DESIGN OF MACHINERY SOLUTION MANUAL 2-1-1 ! PROBLEM 2-1 Statement: Find three (or other number as assigned) of the following common devices. Sketch careful kinematic diagrams and find their total degrees of freedom. a. An automobile hood hinge mechanism b. An automobile hatchback lift mechanism c. An electric can opener d. A folding ironing board e. A folding card table f. A folding beach chair g. A baby swing h. A folding baby walker i. A drafting machine j. A fancy corkscrew k. A windshield wiper mechanism l. A dump-truck dump mechanism m. A trash truck dumpster mechanism n. A station wagon tailgate mechanism o. An automobile jack p. A collapsible auto radio antenna q. A record turntable and tone arm See Mathcad file P0201. Solution: Equation 2.1c is used to calculate the mobility (DOF) of each of the models below. a. An automobile hood hinge mechanism. The hood (3) is linked to the body (1) through two rocker links (2 and 4). Number of links Number of full joints Number of half joints M 3 .( L 1) L J1 J2 2 .J 1 4 4 0 J2 1 BODY 2 4 HOOD 3 M=1 b. An automobile hatchback lift mechanism. The hatch (2) is pivoted on the body (1) and is linked to the body by the lift arm, which can be modeled as two links (3 and 4) connected through a translating slider joint. HATCH Number of links L 4 Number of full joints Number of half joints M 3 .( L 1) J1 J2 2 .J 1 4 0 1 2 3 4 1 BODY J2 M=1 c. An electric can opener has 2 DOF. d. A folding ironing board. 2nd Edition, 1999 DESIGN OF MACHINERY SOLUTION MANUAL 2-1-2 The board (1) itself has one pivot (full) joint and one pin-in-slot sliding (half) joint. The two legs (2 and 3) have a common pivot. One leg connects to the pivot joint on the board and the other to the slider joint. Number of links Number of full joints Number of half joints M 3 .( L 1) L J1 J2 2 .J 1 3 2 1 J2 3 1 2 M=1 e. f. A folding card table has 7 DOF: One for each leg, 2 for location in xy space, and one for angular orientation. A folding beach chair. The seat (3) and the arms (6) are ternary links. The seat is linked to the front leg(2), the back (5) and a coupling link (4). The arms are linked to the front leg (2), the rear leg (1), and the back (5). Links 1, 2, 4, and 5 are binary links. The analysis below is appropriate when the chair is not fully opened. When fully opened, one or more links are prevented from moving by a stop. Subtract 1 DOF when forced against the stop. Number of links Number of full joints Number of half joints M 3 .( L 1) L J1 J2 2 .J 1 6 7 0 J2 4 1 3 5 6 2 M=1 g. A baby swing has 4 DOF: One for the angular orientation of the swing with respect to the frame, and 3 for the location and orientation of the frame with respect to a 2-D frame. A folding baby walker has 4 DOF: One for the degree to which it is unfolded, and 3 for the location and orientation of the walker with respect to a 2-D frame. A drafting machine has 4 DOF: Two for the xy position of the head with respect to the board, one for the rotation angle about an axis perpendicular to the board, and one for the angular position about an axis parallel to the board (the head can be down on the board or rotated up, off of the board). A fancy corkscrew has 2 DOF: The screw can be rotated and the arms rotate to translate the screw. A windshield wiper mechanism has 1 DOF: The position of the wiper blades is defined by a single input. A dump-truck dump mechanism has 1 DOF: The angle of the dump body is determined by the length of the hydraulic cylinder that links it to the body of the truck. h. i. j. k. l. m. A trash truck dumpster mechanism has 2 DOF: These are generally a rotation and a translation. n. o. A station wagon tailgate mechanism has 2 DOF: This assumes that there are two, independent, parts. An automobile jack has 4 DOF: One is the height of the jack and the other 3 are the position and orientation of the jack with respect to a 2-D frame. A collapsible auto radio antenna has as many DOF as there are sections, less one. A record turntable and tone arm has 3 DOF: Angular orientation of the record, and rotation of the arm about two axes. p. q. 2nd Edition, 1999 DESIGN OF MACHINERY SOLUTION MANUAL 2-2-1 ! 1. PROBLEM 2-2 Statement: Solution: How many DOF do you have in your wrist and hand combined? See Mathcad file P0202. Holding the palm of the hand level and facing toward the floor, the hand can be rotated about an axis through the wrist that is parallel to the floor (and perpendicular to the forearm axis) and one perpendicular to the floor (2 DOF). The wrist can rotate about the forearm axis (1 DOF). Each finger (and thumb) can rotate up and down and side-to-side about the first joint. Additionally, each finger can rotate about each of the two remaining joints for a total of 4 DOF for each finger (and thumb). Adding all DOF, the total is Wrist 1 Hand 2 Thumb 4 Fingers 4x4 16 TOTAL 23 2. 3. 2nd Edition, 1999 b. Your knuckle See Mathcad file P0203. d. Your ankle c. 3 DOF: Three rotations about mutually perpendicular axes. Your shoulder. Your knee. 1 DOF: A rotation about an axis parallel to the ground. Your knuckle. 2 DOF: Two rotations about mutually perpendicular axes. c. 1999 . Your knee b. 3 DOF: Three rotations about mutually perpendicular axes. Your shoulder d. Your hip. 3 DOF: Three rotations about mutually perpendicular axes. e 2nd Edition. Your ankle. Solution: a.DESIGN OF MACHINERY SOLUTION MANUAL 2-3-1 ! PROBLEM 2-3 Statement: How many DOF do the following joints have? a. Your hip e. The pen in an x-y plotter. Thus. the depth of the ship with respect to mean sea level is constant. e. f. d. c. 1999 .. In addition. since it probably needs to be oriented with respect to the earth. Thus. b. for an earth-centered frame. it has 6 DOF.DESIGN OF MACHINERY SOLUTION MANUAL 2-4-1 ! PROBLEM 2-4 Statement: How many DOF do the following have in their normal environment? a. d. yaw. however that is not strictly true. e. for all practical purposes. A submerged submarine An earth-orbiting satellite A surface ship A motorcycle The print head on a 9-pin dot matrix computer printer The pen in an x-y plotter Solution: a. See Mathcad file P0201. a submarine has 6 DOF: 3 linear coordinates and 3 angles. For a given position. it will have some heading angle (turning a corner) and roll angle (if turning). A motorcycle. But. c. Using a coordinate frame attached to earth. The print head carrier has 1 DOF and each pin has 1 DOF for a total of 10 DOF. A surface ship. 2nd Edition. If the satellite was just a particle it would have 3 DOF. b. An earth-orbit satellite. a ship can also have pitch. sun. and roll angles. the motorcycle's position is given by two coordinates. there are 4 DOF. There is no difference between a submerged submarine and a surface ship. for a total of 3 DOF . One might argue that. etc. A submerged submarine. or an inertial coordinate frame. The pen holder has 2 DOF (x and y) and the pen may be either up or down. f. both have 6 DOF. At an intersection. The print head on a 9-pin dot matrix computer printer. A ship's position is generally given by two coordinates (longitude and latitude). a surface ship has 5 DOF. 2nd Edition. They are force closed by ligaments that hold them together. 1999 .DESIGN OF MACHINERY SOLUTION MANUAL 2-5-1 ! PROBLEM 2-5 Statement: Solution: Are the joints in Problem 2-3 force closed or form closed? See Mathcad file P0205. None are geometrically closed. Pure rotation. b. A windmill. A conventional "double-hung" window. The pen in an XY plotter h. a. Pure translation for the frame. A hockey puck on the ice g. c. d. A bicycle (in the vertical plane. Pure translation. i. not turning) c. complex planar motion for the wheels.DESIGN OF MACHINERY SOLUTION MANUAL 2-6-1 ! PROBLEM 2-6 Statement: Describe the motion of the following items as pure rotation. not turning). Pure rotation. or complex planar motion. A "casement" window See Mathcad file P0206. h. The pen in an XY plotter. 2nd Edition. A bicycle (in the vertical plane. Pure translation. f. e. The keys on a computer keyboard e. Pure translation. The print head in a computer printer. The hand of a clock f. The print head in a computer printer i. The keys on a computer keyboard. Complex planar motion. A "casement" window. A windmill b. Solution: a. A hockey puck on the ice. pure translation. Pure translation. g. 1999 . Complex planar motion. A conventional "double-hung" window d. The hand of a clock. ( L 1) L J1 J2 2 .DESIGN OF MACHINERY SOLUTION MANUAL 2-7-1 ! PROBLEM 2-7 Statement: Solution: 1. See Figure P2-1 and Mathcad file P0207.1c (Kutzbach's modification) to calculate the mobility. 1999 .( L 1) L J1 J2 2 .J 1 6 7 1 J2 1 1 3 6 3 5 2 4 M=0 (a) b. a. Number of links Number of full joints Number of half joints M 3 . Calculate the DOF of the linkages shown in Figure P2-1.J 1 4 4 0 J2 2 3 1 M=1 (c) 1 7 1 d.( L 1) L J1 J2 2 . Number of links Number of full joints Number of half joints M 3 .J 1 7 7 1 J2 1 6 5 2 3 4 M=3 (d) 2nd Edition. Use equation 2. Number of links Number of full joints Number of half joints M 3 .J 1 3 2 1 J2 1 2 1 M=1 (b) 4 c. Number of links Number of full joints Number of half joints M 3 .( L 1) L J1 J2 2 . Number of links Number of full joints Number of half joints M 3 . See Figure P2-1 and Mathcad file P0208.( L 1) L J1 J2 2 . or preloaded structures.( L 1) L J1 J2 2 . Number of links Number of full joints Number of half joints M 3 . structures.( L 1) L J1 J2 2 .5 on page 32 of the text to classify the linkages.( L 1) L J1 J2 2 .J 1 6 7 1 J2 1 2 4 3 5 6 a. Number of links Number of full joints Number of half joints M 3 .J 1 7 7 1 J2 1 6 5 2 3 4 M=3 Mechanism (d) 2nd Edition. Number of links Number of full joints Number of half joints M 3 .J 1 4 4 0 J2 2 3 1 M=1 Mechanism (c) 1 7 1 d.1c (Kutzbach's modification) to calculate the mobility and the definitions in Section 2. 1999 . PROBLEM 2-8 Statement: Solution: Identify the items in Figure P2-1 as mechanisms.J 1 3 2 1 J2 1 2 1 M=1 Mechanism (b) 4 c.DESIGN OF MACHINERY SOLUTION MANUAL 2-8-1 ! 1. M=0 Structure 1 3 (a) b. Use equation 2. The mechanism in Figure P2-1a has mobility: Number of links Number of full joints Number of half joints M 3 ." One way to do this is to replace one of the pin joints with a pin-in-slot joint such as that shown in Figure 2-3c.( L 1) L J1 J2 2 . but this will increase the DOF by one. See Figure P2-1a and Mathcad file P0209. Use rule 2 on page 40.J 1 6 7 1 6 3 5 2 4 1 1 J2 M=0 2.J 1 6 6 2 J2 6 3 5 2 4 1 1 M=1 2nd Edition. Use linkage transformation on the linkage of Figure P2-1a to make it a 1-DOF mechanism.DESIGN OF MACHINERY SOLUTION MANUAL 2-9-1 ! PROBLEM 2-9 Statement: Solution: 1. Choosing the joint between links 2 and 4. which states: "Any full joint can be replaced by a half joint. we now have mobility: Number of links Number of full joints Number of half joints M 3 .( L 1) L J1 J2 2 . 1999 . which states: "Removal of a link will reduce the DOF by one.J 1 7 7 1 J2 1 2 6 5 3 4 M=3 2. 7 1 The mechanism in Figure P2-1d has mobility: Number of links Number of full joints Number of half joints M 3 . 1999 . See Figure P2-1d and Mathcad file P0210.( L 1) L J1 J2 2 . Use rule 3 on page 40." One way to do this is to remove link 7 such that link 6 pivots on the fixed pin attached to the ground link (1).( L 1) L J1 J2 2 . Use linkage transformation on the linkage of Figure P2-1d to make it a 2-DOF mechanism.DESIGN OF MACHINERY SOLUTION MANUAL 2-10-1 ! PROBLEM 2-10 Statement: Solution: 1.J 1 6 6 1 J2 1 2 3 4 1 6 5 M=2 2nd Edition. We now have mobility: Number of links Number of full joints Number of half joints M 3 . P P 3 .H H 4 . Note that the number of links must be odd to have an even DOF (see Eq.P P H 4 .P 4 .P 4 . For L 7 2 T 2 .Q 3 .H 1 T B Case 2a: H 0 4 9 Case 2b: P 1 1 B Case 2c: P 0 4 9 Case 2c1: Q 2 T B Case 2c2: Q 1 T B Case 2c3: Q 0 T B 0 L T B T L T B 4 9 4 9 4 9 T 2 .6 with DOF = 2 and iterate the solution for valid combinations. 1999 .P Q 4 .P H 4 . Use number synthesis to find all the possible link combinations for 2-DOF.Q T 2 .H 0 T B Case 2: Q 1 T B 2 L 2 L 2 .Q T 2 .H 0 T Q P H B 5 3. up to 9 links.Q T 2 .Q 2 . For L 5 0 T 2 .Q T 2 .Q T Q Q 3 . For L 9 4 T 2 .4).P 3 .3a and 2. see Figure 2-7).Q T 2 .Q 3 . 2.H 4 . using only revolute joints. Use equations 2.H M 2 2.P P Q H 0 T 1 Q H 0 P 0 Case 1: B=8 B=7 2nd Edition.Q 3 . to hexagonal order.Q 3 .Q T 2 .Q T Q Q Q Q T=0 B=7 T=2 B=6 T=4 B=5 Q P Q Q P 3 . The smallest possible 2-DOF mechanism is then 5 links since three will give a structure (the delta triplet. L B T Q P H L L 3 5 M T T 2 . See Mathcad file P0211.DESIGN OF MACHINERY SOLUTION MANUAL 2-11-1 ! PROBLEM 2-11 Statement: Solution: 1.H H 0 P 0 Case 1: T=2 B=5 T=0 B=6 4. we expect that the 16 valid isomers are distributed among the five link sets and that there will be fewer than 16 isomers among the four link sets listed above. See Mathcad file P0212. Four binary and four ternary links. Six binaries. 2. 1. (2) A number is placed within each circle (the "valence" number) to describe the type (ternary. (4) Numbers (0. Therefore. Solution: 1. Table 2-2 shows that there are five possible link sets. 2. One method that is helpful in finding isomers is to represent the linkage in terms of molecules as defined in Franke's Condensed Notations for Structural Synthesis. A summary of the rules for obtaining Franke's molecules follows: (1) The links of order greater than 2 are represented by circles. etc. c. one ternary. b. quaternary. 1999 . 0 3 1 1 3 0 1 3 1 3 8 5 6 3 2 1 4 7 2nd Edition. a. In this case. Table 2-3 on page 38 lists 16 possible isomers for an eightbar chain. Six binaries and two quaternary links.DESIGN OF MACHINERY SOLUTION MANUAL 2-12-1 ! PROBLEM 2-12 Statement: Find all of the valid isomers of the eightbar 1-DOF link combinations in Table 2-2 (p. The number of straight lines emanating from a circle must be equal to its valence number. etc. (3) The circles are connected using straight lines. Five binaries. (5) There is one-to-one correspondence between the molecule and the kinematic chain that it represents.) are placed on the straight lines to correspond to the number of binary links used in connecting the higher order links. However. and one quaternary link. Four binary and four ternary links. four of which are listed above. 36) having: a. two ternaries. Draw 4 circles with valence numbers of 3 in each. d. and one pentagonal link.) of link. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly three lines emanating from each circle and the total of the numbers written on the lines is exactly equal to 4. there are three valid isomers as depicted by Franke's molecules and kinematic chains below. Five binaries.DESIGN OF MACHINERY SOLUTION MANUAL 2-12-2 8 0 3 1 2 3 0 1 3 0 3 5 6 3 1 2 4 7 8 0 3 2 0 3 0 2 3 0 3 5 4 6 3 1 2 7 The mechanism shown in Figure P2-5b is the same eightbar isomer as that depicted schematically above. and one quaternary link. there are five valid isomers as depicted by Franke's molecules and kinematic chains below. and the total of the numbers written on the lines is exactly equal to 5. 1999 . two ternaries. In this case. b. 3 2 1 0 0 3 5 7 4 3 6 2 4 8 1 2 2nd Edition. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly three lines emanating from each circle with valence of three and four lines from the circle with valence of four. Draw 2 circles with valence numbers of 3 in each and one with a valence number of 4. Draw 2 circles with valence numbers of 4 in each. 2nd Edition. In this case. there are two valid isomers as depicted by Franke's molecules and kinematic chains below. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly four lines emanating from each circle and the total of the numbers written on the lines is exactly equal to 6. 1999 .DESIGN OF MACHINERY SOLUTION MANUAL 2-12-3 5 3 0 2 4 1 0 2 3 6 7 4 3 8 1 2 5 3 1 2 4 0 1 1 3 3 4 2 1 7 6 8 5 3 1 1 4 1 0 2 3 6 4 3 7 1 2 8 5 3 1 1 4 1 1 1 3 6 3 4 2 1 8 7 c. Six binaries and two quaternary links. 2nd Edition. 1999 . Six binaries.DESIGN OF MACHINERY SOLUTION MANUAL 2-12-4 0 2 4 2 2 4 7 5 6 4 3 8 1 2 1 1 4 2 2 4 4 7 5 6 8 1 3 2 d. and one pentagonal link. one ternary. There are no valid implementations of 6 binary links with 1 pentagonal link. There are no 2-DOF joints in either mechanism. The mechanism in Figure 2-14a has mobility: Number of links Number of full joints Number of half joints M 3 .J 1 6 7 0 2 5 B 6 J2 1 M=1 2.( L 1) L J1 J2 2 . 1999 . provided that at least two revolute joints remain in the loop. Use rule 1 on page 40. A 3 4 5 2 6 B 1 2nd Edition. A 3 4 Solution: 1. See Figure 2-14a and Mathcad file P0213. The number of links and 1-DOF joints remains the same." One way to do this is to replace pin joints at A and B with translating full slider joints such as that shown in Figure 2-3b. 45).DESIGN OF MACHINERY SOLUTION MANUAL 2-13-1 ! PROBLEM 2-13 Statement: Use linkage transformation to create a 1-DOF mechanism with two sliding full joints from a Stephenson's sixbar linkage as shown in Figure 2-14a (p. Note that the sliders are attached to links 3 and 5 in such a way that they can not rotate relative to the links. which states: "Revolute joints in any loop can be replaced by prismatic joints with no change in DOF of the mechanism. use rule 1 on page 40. 3 5 Solution: 1. See Figure 2-14a and Mathcad file P0213. To get the sliding full joint. 1999 . To get the half joint.DOF joints remains the same. which states: "Revolute joints in any loop can be replaced by prismatic joints with no change in DOF of the mechanism.J 1 5 5 1 2 4 1 3 5 J2 M=1 1 2nd Edition." One way to do this is to remove link 6 (and its two nodes) and insert a half joint between links 5 and 1.( L 1) L J1 J2 2 . Note that the slider is attached to link 3 in such a way that it can not rotate relative to the link. use rule 4 on page 40. provided that at least two revolute joints remain in the loop. which states: "The combination of rules 2 and 3 above will keep the original DOF unchanged. The number of links and 1.DESIGN OF MACHINERY SOLUTION MANUAL 2-14-1 ! PROBLEM 2-14 Statement: Use linkage transformation to create a 1-DOF mechanism with one sliding full joint a a half joint from a Stephenson's sixbar linkage as shown in Figure 2-14b (p." One way to do this is to replace pin joint links 3 and 5 with a translating full slider joint such as that shown in Figure 2-3b. 45). 3 4 2 5 6 1 3. Number of links Number of full joints Number of half joints M 3 .( L 1) L J1 J2 2 . The mechanism in Figure 2-14b has mobility: Number of links Number of full joints Number of half joints M 3 .J 1 6 7 0 J2 4 2 6 M=1 1 2. DESIGN OF MACHINERY SOLUTION MANUAL 2-15-1 ! PROBLEM 2-15 Statement: Calculate the Grashof condition of the fourbar mechanisms defined below. c.5 4. a. 9 . c. 6 ) = "Grashof" This is a special case Grashof since the sum of the shortest and longest is equal to the sum of the other two link lengths. b. 7 ) = "Grashof" Condition( 2 . Condition( S . 2 2 2 4. 4. 1999 . See Mathcad file P0215 Use inequality 2. 7 ) = "non-Grashof" Condition( 2 . Link lengths are in inches (or double given numbers for centimeters).0 . L .5 3. a. 9 . 2nd Edition. 8 . 4. Build cardboard models of the linkages and describe the motions of each inversion. b.0 7 7 6 9 9 8 Solution: 1.5 . P . 3.5 .8 to determine the Grashof condition. Q ) SL PQ S P L Q return "Grashof" if SL PQ return "non-Grashof" otherwise Condition( 2 . To minimize variation of speed with load variation. Those with this characteristic include series-wound. Speed-controlled DC motors will maintain accurate constant speed regardless of load variations. b. Motors with high starting torque are suited to drive large inertia loads.DESIGN OF MACHINERY SOLUTION MANUAL 2-16-1 ! PROBLEM 2-16 Statement: Which type(s) of electric motor would you specify a. To drive a load with large inertia. Motors with flat torque-speed curves (in the operating range) will minimize variation of speed with load variation. and shunt-wound DC motors. 1999 . See Mathcad file P0216. c. and synchronous and capacitor-start AC motors. To maintain accurate constant speed regardless of load variations. and capacitor-start AC motors. b. compound-wound. Solution: a. 2nd Edition. Those with this characteristic include shunt-wound DC motors. b. rotation and translation. Describe the difference between a cam-follower (half) joint and a pin joint. The pin joint also captures its lubricant in the annulus between pin and bushing while the cam-follower joint squeezes its lubricant out of the joint. A cam-follower joint has 2 DOF. See Mathcad file P0217.DESIGN OF MACHINERY SOLUTION MANUAL 2-17-1 ! PROBLEM 2-17 Statement: Solution: 1. 1999 . A pin joint has one rotational DOF. 2nd Edition. Solution: 2nd Edition. Make a cardboard model. Analyze it with a free-body diagram.DESIGN OF MACHINERY SOLUTION MANUAL 2-18-1 ! PROBLEM 2-18 Statement: Examine an automobile hood hinge mechanism of the type described in Section 2.14. Sketch it carefully. Calculate its DOF and Grashof condition. 1999 . Solution of this problem will depend upon the specific mechanism modeled by the student. Describe how it keeps the hood up. Measure it and sketch it to scale. Are there any positions in which it loses stability? Why? Solution of this problem will depend upon the specific mechanism modeled by the student. 1999 . Analyze it with a free-body diagram. Solution: 2nd Edition. Calculate its DOF and Grashof condition. Make a cardboard model. Describe how it keeps itself stable. Sketch it carefully.DESIGN OF MACHINERY SOLUTION MANUAL 2-19-1 ! PROBLEM 2-19 Statement: Find an adjustable arm desk lamp of the type shown in Figure P2-2. The secondary boom (link 7) is rotated about the main boom at pivot B by hydraulic cylinder 5-6. 4. There are 3 translating joints and 12 pin joints. link 7 is pentagonal. 2. Its position is controlled by hydraulic cylinder 8-9 acting on the input link (10). link 7 is pentagonal. and the remainder are binary. Number of links Number of full joints Number of half joints M 3 . Number of links Number of full joints Number of half joints M 3 . There are 2 translating joints and 9 pin joints.10. b.DESIGN OF MACHINERY SOLUTION MANUAL 2-20-1 ! PROBLEM 2-20 Statement: Solution: 1.1c (Kutzbach's modification) to calculate the mobility. The bucket is the output link of a fourbar (links 7.J 1 9 11 0 J2 M=2 9 8 6 5 4 1 2 3 Link 1 is ternary.( L 1) L J1 J2 2 . a. The bucket position is determined by hydraulic cylinder 8-9. and 12).11. Make kinematic sketches. and the remainder are binary. define the types of all the links and joints. All joints are full. Use equation 2. 5. 1999 . Hydraulic cylinder 5-6 is the input to the basic linkage and controls the position of the main boom (link 3). The main boom (link 2) is rotated about fixed pivot A by hydraulic cylinder 3-4. and 6) is a Stephenson's sixbar.J 1 12 15 0 B C 6 7 8 J2 M=3 5 4 2 1 A 3 10 11 12 9 Link 2 is a quaternary.( L 1) L J1 J2 2 . The basic linkage (1. 7 2nd Edition. 3. See Figure P2-3 and Mathcad file P0220. All joints are full. and determine the DOF of the mechanisms shown in Figure P2-3. See Figure P2-4 and Mathcad file P0221. The input is link 2 and the output is link 4. The cross-hatched pivot pins at A and D are attached to the ground link (1). which in this case is the wheel W with a pin at B. 1999 . This is a 3-cylinder. a. The pistons (sliders) at C.( L 1) L J1 J2 2 .1c (Kutzbach's modification) to calculate the mobility. D. B Number of links Number of full joints Number of half joints M 3 . This is a fourbar linkage.J 1 8 10 0 A D 7 4 2 B 3 C 6 5 8 E J2 1 M=1 2nd Edition. and BE. B Number of links Number of full joints Number of half joints M 3 . internal combustion engine. The cross-hatched crank-shaft at A is supported by the ground link (1) through bearings.J 1 4 4 2 3 0 J2 A C 4 D M=1 b. PROBLEM 2-21 Statement: Solution: Find the DOF of the mechanisms in Figure P2-4. 4. The input is link 2. rotary.( L 1) L J1 J2 2 . and 5) BC. There are 3 full joints at B. and E drive the output crank (2) through piston rods (couplers 3.( L 1) L J1 J2 2 . and the output is link 4. Use equation 2. The cross-hatched pivot pins at A and D are attached to the ground link (1).DESIGN OF MACHINERY SOLUTION MANUAL 2-21-1 ! 1. Number of links Number of full joints Number of half joints M 3 .J 1 4 4 0 J2 2 A W 3 4 C D M=1 c. This is a basic fourbar linkage. BD. Since the lengths of links 2 and 4 (AB and CD) are the same.J 1 4 4 0 B 3 C W1 D W2 4 1 J2 2 A M=1 e. The input is link 2 (crank) and the output is link 4 (slider block). The input (rocker) is link 2 and the output (rocker) is link 8. the redundant links and their joints are not counted (subtract 2 links and 4 joints from the totals).J 1 4 4 0 J2 1 F 3 E M=1 2 D 2nd Edition.DESIGN OF MACHINERY SOLUTION MANUAL 2-21-2 d. 4. The cross-hatched pivot pin at A is attached to the ground link (1). which in this case is the wheel W1 with a pin at B. The output dyad consists of links 7 and 8. and 5. This is a fourbar linkage. The cross-hatched pivot pins at A. 1999 . G J 4 3 6 F E H 7 K Number of links Number of full joints Number of half joints M 3 . The cross-hatched pivot pins at A and D are attached to the ground link (1).( L 1) L J1 J2 2 . E Number of links Number of full joints Number of half joints M 3 . This is a fourbar linkage with an output dyad. i.( L 1) L J1 J2 2 . 4 Number of links Number of full joints Number of half joints M 3 .J 1 6 7 0 J2 2 I 5 D M=1 C A B 8 f. The input fourbar consists of links 1. the mechanism will have the same motion if they are removed. The input is link 2.e. In the calculation below. link 3.( L 1) L J1 J2 2 . E and J are attached to the ground link (1). This is a fourbar offset slider-crank linkage. and the output is the vertical member on the coupler. the coupler link (3) has curvilinear motion and BC remains parallel to AD throughout the cycle. 2. Links 3 and 6 are redundant. 1999 . The alternate input dyad consists of links 5 and 6. there are two full joints at the pins A and D for a total of 12 joints. Since this mechanism is symmetrical about a vertical centerline. By splitting it into mirror halves about the vertical centerline the DOF is found to be 1. 6 Number of links Number of full joints Number of half joints M 3 . Analyzing the ninebar. The cross-hatched pivot pins at A and G are attached to the ground link (1).DESIGN OF MACHINERY SOLUTION MANUAL 2-21-3 g. Subtract the 3 redundant links and their 5 (6 minus the joint at A) associated joints to determine the mobility of the mechanism. 6 Number of links Number of full joints Number of half joints M 3 . 3. which reduces it to a sixbar. Either links 2. we can split it into two mirrored mechanisms to analyze it. This is a ninebar mechanism with three redundant links. consider 7. The input fourbar consists of links 1.J 1 9 3 12 0 5 C 2 5 A J2 D 1 3 M=1 E 4 2nd Edition. This is a fourbar linkage with an alternate input dyad.( L 1) L J1 J2 2 . and 4. 3 and 5 or links 7. Number of links Number of full joints Number of half joints M 3 .( L 1) L J1 J2 2 .J 1 6 7 0 J2 4 2 E D 5 F A C 6 B 3 G M=1 h.J 1 9 12 0 J2 1 3 D C 2 5 A B 7 8 F 9 M=0 E 4 G The result is that this mechanism is a structure. The input (rocker) is link 2 and the output (rocker) is link 4. 8 and 9 as the redundant links.( L 1) L J1 J2 2 . To analyze it. 8 and 9 are redundant. 2. Scale the diagrams for dimensions. longest link grounded). r 4 = "non-Grashof" This is a Barker Type 5 RRR1 (non-Grashof. 2nd Edition.50 r2 r4 0.09 A C 4 D Condition r 3 . L . and the output is the vertical member on the coupler.8 to determine the Grashof condition and Table 2-4 to determine the Barker classification.50 1. Condition( S . Q ) SL PQ S P L Q return "Grashof" if SL PQ return "non-Grashof" otherwise a. The cross-hatched pivot pins at A and D are attached to the ground link (1). This is a fourbar linkage. The input is link 2 and the output is link 4. r 1 . which in this case is the wheel W with a pin at B. r 4 = "non-Grashof" This is a Barker Type 5 RRR1 (non-Grashof. and d. longest link grounded). the coupler link (3) has curvilinear motion and BC remains parallel to AD throughout the cycle. The cross-hatched pivot pins at A and D are attached to the ground link (1). r1 r3 1. r1 r3 1. r 1 . Use inequality 2.30 2 A W1 E 1 3 C D W2 4 B Condition r 2 . The input is link 2.30 0. Find the Grashof condition and Barker classifications of the mechanisms in Figure P2-4a. parallelogram).DESIGN OF MACHINERY SOLUTION MANUAL 2-22-1 ! PROBLEM 2-22 Statement: Solution: 1. b.41 0. The cross-hatched pivot pins at A and D are attached to the ground link (1).72 1. Since the lengths of links 2 and 4 (AB and CD) are the same. r 3 . 1999 .76 2 B A W 3 4 C D Condition r 2 . and the output is link 4. r 1 . d. r1 r3 1.16 3 B 2 1.97 r2 r4 0.09 r2 r4 1. which in this case is the wheel W1 with a pin at B. link 3. This is a fourbar linkage. See Figure P2-4 and Mathcad file P0222. This is a basic fourbar linkage. r 3 .66 0. P . r 2 . The input is link 2. two equal pairs. r 4 = "Grashof" This is a Barker Type 13 S2X (special case Grashof. b. 50 2 G B 3 4 E D 5 C 6 Using the notation of inequality 2. Scale the diagrams for dimensions. and 5. the mechanism will have the same motion if they are removed. 5. Thus. By inspection. The input (rocker) is link 2 and the output (rocker) is link 8. The alternate input dyad consists of links 5 and 6. e. r1 r3 0. 4 (or 3). both loops are Class III. 2.88 1.e.15. 1 4 F 3 Since this mechanism has a translating joint it cannot be analyzed using the method defined in "Rotability of N-Bar Linkages" starting on page 53 of the text. N 4 LN r2 L1 r4 L2 r1 L3 r3 LN Since L N L 1 = 2. This is a fourbar linkage with an alternate input dyad. E and J are attached to the ground link (1). The cross-hatched pivot pins at A and G are attached to the ground link (1). i. 1999 . and 4. 2. See Figure P2-4 and Mathcad file P0223. Use inequality 2. The input (rocker) is link 2 and the output (rocker) is link 4. The input fourbar consists of links 1. 7 (or 6). this is a a class II mechanism. The second consists of links 1. and g. This is a fourbar linkage with an output dyad. and 8. The cross-hatched pivot pin at A is attached to the ground link (1). Find the rotability of each loop of the mechanisms in Figure P2-4e.15 to determine the rotability of each loop in the given mechanisms.DESIGN OF MACHINERY SOLUTION MANUAL 2-23-1 ! PROBLEM 2-23 Statement: Solution: 1. The input fourbar consists of links 1. 4. The first loop consists of links 1. 2. we see that the sum of the shortest and longest in each loop is equal to the sum of the other two.880 F A L 1>L 2 L 3 . This is a fourbar offset slider-crank linkage. 3. The cross-hatched pivot pins at A.54 0.040 L2 L 3 = 1. There are two loops in this mechanism.00 r2 r4 1. E 2 D g. Links 3 and 6 are redundant. The input is link 2 (crank) and the output is link 4 (slider block). 2nd Edition. The output dyad consists of links 7 and 8. f. and 5. K J 2 I 3 6 H 7 4 F G E 5 D A C 8 B f. 3.J 1 8 10 0 A 3 P 2 O 1 2 4 B C 7 R 8 Q 6 E O' 5 J2 M=1 2nd Edition.( L 1) L J1 J2 2 . 5. Number of links Number of full joints Number of half joints M 3 . Find the DOF of the mechanisms in Figure P2-5. See Figure P2-5 and Mathcad file P0224.J 1 6 7 0 J2 A 1 1 2 C 3 F 6 B 5 E 4 D a. 1999 . and 6) links. Number of links Number of full joints Number of half joints M 3 .( L 1) L J1 J2 2 .1c (Kutzbach's modification) to calculate the mobility. and 6) and 2 ternary (3 and 4) links. 4. 7. 5. Scale the diagrams for dimensions. The inverted U-shaped link at the top of Figure P2-5a is represented here as the binary link FG (6). This is an eightbar linkage with 4 binary (1. The inverted U-shaped link at the top of Figure P2-5b is represented here as the binary link PR (4). Use equation 2. and 8) and 4 ternary (2. binary links are depicted as single lines with nodes at their end points whereas higher order links are depicted as 2-D bars.DESIGN OF MACHINERY SOLUTION MANUAL 2-24-1 ! PROBLEM 2-24 Statement: Solution: 1. 2. This is a sixbar linkage with 4 binary (1. G M=1 b. In the kinematic representations of the linkages below. When clamped to the ice block but before it is picked up (ice grounded).J 1 2 1 0 J2 M=0 1 2 c. and z position and orientation about a vertical axis.( L 1) L J1 J2 2 . 1999 . a. b.J 1 2 1 0 J2 M=1 b.1c (Kutzbach's modification) to calculate the mobility. y. c. When the block is clamped in the tongs another link and two more full joints are added reducing the DOF to zero (the tongs and ice block form a structure). When the person is carrying the ice block with the tongs. Use equation 2. Number of links Number of full joints Number of half joints M 3 .DESIGN OF MACHINERY SOLUTION MANUAL 2-25-1 ! PROBLEM 2-25 Statement: Find the DOF of the ice tongs in Figure P2-6. When operating them to grab the ice block. 2nd Edition. When the block is being carried the system has at least 4 DOF: x. In this case there two links and one full joint and 1 DOF. Solution: 1. Number of links Number of full joints Number of half joints M 3 .( L 1) L J1 J2 2 . See Figure P2-6 and Mathcad file P0225. a. and K are attached to the ground link (1).( L 1) L J1 J2 2 . See Figure P2-7 and Mathcad file P0226. B. the roller rolls without slipping.1c (Kutzbach's modification) to calculate the mobility. This is an eightbar linkage with 8 binary links. The pivot pins at E.e. PROBLEM 2-26 Statement: Solution: Find the DOF of the automotive throttle mechanism shown in Figure P2-7.J 1 8 10 0 J2 A 7 6 B FULL JOINT C 5 4 J 3 2 N P L 8 K M=1 E 2nd Edition. J. i. Number of links Number of full joints Number of half joints M 3 .DESIGN OF MACHINERY SOLUTION MANUAL 2-26-1 ! 1. It is assumed that the joint between the gas pedal and the roller that pivots on joint N is a full joint. Use equation 2. 1999 . Solution: 1. When the screw is turned to give the jack a different height the jack has 1 DOF.( L 1) L J1 J2 2 . The block on the right is not threaded and acts as a bearing.J 1 7 8 2 J2 M=0 2nd Edition. Describe how it works. 5. The block on the left is threaded and acts like a nut. For any given length of link 7 the jack is a structure (DOF = 0). Joints A and B have 2 full joints apiece. 1999 .DESIGN OF MACHINERY SOLUTION MANUAL 2-27-1 ! PROBLEM 2-27 Statement: Sketch a kinematic diagram of the scissors jack shown in Figure P2-8 and determine its DOF. The two blocks at either end of link 7 are an integral part of the link. Both blocks have pins that engage the holes in links 2. See Figure P2-8 and Mathcad file P0227. and 6. 3. Link 7 is a variable length link. The scissors jack depicted is a seven link mechanism with eight full and two half joints (see kinematic diagram below). 4 3 5 A 2 7 B 6 1 Number of links Number of full joints Number of half joints M 3 . Its length is changed by rotating the screw with the jack handle (not shown). the screw (2). the DOF (mobility) is Number of links Number of full joints Number of half joints M 3 . The second arm is present to balance the forces on the assembly but is not necessary from a kinematic standpoint. screw. and two arms with teeth (3).( L 1) L J1 J2 2 .DESIGN OF MACHINERY SOLUTION MANUAL 2-28-1 ! PROBLEM 2-28 Statement: Solution: 1. 2 full joints (sliding joint between the screw and the body. there are 3 links (body. Find the DOF of the corkscrew in Figure P2-9. one of which is redundant.1c.J 1 3 2 1 J2 M=1 2nd Edition. See Figure P2-9 and Mathcad file P0228. and arm). So. 1999 . and pin joint where the arm rotates on the body). The corkscrew is made from 4 pieces: the body (1). Using equation 2. and 1 half joint where the arm teeth engage the screw "teeth". kinematically. 2 A 2 3 1 1 3 4 B 1 4 C 2. The Barker classification scheme requires that we have 4 link lengths. The motion of link 3 can be modeled by a basic fourbar if the half joint at B is replaced with a full pin joint and a link is added to connect B and the fixed pivot that is coincident with the center of curvature of the slot that guides pin B. The beam 2 is driven in oscillation by the piston of the engine. Links 1 and 3 are ternary. shown above by their pitch circles). 1 half pin-in-slot joint (at B). Scaling the links in Figure P2-10.25 0. Sketch a kinematic diagram of this mechanism and determine its DOF. 2nd Edition.( L 1) L J1 J2 2 . 3 full pin joints. Sketch a kinematic diagram of the mechanism. Kutzbach's mobility equation (2. 1999 .15 1. Solution: 1. Use equation 2.1c) Number of links Number of full joints Number of half joints M 3 . what Barker class and subclass is it? See Figure P2-10 and Mathcad file P0229.54 This is a Grashof linkage and the Barker classification is I-4 (type 4) because the shortest link is the output. The planet gear is fixed rigidly to link 3 and its center is guided in the fixed track 1. There are 4 links. The output rotation is taken from the sun gear 4.DESIGN OF MACHINERY SOLUTION MANUAL 2-29-1 ! PROBLEM 2-29 Statement: Figure P2-10 shows Watt's sun and planet drive that he used in his steam engine.80 r2 r4 1. Can it be classified by the Barker scheme? If so. r1 r3 2.1c to determine the DOF (mobility). It is a fourbar linkage with 1 DOF (see below).J 1 4 3 2 J2 M=1 3. The mechanism is shown on the left and a kinematic model of it is sketched on the right. and 1 half joint (at the interface C between the two gears. Its other node is attached through a full pin joint to a third link. Hint: Consider the flexible cable to be a link. 1999 . Number of links Number of full joints Number of half joints M 3 .DESIGN OF MACHINERY SOLUTION MANUAL 2-30-1 ! PROBLEM 2-30 Statement: Figure P2-11 shows a bicycle hand brake lever assembly. Determine its DOF. which drives the slider (link 4). Solution: 1. The brake lever is a binary link that pivots on the ground link.( L 1) L J1 J2 2 .J 1 4 4 0 J2 3 4 1 1 2 CABLE BRAKE LEVER M=1 2nd Edition. The motion of the flexible cable is along a straight line as it leaves the guide provided by the handle bar so it can be modeled as a translating full slider that is supported by the handlebar (link 1). See Figure P2-11 and Mathcad file P0230. Sketch a kinematic diagram of this device and draw its equivalent linkage. When the brake pad contacts the wheel rim we could consider the joint between the pad. therefore. Number of links Number of full joints Number of half joints M 3 . which is rigidly attached to the brake arm and is. Number of links Number of full joints Number of half joints M 3 .( L 1) L J1 J2 2 . to be a half joint. wheel (which is constrained from moving laterally by the frame). However. a. See Figure P2-12 and Mathcad file P0231. Brake pads contacting the wheel rim.DESIGN OF MACHINERY SOLUTION MANUAL 2-31-1 ! PROBLEM 2-31 Statement: Figure P2-12 shows a bicycle brake caliper assembly. Solution: 1. and the frame constitute a structure. we only need to look at one brake arm.( L 1) L J1 J2 2 . kinematically they operate independently and can be analyzed that way. Brake pads not contacting the wheel rim. Therefore.J 1 2 1 1 J2 1 1 HALF JOINT 2 FRAME BRAKE ARM M=0 2nd Edition. Thus.J 1 2 1 0 J2 1 2 FRAME BRAKE ARM M=1 2. there are two links (frame and brake arm) and one full pin joint. 1999 . The rigging of the cable requires that there be two brake arms. Determine its DOF under two conditions. b. Hint: Consider the flexible cable to be a link. a part of link 2. The brake arm (with pad). Sketch a kinematic diagram of this device and draw its equivalent linkage. When the brake pads are not contacting the wheel rim there is a single lever (link 2) that is pivoted on a full pin joint that is attached to the ground link (1). 1999 . When there is no cable in the jaw or before the cable is crimped this is a basic fourbar mechanism with with 4 full pin joints: Number of links Number of full joints Number of half joints L J1 J2 4 4 0 M 3 .( L 1) 2 . P . r 3 .( L 1) 2 . While the cable is clamped the jaws are stationary with respect to each other so that link 4 is grounded along with link 1. L . Number of links Number of full joints Number of half joints 2. leaving only three operational links.92 0. longest link grounded). Use equation 2. the Grashof condition.50 r2 r4 0.1c (Kutzbach's modification) to calculate the mobility. Find the DOF. See Figure P2-13 and Mathcad file P0232. 2nd Edition. Q ) SL PQ S P L Q return "Grashof" if SL PQ return "non-Grashof" otherwise r1 r3 0. L J1 J2 3 3 0 M 3 . r 1 .27 0.J 1 J2 M=1 When there is a cable in the jaw this is a threebar mechanism with with 3 full pin joints. r 4 = "non-Grashof" The Barker classification is II-1 (Type 5) RRR1 (non-Grashof. and the Barker classifications of the mechanism in Figure P2-13.DESIGN OF MACHINERY SOLUTION MANUAL 2-32-1 ! PROBLEM 2-32 Statement: Solution: 1.8 to determine the Grashof condition and Table 2-4 to determine the Barker classification.J 1 J2 M=0 Use inequality 2. Condition( S .60 Condition r 2 . The rightmost cylinder on the conveyor will be pushed off into the cup that is an extension of link 6. the cylinder will roll out of the cup when link 6 is at its extreme counterclockwise position. a Watt's sixbar. The tips of the comb on link 3 will move below the surface of the conveyor and will then move to the right and. and links 1 and 3 have the same length. 2.e." Sketch its kinematic diagram. Describe what it does: Links 1. Meanwhile.J 1 6 7 0 J2 M=1 3. 2nd Edition. This is characteristic of a Watt's sixbar.DESIGN OF MACHINERY SOLUTION MANUAL 2-33-1 ! PROBLEM 2-33 Statement: Figure P2-14 shows a "pick-and-place" mechanism in combination with a "walking beam. Solution: 1. determine its DOF and its type (i. 3 4 2 5 6 3 4 1 5 6 2 2. a Stephenson's sixbar. When the tips of the comb get back up to the surface of the conveyor they will be displaced to the right a distance of about one diameter of the cylinders on the conveyor. Thus. an eightbar. when they start moving to the left again to complete their cycle..1c to calculate the DOF (mobility). and 6 are pivoted. it is always parallel to link 1. eventually.e. 3. but it is also moving downward. It is a Watt's sixbar. The mechanism is shown on the left and a kinematic model of it is sketched on the right. Link 1 is the ground link about which links 2.( L 1) L J1 J2 2 . is it a fourbar. 4. Link 3 (the coupler) moves to the left while in the position shown. the dyad composed of links 5 and 6 will be rocking back and forth as link 2 rotates. or what?) Make a cardboard model of all but the conveyor portion and examine its motions. Use equation 2. Sketch a kinematic diagram of the mechanism (without the conveyor). 1999 . Because links 2 and 4 have the same length. and 4 are a fourbar with link 2 as the input that rotates at constant speed.1c) Number of links Number of full joints Number of half joints M 3 . start moving upward again.. they will have come in contact with the next cylinder down the conveyor to the right and will move it and the others ahead of it to the left. Describe what it does. See Figure P2-14 and Mathcad file P0233. There are two ternary links (1 and 2) that have a common joint and the remaining joints on the ternary links are connected by two binary links. i. Kutzbach's mobility equation (2. Due to the rocking motion of link 6. which is the ground link. link 4 rotates at the same speed as link 2 and link 3 has curvilinear motion. Kutzbach's mobility equation (2. a Stephenson's sixbar. It is a fivebar linkage with 1 DOF (see below). See Figure P2-15 and Mathcad file P0234. 4 full pin joints. Solution: 1. Use equation 2..1c) Number of links Number of full joints Number of half joints M 3 . Sketch a kinematic diagram of the mechanism. 1999 .DESIGN OF MACHINERY SOLUTION MANUAL 2-34-1 ! PROBLEM 2-34 Statement: Figure P2-15 shows a power hacksaw. There are 5 links. a Watt's sixbar. and is shown below. used to cut metal.( L 1) L J1 J2 2 .J 1 5 5 1 J2 M=1 3. Sketch its kinematic diagram. 1 full sliding joint.( L 1) L J1 J2 2 . Use rule 1 (on page 40) to transform the full sliding joint to a full pin joint for no change in DOF. 7 full pin joints.1c) Number of links Number of full joints Number of half joints M 3 .J 1 6 7 0 J2 1 6 1 5 4 3 2 M=1 2nd Edition. or what?) Use reverse linkage transformation to determine its pure revolute-jointed equivalent linkage. The mechanism is shown on the left and a kinematic model of it is sketched on the right. Then use rules 2 and 3 by changing the half joint to a full pin joint and adding a link for no change in DOF. and 1 half joint (at the interface between the hacksaw blade and the pipe being cut). no half joints. 5 4 2 1 1 3 4 1 5 3 2 2. an eightbar.1c to determine the DOF (mobility). is it a fourbar.e. determine its DOF and its type (i. Kutzbach's mobility equation (2. The resulting kinematically equivalent linkage has 6 links. 4 full pin joints. Solution: 1. a Stephenson's sixbar. is it a fourbar. This is a fourbar slider-crank. an eightbar. or what?) Use reverse linkage transformation to determine its pure revolute-jointed equivalent linkage.1c) Number of links Number of full joints Number of half joints M 3 . It is a fourbar linkage with 1 DOF (see below). no half joints.1c to determine the DOF (mobility). There are 4 links. 1 full sliding joint. and is shown below. Sketch its kinematic diagram.e. Use rule 1 (on page 40) to transform the full sliding joint to a full pin joint for no change in DOF. The resulting kinematically equivalent linkage has 4 links. Sketch a kinematic diagram of the mechanism. Use equation 2. 1999 . a Watt's sixbar. determine its DOF and its type (i..J 1 4 4 0 J2 M=1 3. Kutzbach's mobility equation (2. See Figure P2-16 and Mathcad file P0235. 3 full pin joints. and 0 half joints. 4 1 3 2 1 2nd Edition. 4 3 2 2 1 1 4 3 2.( L 1) L J1 J2 2 . The mechanism is shown on the left and a kinematic model of it is sketched on the right.DESIGN OF MACHINERY SOLUTION MANUAL 2-35-1 ! PROBLEM 2-35 Statement: Figure P2-16 shows a manual press used to compact powdered materials. 4 1 4 1 3 3 2 2 INSTANTANEOUS CENTER OF CURVATURE OF CAM SURFACE 1 1 2.( L 1) L J1 J2 2 . See Figure P2-17 and Mathcad file P0236. 1 pure rolling joint and 0 half joints. and 0 half joints.( L 1) L J1 J2 2 . This is a fourbar slider-crank. 1999 .1c) Number of links Number of full joints Number of half joints M 3 .1c to determine the DOF (mobility) of the equivalent mechanism. Show that it has the same DOF as the original mechanism.DESIGN OF MACHINERY SOLUTION MANUAL 2-36-1 ! PROBLEM 2-36 Statement: Sketch the equivalent linkage for the cam and follower mechanism in Figure P2-17 in the position shown.1c to determine the DOF (mobility) of the original mechanism. Use equation 2. Kutzbach's mobility equation (2. 3 full pin joints.1c) Number of links Number of full joints Number of half joints M 3 . There are 4 links. The cam follower mechanism is shown on the left and a kinematically equivalent model of it is sketched on the right. Use equation 2. Solution: 1. 2 full pin joints.J 1 4 4 0 J2 M=1 2nd Edition. 1 full sliding joint. There are 4 links.J 1 4 4 0 J2 M=1 3. Kutzbach's mobility equation (2. 1 full sliding joint.
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