0.9 , r3 = 2.6 , h=0.8, 1 = 6° , 2 3 = 126° , = 120° The position equations cannot be solved directly because there are too many unknowns. However, ' ' ' if we rewrite the equations in terms of r4 and r3 , they can be solved. Here, r4 is perpendicular to ' ' ' r3 , and r3 is measured along r3 from B to the intersection of r4 and r3 . Then the magnitude of r4 '4 = 90° + 3 . After substituting '4 = 90° + 3 into the equations for position and is h = 0.8. Also, rewriting the equations, we get: r1 cos r1sin 1 1 r2 cos r2 sin 3, 2 2 ' = r4 sin ' 3 + r3 cos 3 3 + r' sin 3 3 (1) (2) = r'4 cos To eliminate square Eqs. (1) and (2) and add the results. This gives: - 194 - 2 2 r1 + r2 r' 2 2r1r2 cos( 4 1 2) = r' 2 3 Substitute or 2 =+ 1 into the equation above. Then 2 ' r3 = r12 + r2 r' 2 2r1r2 cos 4 ' r3 = 1.72 + 0.92 0.82 2(1.7)(0.9)cos120° = 2.14 To solve for sin 3 , use the trigonometric identities 2tan 3= 3 2 3 1 + tan2 1 tan2 2 3 cos 3= 2 3 1 + tan2 2 Let t = tan 2 3 , and substitute the trigonometric identities above into Eq. (1): (3) ' ' A(1 + t 2 ) r3(1 t2 ) r4 (2t) = 0 Where A = r1 cos 1 r2 cos 2 Collecting terms in Eq (3) gives: ' ' ' (A + r3 )t 2 2r4 t + (A r3 ) = 0 The roots are: t= Where ' r4 + r' 2 A2 + r' 2 4 3 ' A + r3 = ±1 . Then, 2 2 2 t1 = 0.8 + 0.8 2.22 + 2.14 = 0.307 2.22 + 2.14 and t1 = 0.8 0.82 2.222 + 2.142 = 0.0597 2.22 + 2.14 for the problem, we must first compute a value of 3 To determine the correct value of value of t using. 3 = 2tan 1 t for each - 195 - Next substitute both values of of 3 satisfying Eq. (2). Then, 3 = 2tan 1 0.307 3 into Eq. (3). The correct value of 3 = 2tan 1 0.0597 will correspond to the value = 34.13° , or Therefore, = 6.83° We know that 3 = 6.83° . = -1. Before solving for 4 , solve for r4 from geometry. Then, ' r4 = r' 2 +(r3 r3 )2 4 = 0.82 + (2.6 2.14)2 = 0.923 and 4 = tan 1 h ' r3 r3 + 3= tan 1 (2.6 0.8 .14 ) + 6.83° = 66.93° 2 r2 ˙ 2 sin 2 r2 ˙ 2 cos 2 To solve for the velocities, substitute ˙ 3 = ˙ 4 , into the equation for velocity, and rewrite in matrix equation form: cos sin or cos6.83° sin6.83° or 3 3 r3 sin 3 + r4 cos 3 ˙3 r = r3 cos 3 + r4 sin 3 ˙ 3 (0.9)(100)sin126° 2.6sin6.83° + 0.923cos6.83° ˙3 r = (0.9)(100)cos126° 2.6cos6.83° + 0.923sin6.83° ˙3 95.21 r ˙3 = ˙3 14.08 To solve for the accelerations, substitute ˙˙3 = ˙˙4 , into the equation for acceleration, and rewrite in matrix equation form: cos sin = or cos6.83° 2.6sin6.83° + 0.923cos6.83° ˙˙3 r sin6.83° 2.6cos6.83° + 0.923sin6.83° ˙˙3 (0.923)( 14.08)2 cos66.93° + (0.9)(100)2 cos126° +2(95.21)( 14.08)sin6.83° + (2.6)( 14.08)2 cos6.83° = (0.923)( 14.08)2 sin66.93° + (0.9)(100)2 sin126° 2(95.21)( 14.08)cos6.83° + (2.6)( 14.08)2 sin6.83° or 132 r3 ˙˙ = ˙˙3 4185 r3 sin 3 + r4 cos 3 ˙r3 ˙ r3 cos 3 + r4 sin 3 ˙˙3 3 2 cos 2 ˙ r4 ˙ 4 4 + r2˙ 2 sin 2 + r ˙ 2 cos 2 r4 ˙ 2 sin 4 r2˙˙2 cos 2 + r2 ˙ 2 sin 4 2 3 2 r3 2 + 2˙ ˙3 sin 3 + r3 ˙3 cos 3 2 r3 2 2˙ ˙3 cos 3 + r3 ˙3 sin 3 To solve the velocity and acceleration of C in link 4: - 196 - r4c = r4(cos 4i + sin 4 j) v4c = r4c = r4 ˙ 4( sin 4i + cos 4 j) ˙ = (0.923)( 14.08)( sin66.93°i + cos66.93°j) = 13 66.93°in / sec a4c = ˙˙4c = r4˙˙4( sin 4i + cos 4 j) + r4 ˙ 2 ( cos 4i sin 4 j) r 4 = (0.923)(4185)( sin66.93°i + cos66.93°j) + (0.923)( 14.08)2( cos66.93°i sin66.93°j) = 3625.5i + 1345.3j = 3867 159.6°in / sec2 Therefore, v4c = 13 66.93°in / sec and a4c = 3867 159.6°in / sec2 Problem 5.12 For the mechanism in the position shown, link 2 is the driver and rotates with a constant angular velocity of 50 rad/s CCW. Write vector loop equations for position, velocity, and acceleration, and solve for the velocity and acceleration of point C on link 3. B 3 d 2 h A φ C 4 = 60˚ d = 0.9" h = 0.8" AB = 1.8" Solution: The vector equation is: r1 + r4 = r2 + r3 In component form of this equation becomes: - 197 - Y B 3 d C 4 r3 2 r2 φ r4 r1 X A r1 cos 1 + r4 cos 4 = r2 cos 2 + r3 cos 3 r1sin 1 + r4 sin 4 = r2 sin 2 + r3 sin 3 Substituting the constant values r1 = r2 cos + r3 sin r4 = r2 sin r3 cos The component equations for velocity are: r1 = ˙2 cos r2 ˙ sin + r3 ˙ cos ˙r 0 = ˙2 sin + r2 ˙ cos + r3 ˙ sin r The component equations for acceleration are: r r1 r ˙˙ = ˙˙2 cos 2˙2 ˙ sin r2˙˙sin 0 = ˙˙2 sin + 2˙2 ˙ cos + r2 ˙˙ cos r r The known input information is: r3 = d = 0.9 r4 = 0.8 , = 60° , ˙ = 50 r2 ˙ 2 cos + r3˙˙ cos r3 ˙ 2 sin ˙ 2 sin + r3˙˙sin + r3 ˙ 2 cos r2 1=0, 4 = 90° , 2 = , and 3= 90° gives Solving for the positions: r2 = r4 + r3 cos = 0.8 + 0.9cos(60°) = 1.44 sin(60°) sin r1 = r2 cos + r3 sin = 1.44cos(60°) + 0.9sin(60°) = 1.5 Solving for the velocities: r2 ˙ cos + r3 ˙ sin sin (1.44)(50)cos60° + (0.9)(50)sin60° = 86.57 = sin60° r1 = ˙2 cos r2 ˙ sin + r3 ˙ cos ˙r = ( 86.57)cos60° (1.44)(50)sin60° + (0.9)(50)cos60° = 83.14 r2 = ˙ - 198 - Solving for the accelerations: r2 ˙ 2 sin + r3˙˙ sin + r3 ˙ 2 cos sin 2 2 = 2( 86.57)(50)cos60° (1.44)(50) sin60° + (0.9)(50) cos60° = 7299 sin60° ˙˙ = ˙˙2 cos 2˙2 ˙ sin r2˙˙sin r2 ˙2 cos + r3˙˙ cos r3 ˙ 2 sin r1 r r = 7299cos60° 2( 86.57)(50)sin60° (1.44)(50)2 cos60° (0.9)(50)2 sin60° = 7398 r2 ˙˙ = Therefore, and v3c = ˙1 = 83.14 in/sec r a3c = ˙r1 = 7398 in / sec2 ˙ Problem 5.13 In the mechanism in shown, link 3 slides on link 2, and link 4 is pinned to link 3 and slides on the frame. If 1 2 = 10 rad/s CCW (constant), use loop equations to find the acceleration of Link 4 for the position defined by = 90˚. B 4 3 2 2r2 ˙ cos + r2˙˙ cos ˙ 1 cm φ A Solution The basic loop equations is: r2 = r1 + r3 In component form this equation becomes: r2 cos r2 sin 2 = r cos 1 + r3 cos 3 1 2 = r sin 1 + r3 sin 3 1 1 Substituting the constant values (r3 = 1, r2 cos r2 sin 2= r 1 2= r 3 = 0, and 3 = 90˚) gives =1 - 199 - Y B 4 3 2 r2 r3 φ A r1 When and 2 X = 90˚, 2 = 1 / 1= 1 2 r2 = 1/ sin r1 = r2 cos = 1• 0 = 0 The equation for the velocities is, r2 = ˙1 + ˙3 ˙rr or in component form, r2 cos ˙ r2 sin ˙ r2 ˙ 2 sin 2 + r2 ˙ 2 cos 2 = ˙1 r 2 =0 2 Substituting the know values gives (1)(10)(1) = ˙1 = 10 r r2 = 0 ˙ The equation for the accelerations is, r r1 r3 ˙˙2 = ˙˙ + ˙˙ or in component form, ˙˙ cos r2 ˙˙ sin r2 2˙2 ˙2 sin r r2 2 + 2˙ ˙ 2 cos 2 r2˙˙2 sin 2 + r2˙˙2 cos 2 2 2 r2 ˙ 2 cos 2 r2 ˙ 2 sin 2 = ˙˙ r1 2 =0 2 Substituting the know values gives r1 ˙˙ = 0 ˙˙ (1)(10)2 = 0 r2 Therefore, 1 aB4 = ˙˙ = 0 cm / sec2 . r1 - 200 - Problem 5.14 For the mechanism in the position shown, the cam (link 2) rotates with an angular velocity of 200 rad/s. Write the vector loop equations for position, velocity, and acceleration and determine the angular velocity and acceleration of the follower (link 3). Use = 60˚ and neglect the follower thickness (i.e., assume that it is zero). 3 C 2 A AD = 6.5 in AB = 1.0 in r = 2.0 in r B φ D ω2 Solution Y 3 C 2 A B r3 r2 r1 r4 D X The vector equation is: r1 + r4 = r2 + r3 In component form, this equation becomes: r1 cos 1 + r4 cos 4 = r2 cos 2 + r3 cos 3 r1sin 1 + r4 sin 4 = r2 sin 2 + r3 sin 3 Now, r1 , r2 , r3 , and 1=0 are constants, and 2 =, 4 = 90° + 3. The component equations for velocity are: r4 cos ˙ r4 sin ˙ r4 ˙ 4 sin 4 + r4 ˙ 4 cos 4 = r2 ˙ 2 sin 2 r3 ˙ 3 sin 4 = r2 ˙ 2 cos 2 + r ˙ 3 cos 3 4 3 3 ˙3 = ˙4 , ˙2 = ˙ . The component equations for acceleration are: - 201 - ˙˙ cos r4 ˙˙ sin r4 2˙4 ˙ 4 sin r r 4 + 2˙4 ˙ 4 cos 4 r4˙˙4 sin ˙ 4 + r4˙ 4 cos 4 4 4 r4 ˙ 2 cos 4 r4 ˙ 2 sin 4 r2˙˙2 sin 2 r2 ˙ 2 cos 2 r3˙˙3 sin 3 r3 ˙ 2 cos 3 2 3 2 2 ˙ ˙ 4 = r2˙ 2 cos 2 r2 ˙2 sin 2 + r3˙ 3 cos 3 r3 ˙ 3 sin 3 4= The known input information is: r1 = 6.5 r2 = 1.0 r3 = 2.0 4 = 60° = 90° + 3 ˙ = 200 into the equation for position and rewrite the (1) (2) To solve for the positions, substitute equation. This gives: r1 r2 cos r2 sin To eliminate 2 2 = r4 sin 3 + r3 cos 3 3 + r3 sin 3 = r4 cos 3, square Eqs. (1) and (2) and add. The result is: 2 2 = r4 2 2 2 r1 + r2 r3 2r1r2 cos Substitute 2 = , into the equation above to get, 2 2 2 r4 = r1 + r2 r3 2r1r2 cos = 6.52 +1.0 2 2.02 2(6.5)(1.0)cos60° = 5.72 To solve for sin 3 , use the trigonometric identities 2tan 3= 3 2 3 1 + tan2 1 tan2 2 3 cos 3= 2 3 1 + tan2 2 Let t = tan 2 3 , and substitute the trigonometric identities above into Eq. (1). This gives: (3) A(1 + t 2 ) r3(1 t2 ) r4 (2t) = 0 where A = r1 r2 cos 2 Collecting terms in Eq (3) gives: (A + r3 )t 2 2r4 t + (A r3 ) = 0 The roots are: t= or r4 + 2 2 r4 A2 + r3 5.72 + = A + r3 5.72 2 62 + 2 2 6 +2 - 202 - t1 = 0.82 and t 2 = 0.61. Where = ±1 . To determine the correct value of of 3 for each value of t using. or 3 = 2tan 1 t 3 = 2tan 1 0.82 for the problem, we must first compute a value = 78.7° and 3 = 2tan 1 0.61 = 62.77° Next substitute both values of 3 into Eq. (3). The correct value of of 3 satisfying Eq. (2). In this problem, 3 = 62.77° . will correspond to the value To solve the velocities, substitute ˙ 3 = ˙ 4 , into the equation for velocity, and rewrite in the matrix equation form: cos sin or cos152.77° sin152.77° or 196.39 r ˙4 ˙ 4 = 1.69 4 4 r4 cos r4 cos 3 + r sin 3 3 3 r3 cos 3 r r2 ˙ 2 sin ˙4 ˙ 4 = r2 ˙ 2 cos 2 2 (1)(200)sin60° 5.72cos62.77° + 2sin62.77° ˙4 r = (1)(200)cos60° 5.72 sin62.77° 2 cos62.77° ˙ 4 To solve the accelerations, substitute ˙˙3 = ˙˙4 , into the equation for acceleration, and rewrite in the matrix equation form: cos sin = or cos152.77° 5.72cos62.77° + 2sin62.77° ˙r4 ˙ sin152.77° 5.72 sin62.77° 2 cos62.77° ˙˙4 2 cos62.77° ( 2( 1.69) 1)(200)2 cos60° 2 = +2(196.39)( 1.69)cos62.77° 5.72( 1.69) sin62.77° 2 sin62.77° (1)(200)2sin60° 2( 1.69) +2(196.39)( 1.69)sin62.77° + 5.72( 1.69)2cos62.77° or r4 1.615 10 4 ˙˙ = ˙˙4 7.10 10 3 Therefore, ˙ 4 = 1.69 rad/sec CW, and ˙˙4 = 7.10 103 rad / sec2 CCW. r r3 sin 3 r4 cos 3 ˙˙4 ˙˙4 r3 cos 3 r4 sin 3 4 ˙2 cos 3 r2˙˙2 sin 2 r2 ˙ 2 cos r3 3 2 r3 ˙ 2 sin 3 + r2˙˙2 cos 2 r2 ˙ 2 sin 3 2 4 r4 ˙ 2 sin 4 ˙ 2 + 2r4 ˙ 4 sin 3 + r4 ˙ 2 cos 4 r 2 + 2˙4 ˙ 4 cos 3 3 3 - 203 - Problem 5.15 In the mechanism shown, link 3 is perpendicular to link 2. Write the vector loop equations for position and velocity. If the angular velocity of link 2 is 100 rad/s CCW, use the vector loop equations to solve for the velocity of point C4 for the position corresponding to = 60˚. B 2 A Solution Y B 2 r2 A The basic loop equations is: r2 = r1 + r3 In component form this equation becomes: r2 cos r2 sin 2 = r cos 1 + r3 cos 3 1 2 = r sin 1 + r3 sin 3 1 1 10" 3 C φ 4 r3 r1 3 C 4 X φ Substituting the constant values (r3 = 10, r2 cos r2 sin When 2 = 0), and 3 = 2 + 90° gives r3 sin 2 = r cos 2 3 2= r 1 2 = 60, r2 = 10 cos60° / sin60° = 5.774 and r1 = r2 cos 2 + r3 sin 2 = 5.774 cos60° + 10sin60° = 11.547 For the velocities, - 204 - r2 = ˙1 + ˙3 ˙rr or in component form, r2 cos ˙ r2 sin ˙ r2 ˙ 2 sin 2 + r2 ˙ 2 cos 2 = ˙1 r3 ˙ 2 cos r = r3 ˙ 2 sin 2 2 2 2 Substituting the know values gives r2 sin60° + (5.774)(100)cos60° = (10)(100)sin60° ˙ or r2 = 1333.40 ˙ (1333.40)cos60° (5.774)(100)sin60° = ˙1 (10)(100)cos60° r or r1 = 666.70 ˙ ˙ Therefore, 1 vC4 = r1 = 666.70 in/sec. Problem 5.16 In the simple, two-link mechanism given, 1 vB2 is 10 in/s to the right. Use the loop equation approach to determine 1 vA2 and 1 2 . A AB = 10 inches 2 30˚ B 1v B 2 Solution r3 = r1 + r2 r3 = r1 + r2 ˙˙˙ In component form r3 cos 3 = r1 cos 1 + r2 cos 2 r3 sin 3 = r1 sin 1 + r2 sin 2 - 205 - A 2 r3 r1 AB = 10 inches r2 30 o θ2 B 1vB 2 Substituting the constant angles, 0 = r1 + r2 cos r3 = r2 sin 2 2 Substituting values r1 = r2 cos 2 = 10cos150 = 8.66 in r3 = r2 sin 2 = 10sin150 = 5 in The velocity components are: 0 = ˙1 ˙ 2 r2 sin r r3 = ˙ 2 r2 cos 2 ˙ 2 Solving for the unknowns, = 10 / [10sin150] = 2 rad CCW =1 sec ˙ 2 r2 cos 2 = 2(10)cos150 = 17.32 in = 1vA 2 r3 = ˙ sec ˙ 2 = ˙1 / r2 sin r 2 2 Problem 5.17 In the mechanism below, the angular velocity of link 2 is 100 rad/s CCW and the dimensions of various links are given. Use loop equations to find the position and velocity of point D on link 3 when 2 is 90˚. B 3 2 A θ2 C 4 D AB = 1.75 in AC = 2.5 in BD = 5 in - 206 - Solution θ3 B 2 3 θ2 A r3 C r2 r1 4 D Loop equation: r1 = r2 + r3 In component form, r1 cos 1 = r2 cos 2 + r3 cos 3 r1sin 1 = r2 sin 2 + r3 sin 3 Noting that 1 = 0, these equations can be simplified to: r1 = r2 cos 2 + r3 cos 3 0 = r2 sin 2 + r3 sin 3 To solve for r3 , eliminate Then, r1 r2 cos 2 = r3 cos r2 sin 2 = r3 sin 3 and 2 2 r1 2r1r2 cos 2 + r2 cos2 2 2 r2 sin2 2 = r3 sin2 3 2 2 = r cos2 3 3 3 3 by isolating 3 in each equation, squaring both equations and adding. and 2 r1 2r1r2 cos 2 2 2 + r2 (cos2 2 + sin2 2 ) = r3 (cos2 3 + sin2 3) Because sin2 + cos2 = 1 , this equation becomes, or 2 r1 2r1r2 cos 2 r3 = r1 2 2 + r2 2 = r3 2 2 + r2 2r r2 cos 1 For the values given in the problem, - 207 - r3 = 2.52 2(2.5)(1.75)cos90 + 1.752 = 3.052in. and tan or tan 3= 3= r2 sin 2 r1 r2 cos 2 r2 sin 2 = (1.75)sin90 = 0.70 r1 r2 cos 2 2.5 1.75cos90 Therefore, 3= 34.99˚ To find the position of Point D let r4 = BD. Then, or rD = r2 + r4 rD = (r2 cos 2 + r4 cos 3)i + (r2 sin 2 + r4 sin 3 )j Substituting numbers, rD = [1.75cos90 + 5cos( 34.99)]i + [1.75sin90 + 5sin( 34.99)]j = 4.096 i 1.117j = 4.246 15.25˚ The velocity equation is given by: r1 = r2 + ˙3 ˙˙r And in component form, r 0 = r2 ˙ 2 sin 2 + ˙3 cos 3 r3 ˙ 3 sin 3 ˙ 2 cos 2 + ˙3 sin 3 + r3 ˙ 3 cos 3 0 = r2 r r r2 ˙ 2 sin 2 = ˙3 cos 3 r3 ˙ 3 sin 3 r2 ˙ 2 cos 2 = ˙3 sin 3 + r3 ˙ 3 cos r or 3 Substituting numbers for this problem, 1.75(100)sin90 = ˙3 cos( 34.99) 3.052 ˙ 3 sin( 34.99) r 1.75(100)cos90 = ˙3 sin( 34.99) + 3.052˙ 3 cos( 34.99) r or 0.819 1.750 r3 175 ˙ 175 = 0.819˙3 +1.750 ˙ 3 r = or ˙3 0.573 2.500 ˙ 3 0 0 = 0.573˙3 + 2.500 r Using Cramer's Rule, - 208 - 175 1.750 0 2.500 r3 = 0.819 1.750 = 437.5 = 143.4 in / sec ˙ 3.051 0.573 2.500 and 0.819 175 ˙ 3 = 0.573 0 = 100.28 = +32.87 rad / sec (CCW) 0.819 1.750 3.051 0.573 2.500 Now considering Point D, rD = ( r2 ˙ 2 sin ˙ or rD = [1.75(100)sin90 5(32.87)sin( 34.99)]i + [1.75(100)cos90 + 5(32.87)cos( 34.99)]j = 80.78i +134.6 j = 156.98 120.97˚ Therefore, and rD3 = 4.096 i-1.117 j = 4.246 1v D3 2 r4 ˙ 3 sin 3)i + (r2 ˙ 2 cos 2 + r3 ˙3 cos 3 )j -15.25˚ 120.97˚ = -80.78 i + 134.6 j = 156.98 Problem 5.18 In the Scotch Yoke mechanism shown, 1 2 = 10 rad/s, 1 2 = 100 rad/s2 , and O2 A = 20 inches. Determine 1 vA4 and 1 aA4 using loop equations. 2 = 60˚. Also, length 3 2 ω2 O2 1 θ2 A 4 B - 209 - Solution 3 2 A r2 O2 1 r3 θ2 r1 4 B For the vector loop given, r2 = r1 + r3 and in component form r2 cos r2 sin 2 = r cos 1 + r3 cos 3 1 2 = r sin 1 + r3 sin 3 1 We know that 1 = 0, 2 = 60˚, and 3 = 90˚ 2. The variables are r1 , r3 , and position equations. r2 cos r2 sin 2= r 1 2= r 3 Substituting in the know constants gives the following for the For the given input values (r2 = 20 and The velocity equations are: r2 = ˙1 + ˙3 ˙rr and r2 cos ˙ r2 sin r2 ˙ 2 sin 2 + r2 ˙ 2 cos 2 2 = 60˚) it is clear that r1 = 10 and r3 = 17.32. = ˙1 r r 2 = ˙3 2 Simplifying: - 210 - r3 = r2 ˙ 2 cos 2 = 20(10)cos60 = 100 in / sec ˙ r1 = r2 ˙ 2 sin 2 = 20(10)sin60 = 173.2 in / sec ˙ The acceleration equations are: r r1 r3 ˙˙2 = ˙˙ + ˙˙ and ˙˙ = r2˙˙2 sin r1 ˙˙ = r2˙˙2 cos r3 2 2 r2 ˙ 2 cos 2 r2 ˙2 sin 2 2 2 These equations simplify to: ˙˙ = r2˙˙2 sin 2 r2 ˙ 2 cos 2 = 20(100)sin60 20(10)2 cos60 = 2732 in / sec2 r1 2 ˙˙ = r2˙˙2 cos 2 r2 ˙2 sin 2 = 20(100)cos60 20(10)2 sin60 = 732 in / sec2 r3 2 1v 1a A4 = -173.2 in/s -2732 in/s2 A4 = Problem 5.19 Use loop equations to determine the velocity and acceleration of point B on link 4. The angular velocity of link 2 is constant at 10 rad/s counterclockwise. 4 r1 = 10 cm φ = 30˚ θ 2 = 60˚ r2 B φ r3 θ3 O3 θ2 1ω2 O2 r1 Solution Position Analysis The basic vector loop equation is r1 + r3 = r2 and - 211 - 4 r2 θ2 1ω2 B φ r3 θ3 O2 r1 O3 3= 2+ In component form, r1 cos0 + r3 cos 3 = r2 cos 2 r1sin 0 + r3 sin 3 = r2 sin 2 r1 + r3 cos 3 = r2 cos r3 sin 3 = r2 sin 2 2 or In the position shown, 2 = 60 . Therefore, by geometry, the triangle is a 30-60 right triangle. Therefore, r2 = r1 / cos and r3 = r1 / sin 2 = 10 / sin(60˚) = 17.321 2 2 = 60 . Therefore, 3 = 90 . Therefore, = 10 / cos(60˚) = 20 Velocity Analysis The velocity loop equation is r1 + r3 = r2 ˙˙˙ In component form, r r1 + ˙3 cos 3 r3 ˙ 3 sin 3 = ˙2 cos 2 r2 ˙ 2 sin ˙r ˙3 cos 3 = ˙2 sin 2 + r2 ˙2 cos 2 r r3 sin 3 + r3 ˙ and ˙3 = ˙2 Therefore, r3 cos ˙ r3 sin ˙ 3 3 2 r ˙2 cos 2 = r2 ˙ 2 sin 2 + r3 ˙ 3 sin 3 = ˙ 2[r3 sin 3 r2 sin 2] r ˙2 sin 2 = r2 ˙ 2 cos 2 r3 ˙ 3 cos 3 = ˙2[r2 cos 2 r3 cos 3 ] - 212 - or r3 cos90 ˙2 cos60 = 10[17.321sin90 20sin60] ˙ r r r3 sin90 ˙2 sin60 = 10[20cos60 17.321cos90] ˙ or and 0.5 ˙2 = 0.0 r r r3 0.866˙2 = 100 ˙ r2 = 0.0 ˙ r3 = 100 ˙ Then, ˙˙ ˙ vB4 = r1 + r3 = (r3 cos Substituting in numbers, vB4 = ( 173.21,100) = 200 in / sec 149.98˚ Acceleration Analysis The basic acceleration loop equation is r1 r r ˙˙ + ˙˙3 = ˙˙2 In component form, ˙˙ cos 3 r3˙˙3 sin r3 ˙˙ sin 3 + r3˙˙3 cos r3 and ˙˙3 = ˙˙2 = 0 Then, ˙˙ cos r3 ˙˙ sin r3 3 3 3 r3 ˙ 3 sin 3, ˙3 sin r 3 + r3 ˙ 3 cos 3 ) 2˙3 ˙ 3 sin r r3 3 + 2˙ ˙ 3 cos 3 3 3 r3 ˙ 2 cos 3 r3 ˙ 2 sin 3 r2 ˙˙2 sin ˙ r2 3 = ˙˙ sin 2 + r2˙ 2 cos r ˙ 3 = ˙2 cos 2 2˙2 ˙ 2 sin r r 2 + 2˙2 ˙ 2 cos 2 2 2 r2 ˙ 2 cos 2 r2 ˙ 2 sin 2 2 2 ˙˙ cos 2 = 2r2 ˙ 2 sin 2 r2 ˙ 2 cos 2 + 2˙3 ˙ 3 sin 3 + r3 ˙ 2 cos r2 ˙ r 2 3 ˙ 2 cos 2 r2 ˙2 sin 2 2˙3 ˙ 3 cos 3 + r3 ˙2 sin 3 ˙˙2 sin 2 = 2˙2 r r r 2 3 3 Substituting in numbers, r r r ˙˙ cos90 ˙˙2 cos60 = 2˙2 ˙ 2 sin60 r2 ˙ 2 cos60 + 2˙3 ˙3 sin90 + r3 ˙ 2 cos90 r3 2 3 ˙˙ sin90 ˙˙ sin60 = 2˙2 ˙2 cos60 r2 ˙ 2 sin60 2˙3 ˙ 3 cos90 + r3 ˙ 2 sin90 r3 r2 r r 2 3 r 0.5˙r2 = 0.5(20)(100) + 2(100)(10) = 1000 ˙˙2 = 2000 in / sec2 ˙ r2 r r3 ˙˙ 0.866˙˙ = 0.886(20)(100) + 17.321(100) = 0 ˙˙3 = (0.866)2000 = 1732 in / sec2 aB4 = ˙˙ + ˙˙3 r1 r = (˙r3 cos 3 r3˙˙3 sin ˙ 2˙3 ˙ 3 sin r r3 ˙ 2 cos 3, ˙˙3 sin r 3 ˙ r 3 + r3˙ 3 cos 3 + 2˙3 ˙ 3 cos 3 r3 ˙ 2 sin 3 or 3 3 3 ) - 213 - Substituting in numbers, r r 100)10, 1732 17.321(100)) = ( 2000, 3464) aB4 = ( 2˙3 ˙3, ˙˙3 r3 ˙ 2 ) = ( 2( 3 vB4 = 200 in / s aB4 = 4000 in / s2 Problem 5.20 The oscillating fan shown below is to be analyzed as a double rocker. The fan is link 2, the motor shaft is connected to link 3, and link 4 is connected from the coupler to the frame. The actual input of the mechanism is the coupler, and 2 3 that is a constant 956 (rad/s) in the counterclockwise direction. Compute the angular velocity and angular acceleration of link 2 if = 120˚, AD = 0.75 in, AB = DC = 3.0 in, BC = 0.50 in. 2 ω 3 B 3 C 2 4 θ A D Solution The basic loop equations is: r1 + r4 = r2 + r3 In component form, this equation becomes: r1 cos 1 + r4 cos 4 = r2 cos 2 + r3 cos 3 r1sin 1 + r4 sin 4 = r2 sin 2 + r3 sin 3 - 214 - B 3 r3 C r4 4 2 r2 Y θ A Substituting the constant values ( 1 = 0 ) gives 3 D r1 X r1 + r4 cos 4 = r2 cos 2 + r3 cos r4 sin 4 = r2 sin 2 + r3 sin 3 To solve for 3 and 4 , rewrite the equations above 3 r1 + r4 cos 4 r2 cos 2 = r3 cos r4 sin 4 r2 sin 2 = r3 sin 3 Now square both equations and add. The resulting equation is: 2r4 (r1 r2 cos 2 )cos Let A = 2r4(r1 r2 cos 2) = 2 3 (0.75 3cos120°) = 13.5 B = 2r4 r2 sin 2 = 2 3 3sin120° = 15.588 2 2 2 C = (r1 r2 cos 2 )2 + r2 sin2 2 + r4 r3 = (0.75 3 cos120°)2 + 32 sin2 120° + 32 0.52 = 20.563 4 2r4r2 sin 2 sin 4 2 + (r1 r2 cos 2 )2 + r2 sin2 2 2 + r4 2 r3 = 0 Then rewrite the equation above A cos 4 + Bsin 4 + C =0 (1) Simplify using the following trigonometric itentities sin 2tan 4 4 = 2 4 1 + tan2 2 - 215 - cos 1 tan2 4 4 = 2 4 1 + tan2 2 Let t = tan 2 4, and substitute the trigonometric identities into Eq. (1). Then: A(1 t 2) + B(2t) + C(1 + t 2 ) = 0 Collecting terms gives: (C A)t 2 + 2Bt + (C + A) = 0 The roots are: t= B+ B2 C2 + A2 15.588 + = CA ( 15.588)2 20.5632 + 13.52 15.588 ± 1.549 = 7.063 20.563 13.5 Determine the sign for the square root that corresponds to this problem. t1 = 2.426 ; t 2 = 1.988 ; 4 4 = 2tan 1 2.426 = 135.197° = 2tan 11.988 = 126.586° 4 According to the problem figure, r4 sin 4 3 = sin 1 r3 r2 sin 2 = 126.586° is the correct root. Then = sin 1 3sin126.586° 3sin120° = 22.233° 0.5 The component form for the velocities are: r4 ˙ 4 sin 4 = r2 ˙ 2 sin 2 r3 ˙ 3 sin 3 r4 ˙ 4 cos 4 = r2 ˙ 2 cos 2 + r3 ˙ 3 cos 3 Substituting ˙ 3 = ˙ 2 + 2 3 , into the equations above and simplifying, we get r4 ˙ 4 sin 4 + (r2 sin 2 + r3 sin 3)˙ 2 = r3 (2 3) sin 3 r4 ˙ 4 cos 4 (r2 cos 2 + r3 cos 3) ˙2 = r3 (2 3 ) cos 3 Substituting the pertinent values into the equations gives 3sin126.586° 3cos126.586° or ˙4 ˙2 = 486.365 411.362 (3sin120° + 0.5sin( 22.233°)) ˙ 4 0.5 956sin( 22.233°) = (3cos120° + 0.5cos( 22.233°)) ˙ 2 0.5 956cos( 22.233°) The component equations for acceleration are: - 216 - 2 r4˙˙4 sin 4 r4 ˙ 2 cos 4 = r2 ˙˙2 sin 2 r2 ˙2 cos 2 r3˙˙3 sin 3 r3 ˙3 cos 4 2 r4 ˙˙4 cos 4 r4 ˙ 2 sin 4 = r2˙˙2 cos 2 r2 ˙ 2 sin 2 + r3˙˙3 cos 3 r3 ˙ 2 sin 3 4 2 3 3 Substituting ˙˙3 = ˙˙2 , into the equations above and rewriting, we get r4˙˙4 sin 4 + (r2 sin 2 + r3 sin 3)˙˙2 = r4 ˙ 2 cos 4 r4 ˙˙4 cos 4 (r2 cos 2 + r3 cos 3)˙˙2 = r4 ˙ 2 sin 4 4 4 r2 ˙ 2 cos 2 r3 ˙2 cos 3 2 3 r2 ˙ 2 sin 2 r3 ˙ 2 sin 3 2 3 Substituting the pertinent values into the equations gives 3sin126.586° (3sin120° + 0.5sin( 22.233°)) ˙˙4 3cos126.586° (3cos120° + 0.5cos( 22.233°)) ˙˙2 3( 486.365)2 cos126.586° 3( 411.362)2 cos120° 0.5( 411.362 + 956)2 cos( 22.233°) = 3( 486.365)2 sin126.586° 3( 411.362)2 sin120° 0.5( 411.362 + 956)2 sin( 22.233°) or ˙˙4 ˙˙2 = 4.244 105 5.516 105 Therefore ˙ 2 = 411.362 rad/sec CW; and ˙˙2 = 5.516 10 5 rad / sec2 CW. Problem 5.21 The rear motorcycle suspension can be analyzed as an inverted slider-crank mechanism. The frame of the motorcycle is link 1, the tire assemble is attached to link 2 at point C. The shock absorber is links 3 and 4. As the bicycle goes over a bump in the position shown, the angular velocity of link 2 relative to the frame is 1 2 is 5 (rad/s), and the angular acceleration is 1 2 is 45 (rad/s2 ), both in the clockwise direction. Compute the angular velocity and angular acceleration of link 3 for the position defined by = 187˚. (-9.262, 10.728) A 3 4 14.17" Y 4.7˚ θ B C 2 19.27" D 1 ω 2 X - 217 - Solution A 3 r1 Y r3 B C 4 2 r2 D X The basic loop equations is: r2 = r1 + r3 In component form this equation becomes: r2 cos r2 sin 2 = r cos 1 + r3 cos 3 1 2 = r sin 1 + r3 sin 3 1 The known values are: r1 = 14.17 ; r2 = 19.27 ; To solve for r2 cos r2 sin 2 2 3 1 = 180° + tan 1(10.728 / ( 2 9.262)) = 130.806° = + 4.7° = 187° + 4.7° = 191.7° and r3 , rewrite the equations above as r1 cos r1 sin 1 = r3 cos 3 1 = r3 sin 3 (1) (2) Divide Eq. (2) by Eq. (1), we get r2 sin 2 3 = 180° + tan 1 r2 cos 2 = 236.708° Then, r3 = r2 sin sin 2 r1 sin 1 = 180° + tan 1 19.27sin191.7° 14.17sin130.806° 19.27cos191.7° 14.17 cos130.806° r1 cos 1 r1 sin 3 1 = 19.27sin191.7° 14.17sin130.806° = 17.507 sin236.708° The component equations for the velocities are: r r2 ˙ 2 sin 2 = ˙3 cos 3 r3 ˙3 sin 3 r2 ˙ 2 cos 2 = ˙3 sin 3 + r3 ˙ 3 cos 3 r - 218 - Substitute the relevant values cos236.708° sin236.708° or r ˙3 ˙3 = 19.27 ( 5)sin191.7° 17.507sin236.708° ˙3 r ˙ 3 = 19.27 ( 5)cos191.7° 17.507cos236.708° 68.128 3.891 The component equations for acceleration are: r3 r r2˙˙2 sin 2 r2 ˙ 2 cos 2 = ˙˙ cos 3 2˙3 ˙ 3 sin 3 r3˙˙3 sin 3 r3 ˙ 2 cos 3 2 3 2 sin 2 r2˙˙2 cos 2 r2 ˙ 2 = ˙˙ sin 3 + 2˙3 ˙ 3 cos 3 + r3˙˙3 cos 3 r3 ˙3 sin 3 r3 r 2 Rewrite the equations to put the unknowns on the same side of the equations r r3 r2˙˙2 sin 2 r2 ˙ 2 cos 2 + 2˙3 ˙3 sin 3 + r3 ˙ 2 cos 3 = ˙˙ cos 3 r3˙˙3 sin 3 2 3 2 sin 2 sin r2˙˙2 cos 2 r2 ˙ 2 r r ˙ 2 2˙3 ˙ 3 cos 3 + r3 ˙ 3 3 = ˙3 sin 3 + r ˙ 3 cos 3 3˙ Substitute the pertinent values cos236.708° 17.507sin236.708° ˙˙3 r ˙˙3 sin236.708° 17.507cos236.708° 19.27( 45)sin191.7° 19.27( 5)2 cos191.7° 2 = +2( 68.128)( 3.891)sin236.708° + 17.507( 3.891) cos236.708° 2 sin191.7° 19.27( 45)cos191.7° 19.27( 5) 2( 68.128)( 3.891)cos236.708° + 17.507( 3.891)2 sin236.708° or r3 ˙˙ ˙˙3 = 688.691 45.842 Therefore ˙ 3 = 3.891 rad/sec, CW; and ˙˙3 = 45.842 rad / sec2 , CW. - 219 - Problem 5.22 The door-closing linkage shown is to be analyzed as a slider-crank linkage. Link 2 is the door, and links 3 and 4 are the two links of the door closer. Assume that the angular velocity of the door (link 2) is a constant at 3.71 radians per second clockwise. Compute the angular velocity and angular acceleration of link 4 if the dimensions are as follows: Coordinates of D (-2.5, -3.0) AB = 17.0 inches 1ω 2 y A 6˚ 2 Door C B 3 x D 4 Solution Y r2 r1 D The basic loop equations is: r1 = r2 + r3 In component form, this equation becomes: r1 cos 1 = r2 cos 2 + r3 cos 3 r1sin 1 = r2 sin 2 + r3 sin 3 The known values are: r1 = 2.52 + 3.02 = 3.905 ; r2 = 17.0 ; To solve for r1 cos 1 3 1= 2 2 r3 3 C B X A 4 tan 1(3.0 / 2.5) = 50.194° = 174° and r3 , rewrite the equations above r2 cos 2 = r3 cos 3 (1) - 220 - r1sin 1 r2 sin 2 = r3 sin 3 (2) Dividing Eq. (2) by (1), we get 3= tan 1 1 r sin 1 r1 cos 1 1 r2 sin r2 cos 2 2 2 = tan 1 3.905sin50.194° 17sin174° = 3.606° 3.905cos50.194° 17cos174° r3 = r1 sin sin r2 sin 3 = 3.905sin50.194° 17 sin174° = 19.445 sin3.606° The component equations for the velocities are: r r2 ˙ 2 sin 2 = ˙3 cos 3 r3 ˙ 3 sin 3 r2 ˙ 2 cos 2 = ˙3 sin 3 + r3 ˙ 3 cos r Substitute the relevant values cos3.606° sin3.606° or r ˙3 ˙3 = 19.445sin3.606° ˙3 r = 19.445cos3.606° ˙ 3 17 ( 3.71)sin174° 17 ( 3.71)cos174° 3 10.525 3.198 The component equations for the acceleration are: 2 r3 r r2˙˙2 sin 2 + r2 ˙ 2 cos 2 = ˙˙ cos 3 2˙3 ˙ 3 sin 3 r3˙˙3 sin 3 r3 ˙3 cos 3 2 r2˙˙2 cos 2 + r2 ˙ 2 sin 2 = ˙˙ sin 3 + 2˙3 ˙ 3 cos 3 + r3˙˙3 cos 3 r3 ˙ 2 sin 3 r3 r 2 3 Rewrite the equations to isolate the unknowns on the same side of the equations ˙ r r2˙˙2 sin 2 + r2 ˙ 2 cos 2 + 2r3 ˙ 3 sin 3 + r3 ˙ 2 cos 3 = ˙˙3 cos 3 r3˙˙3 sin 3 2 3 ˙˙2 cos 2 + r2 ˙ 2 sin 2 2˙3 ˙3 cos 3 + r3 ˙ 2 sin 3 = ˙˙ sin 3 + r3˙˙3 cos 3 r2 r r3 2 3 Substituting the pertinent values gives cos3.606° 19.445sin3.606° ˙r3 ˙ sin3.606° 19.445cos3.606° ˙˙3 17( 3.71)2 cos174° + 2( 10.525)( 3.198)sin3.606° + (19.445)( 3.198)2 cos3.606° = 17( 3.71)2 sin174° 2( 10.525)( 3.198)cos3.606° + (19.445)( 3.198)2 sin3.606° or r3 ˙˙ ˙˙3 = 31.842 1.454 Therefore ˙ 3 = 3.198 rad/sec, CW; and ˙˙3 = 1.454 rad / sec2 , CW. - 221 - Problem 5.23 The general action of a person who is doing pushups can be modeled as a four-bar linkage as shown below. The floor is the base link, and link 4 is the back and legs. Link 2 is the forearm, and link 3 is the upper arm. For the purposes of analysis, the motion that is controlled is the motion of link 3 relative to link 2 (elbow joint). Assume that 2 3 is a constant 6.0 rad/s in the counterclockwise direction. Compute the angular velocity and angular acceleration of link 4 if link 2 is oriented at 45˚ to the horizontal. C 2ω 14.0" 3 57.7" 4 D 3 2 A B 45˚ 12.0" 52.0" Solution Y C r3 A B r4 D r2 r1 X The basic loop equations is: r1 + r4 = r2 + r3 In component form, this equation becomes: r1 cos 1 + r4 cos 4 = r2 cos 2 + r3 cos 3 r1sin 1 + r4 sin 4 = r2 sin 2 + r3 sin 3 Substituting the constant values ( 1 = 0) gives 3 r1 + r4 cos 4 = r2 cos 2 + r3 cos r4 sin 4 = r2 sin 2 + r3 sin 3 To solve for 3, 4, rewrite the equations above by isolating the terms which contain 3 3. Then, r1 + r4 cos 4 r2 cos 2 = r3 cos r4 sin 4 r2 sin 2 = r3 sin 3 - 222 - Square both sides of both equations and add the results. This gives 2r4 (r1 r2 cos 2 )cos Let A = 2r4(r1 r2 cos 2) = 2 57.7 (52.0 12.0cos45°) = 5021.6 B = 2r4 r2 sin 2 = 2 57.7 12.0 sin45° = 979.2 2 2 2 C = (r1 r2 cos 2 )2 + r2 sin2 2 + r4 r3 = (52.0 12.0cos45°)2 + 12.02 sin2 45° + 57.72 14.02 = 5098.8 Equation (1) can now be simplified as A cos 4 + Bsin 4 + C 4 2r4r2 sin 2 sin 4 2 + (r1 r2 cos 2 )2 + r2 sin2 2 2 + r4 2 r3 = 0 (1) =0 (2) Simplify the resulting equation using sin 4 = 2t 2 and cos 1+ t 4 2 = 1 t2 . 1+ t where t = tan 4 . Then, Eq. (2) becomes 2 A(1 t 2) + B(2t) + C(1 + t 2 ) = 0 (3) Collecting terms in Eq. (3) gives: (C A)t 2 + 2Bt + (C + A) = 0 The roots are: t= where b = B+ B2 C2 + A2 979.2 + = CA ( 979.2)2 5098.82 + 5021.62 5098.8 5021.6 = ±1 . Then, 4 t1 = 18.142 which gives t 2 = 7.226 which gives 4 = 2tan 1 18.142 = 173.69° = 2tan 1 7.226 = 164.242° 4 According to the picture shown, the appropriate value is r4 sin 4 3 = sin 1 r2 sin 2 = 164.242° . Then, r3 1 57.7sin164.242° 12.0sin45° = 149.12° = sin 14.0 The component form of velocity equations is: r4 ˙ 4 sin 4 = r2 ˙ 2 sin 2 r3 ˙ 3 sin 3 r4 ˙ 4 cos 4 = r2 ˙ 2 cos 2 + r3 ˙ 3 cos 3 - 223 - When we substitute ˙ 3 = ˙ 2 + 2 3 , into the equations above and rewrite, we get r4 ˙ 4 sin 4 + (r2 sin 2 + r3 sin 3)˙ 2 = r32 3 sin 3 r4 ˙ 4 cos 4 (r2 cos 2 + r3 cos 3) ˙2 = r3 2 3 cos 3 Substituting the pertinent values into the equations and rewriting them in matrix form gives 57.7sin164.242° 57.7cos164.242° or ˙4 ˙2 = 1.2 1.55 ˙4 12.0sin45° +14.0 sin149.12° 14.0 6.0sin149.12° = (12.0cos45° + 14.0cos149.12°) ˙ 2 14.0 6.0cos149.12° The component equations for the accelerations are: 2 r4˙˙4 sin 4 r4 ˙ 2 cos 4 = r2 ˙˙2 sin 2 r2 ˙2 cos 2 r3˙˙3 sin 3 r3 ˙3 cos 4 2 r4 ˙˙4 cos 4 r4 ˙ 2 sin 4 = r2˙˙2 cos 2 r2 ˙ 2 sin 2 + r3˙˙3 cos 3 r3 ˙ 2 sin 3 4 2 3 3 Substitute ˙˙3 = ˙˙2 into the equations above and rewrite. This gives r4˙˙4 sin 4 + (r2 sin 2 + r3 sin 3)˙˙2 = r4 ˙ 2 cos 4 r4 ˙˙4 cos 4 (r2 cos 2 + r3 cos 3)˙˙2 = r4 ˙ 2 sin 4 4 4 r2 ˙ 2 cos 2 r3 ˙2 cos 3 2 3 r2 ˙ 2 sin 2 r3 ˙ 2 sin 3 2 3 Now substitute the pertinent values into the equations to get 57.7sin164.242° (12.0sin45° +14.0 sin149.12°) 57.7cos164.242° (12.0cos45° + 14.0cos149.12°) 2 cos164.242° 12.0( 1.55)2 cos45° 14.0( 57.7(1.2) = 57.7(1.2)2 sin164.242° 12.0( 1.55)2 sin45° 14.0( or ˙˙4 3.29 ˙˙2 = 12.07 Therefore ˙ 4 = 1.2 rad/s, CCW and ˙˙4 = 3.29rad / s2 , CCW. ˙˙4 ˙˙2 1.55 + 6.0)2 cos149.12° 1.55 + 6.0)2 sin149.12° - 224 - Problem 5.24 A carousel mechanism can be modeled as an inverted slider-crank mechanism as shown. Point D is the location of the saddle on the horse. Assume that the angular velocity of the driver (Link 2) is a constant 2 rad/s counterclockwise. Compute the velocity and acceleration of D3 in the position shown if AB = 8.0 in, BC = 96.0 in, and BD = 54 in. 2 A 1ω2 B 45˚ 3 D 4 C Solution The basic loop equations is: r1 = r2 + r3 In component form, this equation becomes: r1 cos 1 = r2 cos 2 + r3 cos 3 r1sin 1 = r2 sin 2 + r3 sin 3 r2 = 8.0 ; r3 = 96.0 ; 1= 2 = 45° 3 = cos 1( r2 cos 2 / r3) = cos 1( 8cos( 45°) / 96) = 93.378° 90° 2 + r3 sin sin 1 3 r1 = r2 sin = 8sin( 45°) + 96sin( 93.378°) = 101.490 sin( 90°) - 225 - In component form, the velocities are: r r2 ˙ 2 sin 2 = ˙3 cos 3 r3 ˙ 3 sin 3 r2 ˙ 2 cos 2 = ˙3 sin 3 + r3 ˙ 3 cos r Substitute the relevant values Y A 3 r2 2 B X 3 r1 r3 D 4 C cos( 93.378°) sin( 93.378°) or 11.964 r ˙3 = ˙3 0.111 96 sin( 93.378°) ˙3 r = 96cos( 93.378°) ˙ 3 8(2)sin( 45°) 8(2)cos( 45°) To solve for vD 3 , rC /D = rC/ D(cos 3i + sin 3 j) r vD 3 = rC /D = ˙C / D(cos 3i + sin 3 j) rCD ˙ 3( sin 3i + cos 3 j) ˙ = [( 11.964)cos( 93.378°) + 42( 0.111)sin( 93.378°)]i + [( 11.964)sin( 93.378°) 42( 0.111)cos( 93.378°)]j = 5.359i + 11.669j = 12.841 65.333° Therefore vD 3 = 12.841 65.333° in/sec. - 226 - The component equations for acceleration are: 2 r3 r r2˙˙2 sin 2 + r2 ˙ 2 cos 2 = ˙˙ cos 3 2˙3 ˙ 3 sin 3 r3˙˙3 sin 3 r3 ˙3 cos 3 2 r2˙˙2 cos 2 + r2 ˙ 2 sin 2 = ˙˙ sin 3 + 2˙3 ˙ 3 cos 3 + r3˙˙3 cos 3 r3 ˙ 2 sin 3 r3 r 2 3 Rewriting the equations ˙ r r2˙˙2 sin 2 + r2 ˙ 2 cos 2 + 2r3 ˙ 3 sin 3 + r3 ˙ 2 cos 3 = ˙˙3 cos 3 r3˙˙3 sin 3 2 3 2 sin 2 sin ˙ r2˙˙2 cos 2 + r2 ˙ 2 r r ˙ 2 2˙3 ˙3 cos 3 + r3 ˙ 3 3 = ˙3 sin 3 + r3˙ 3 cos 3 Substituting the pertinent values cos( 93.378°) sin( 93.378°) 8(2)2 cos( = 8(2)2 sin( or 22.444 r3 ˙˙ ˙˙3 = 0.277 96 sin( 93.378°) ˙˙3 r ˙˙3 96cos( 93.378°) 45°) + 2(11.964)( 0.111)sin( 93.378°) + 96( 0.111)2 cos( 93.378°) 45°) 2(11.964)( 0.111)cos( 93.378°) + 96( 0.111)2 sin( 93.378°) To solve for aD 3 , aD 3 = ˙˙CD = ˙˙ D (cos 3i + sin 3 j) 2˙CD ˙ 3( sin 3i + cos 3 j) r rC r 2 (cos 3i + sin 3 j) rCD˙˙3( sin 3i + cos 3 j) + rCD ˙ 3 = (22.444)(cos( 93.378°)i + sin( 93.378°) j) 2(11.964)( 0.111)( sin( 93.378°)i + cos( 93.378°)j) 42(0.277)( sin( 93.378°)i + cos( 93.378°) j) +42( 0.111)2 (cos( 93.378°)i + sin( 93.378°)j) = 7.669i + 22.411j = 23.687 108.891° Therefore aD 3 = 23.687 108.891° in / sec2 . Problem 5.25 The shock absorber mechanism on a mountain bicycle is a four-bar linkage as shown. The frame of the bike is link 1, the fork and tire assembly is link 3, and the connecting linkage are links 2 and 4. As the bicycle goes over a bump in the position shown, the angular velocity of link 2 relative to the frame is 2 is 205 (rad/s) CW, and the angular acceleration is 2 is 60 (rad/s2 ) CW. Compute the angular velocity and angular acceleration of link 3 for the position shown - 227 - 201˚ A B 2 1ω 2 61˚ 1.55" 1 D 1.88" 1.64" 3 4 C 2.22" Solution r2 B 2 A 1ω2 1 r1 D r3 3 4 C r4 The basic loop equations is: r1 + r2 = r3 + r4 In component form, this equation becomes: r1 cos 1 + r2 cos 2 = r3 cos 3 + r4 cos 4 r1sin 1 + r2 sin 2 = r3 sin 3 + r4 sin 4 The known values are: r1 = 1.55" 1 = 119° r2 = 1.88" 2 r3 = 1.64" r4 = 2.22" = 201° in both equations. Then, To solve for 3 , 4 , rewrite the equations above to isolate 4 - 228 - r1 cos 1 + r2 cos 2 r3 cos 3 = r4 cos 4 r1sin 1 + r2 sin 2 r3 sin 3 = r4 sin 4 Square both sides of both equations and add the results. This gives 2r3(r1 cos 1 + r2 cos 2)cos 3 + 2r3 (r1 sin 1 + r2 sin 2 )sin 3 2 2 +r4 r3 (r1 cos 1 + r2 cos 2 )2 (r1 sin 1 + r2 sin 2 )2 = 0 To simplify the equation, let A = 2r3 (r1 cos 1 + r2 cos 2 ) = 2(1.64)(1.55cos119° + 1.88cos201°) = 8.222 B = 2r3(r1 sin 1 + r2 sin 2 ) = 2(1.64)(1.55sin119°+ 1.88sin201°) = 2.237 2 2 C = r4 r3 (r1 cos 1 + r2 cos 2)2 (r1 sin 1 + r2 sin 2 )2 = 2.222 1.642 (1.55cos119° + 1.88cos201°)2 (1.55sin119° + 1.88sin201°)2 = 4.509 Now rewrite Eq. (2) as A cos To solve for t = tan 2 3 3 + Bsin 3 + C (1) (2) =0 sin 3= 3 , replace the trignometric functions with 2t and cos 1 + t2 3= 1 t 2 where 1 + t2 . Then (3) A(1 t 2) + B(2t) + C(1 + t 2 ) = 0 Collecting terms in Eq. (3) gives: (C A)t 2 + 2Bt + (C + A) = 0 The roots are: t= where B+ B2 C2 + A2 2.237 + = CA 2.2372 ( 4.509)2 + ( 8.222)2 4.509 + 8.222 = ±1 . The two possible solutions are 3 = 2tan 11.345 3 = 2tan 1( t1 = 1.345 which gives t 2 = 2.550 which gives = 106.7° 2.550) = 137.2° 3 = 106.7° . According to the picture shown, the correct value is tan and 4 Then from Eq. (1), = r1 sin 1 + r2 sin r1 cos 1 + r2 cos 2 2 r3 sin r3 cos 3 3 - 229 - 4 r3 sin 3 r3 cos 3 1.55sin119° +1.88sin 201° 1.64sin106.7° = 156.4° = tan 1 1.55cos119° +1.88 cos201° 1.64cos106.7° = tan 1 2 1 r sin 1 + r sin 2 r1 cos 1 + r2 cos 2 [ [ The component form of velocity equations is: r2 ˙ 2 sin r2 ˙ 2 cos 2 = r3 ˙ 3 sin 3 + r4 ˙ 4 sin 4 2 = r3 ˙ 3 cos 3 + r4 ˙ 4 cos 4 Substitute the pertinent values into the velocity equations to get 1.64sin106.7° 2.22sin( 156.4°) ˙ 3 1.88( 205)sin201° = 1.64cos106.7° 2.22cos( 156.4°) ˙ 4 1.88( 205)cos201° or ˙3 ˙4 = 10.777 174.393 The component equations for acceleration is: r2˙˙2 sin 2 + r2 ˙ 2 cos 2 ˙˙2 cos 2 r2 ˙ 2 sin r2 2 r3 ˙ 2 cos 3 ˙ 2 sin 2+r 3 3 2 r4 ˙2 cos 4 = r3˙˙3 sin 3 + r4 ˙˙4 sin 4 ˙ 2 sin 4 = r3˙˙3 cos 3 + r4 ˙˙4 cos 3 + r4 4 3 4 4 Substitute the pertinent values into the acceleration equations to get 1.64sin106.7° 2.22sin( 156.4°) ˙˙3 1.64cos106.7° 2.22cos( 156.4°) ˙˙4 1.88( 60)sin201° + 1.88( 205)2 cos201° 1.64( 10.777)2 cos(106.7°) 2.22( 174.393)2 cos( 156.4°) = 1.88( 60)cos201° 1.88( 205)2 sin201° + 1.64( 10.777)2 sin(106.7°) +2.22( 174.393)2 sin( 156.4°) or ˙˙3 ˙˙4 = 7026.414 860.77 Therefore ˙ 3 = 10.777 rad/s, CW and ˙˙3 = 7026.4 rad / s2 , CW. - 230 - ...
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This note was uploaded on 02/20/2011 for the course MEC 411 taught by Professor Shudong during the Winter '11 term at Ryerson.

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