# Mechanical Technology Grade 12 NSC June 2024 Memo (English)

> Grade 12 Mechanical Technology memorandum from the NSC June Exam 2024 (English), 22 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-mechanical-technology-2024-nsc-june-exam-memorandum-english/da6586d5-1acb-4f61-93d7-ce97c2c85a2a
- Updated: 2026-09-28
- Subject: Mechanical Technology
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2024
- Exam period: NSC June Exam
- Curriculum: CAPS
- Pages: 22

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## About this document

Mechanical Technology · Grade 12 · NSC June Exam · 2024 · English. Memorandum, 22 pages. Read online or download the PDF.

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### Page 1

SENIOR CERTIFICATE EXAMINATIONS/
         NATIONAL SENIOR CERTIFICATE EXAMINATIONS

      MECHANICAL TECHNOLOGY: FITTING AND MACHINING

                                  MAY/JUNE 2024

                             MARKING GUIDELINES

                                                                     MARKS: 200

                     These marking guidelines consist of 22 pages.

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Mechanical Technology: Fitting and Machining     2                 DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 1: MULTIPLE-CHOICE QUESTIONS (GENERIC)

1.1       A ✓                                                                            (1)

1.2       B ✓                                                                            (1)

1.3       A ✓                                                                            (1)

1.4       D ✓                                                                            (1)

1.5       C ✓                                                                            (1)

1.6       C ✓                                                                            (1)
                                                                                         [6]

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Mechanical Technology: Fitting and Machining     3                      DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 2: SAFETY (GENERIC)

2.1       First aid:
          • When illness occurs. ✓
          • When an injury is sustained. ✓
          • When an accident occurs. ✓
                                                                          (Any 2 x 1)           (2)

2.2       Bench grinder:
           A. A fire extinguisher should be readily available. ✓
           B. Safety glasses must be worn. ✓
           C. Maximum grinding wheel speed. ✓
           D. Maximum distance between tool rest and grinding wheel. ✓                          (4)

2.3       Drill press:
          • Never try to stop/hold the work piece by hand when the drill bit get stuck
              during drilling. ✓
          • Don’t force a drill bit into the work piece. ✓
          • Keep loose clothing and hair away from revolving parts. ✓
          • Never leave the machine running if it is unattended. ✓
          • Use a brush or wooden rod to remove chips from the drill. ✓
          • Do not put hands near moving parts. ✓
          • Never clean or adjust the machine while it is in motion. ✓
          • Never try to stop the drill/chuck by hand. ✓
                                                                          (Any 2 x 1)           (2)

2.4       Surface grinder:
          • Never clean or adjust the machine while it is in motion. ✓
          • Know how to stop the machine in an emergency. ✓
          • Do not use excessive force when grinding the work piece. ✓
          • Immediately report any dangerous defects of the machine. ✓
          • Stop using defective machinery until it has been repaired by a qualified
             person. ✓
          • Ensure that the grinding wheel is not submerged in coolant. ✓
          • Never leave the machine running if it is unattended. ✓
          • Do not put hands near moving parts. ✓
                                                                         (Any 2 x 1)          (2)
                                                                                            [10]

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Mechanical Technology: Fitting and Machining     4                 DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 3: MATERIALS (GENERIC)

3.1       Critical temperature:

          3.1.1      Hardening:
                     Above ✓                                                              (1)

          3.1.2      Tempering:
                     Below ✓                                                              (1)

          3.1.3      Normalising:
                     Above ✓                                                              (1)

3.2       Machining test:
          • The chips heating colour ✓
          • The chips curl ✓                                                              (2)

3.3       Material tests:
          • Sound test ✓
          • Bending test ✓
          • Filing test ✓
          • Hardness test ✓
          • Density test ✓
          • Weight measurement ✓
          • Magnetic test ✓
          • Visual inspection/observation ✓
          • Scratch test ✓
                                                                   (Any 3 x 1)            (3)

3.4       Quenching methods:
          • Carburising ✓
          • Nitriding ✓
          • Cyaniding ✓
                                                                   (Any 2 x 1)            (2)

3.5       Heat treatment temperature:
          • Pyrometer ✓
          • Crayons ✓
          • Visually ✓
          • Magnet ✓
                                                                   (Any 1 x 1)            (1)

3.6       Heat-treatment steps:
          • Heat the metal. ✓
          • Soak the metal. ✓
          • Cool the metal. ✓                                                           (3)
                                                                                       [14]

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Mechanical Technology: Fitting and Machining     5                 DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 4: MULTIPLE-CHOICE QUESTIONS (SPECIFIC)

4.1       B✓                                                                              (1)

4.2       C✓                                                                              (1)

4.3       B✓                                                                              (1)

4.4       C✓                                                                              (1)

4.5       D✓                                                                              (1)

4.6       D✓                                                                              (1)

4.7       C✓                                                                              (1)

4.8       A✓                                                                              (1)

4.9       B✓                                                                              (1)

4.10      A✓                                                                              (1)

4.11      B✓                                                                              (1)

4.12      A✓                                                                              (1)

4.13      A✓                                                                              (1)

4.14      D✓                                                                            (1)
                                                                                       [14]

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Mechanical Technology: Fitting and Machining     6                              DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 5: TERMINOLOGY (LATHE AND MILLING MACHINE) (SPECIFIC)

5.1       Advantages of tailstock set-over method:
          • The automatic feed of the machine can be used. ✓
          • Good finish. ✓
          • Long tapers can be cut. ✓
          • Accurate tapers can be cut. ✓
          • Experience less operator fatigue. ✓
                                                                                (Any 2 x 1)             (2)

5.2       Taper calculations:

          5.2.1      Tool:
                     Boring bar. ✓                                                                      (1)

          5.2.2      Diameter of taper:

                                                                              x
                                θ D -d                             tan3,5 =
                             tan =                                          155 
                                2 2l                                   x = tan3,5  155 
                               7 46 − d ✓
                             tan =                                       = 9,46 mm 
                                2 2  155              OR

                             ✓                                         d = 46 − 2 x 
                      310 tan3, 5 = 46 ✓
                                        −d
                                 d =46 − 18, 96                          = 46 − 2(9,46 ) 
                                 d =27, 04 mm ✓                          = 27,04 mm                    (6)

5.3       Calculation of parallel key:

          5.3.1                D
                      Width =
                               4
                               82 ✓
                             =
                                4
                             = 20, 50 mm ✓                                                              (2)

          5.3.2                     D
                      Thickness =
                                    6
                                    82 ✓
                                  =
                                     6
                                  = 13, 67 mm ✓                                                         (2)

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          5.3.3       Lenght = 1,5  diameter of shaft
                             = 1,5  82 ✓
                             = 123 mm ✓                                                   (2)

5.4       Advantages of gang milling:
          • Several surfaces can be milled simultaneously. ✓
          • Saving time. ✓
          • Makes production more effective. ✓
          • Fewer cutters need to be changed. ✓
                                                                   (Any 3 x 1)          (3)
                                                                                       [18]

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Mechanical Technology: Fitting and Machining     8                                DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 6: TERMINOLOGY (INDEXING) (SPECIFIC)

6.1       Gear calculations:

          6.1.1      Number of teeth:
                               PCD
                      Module =
                                 T
                               PCD ✓
                            T=
                                m
                               156
                             =      ✓
                                3
                             = 52 teeth ✓                                                                 (3)

          6.1.2      Dedendum:
                     Dedendum = 1,157 (m)                               = 1, 25 ( m)
                                  = 1,157  3 ✓           OR            = 1, 25  3 ✓
                                  = 3, 47 mm ✓                          = 3, 75 mm ✓                      (2)

          6.1.3      Outside diameter:
                     OD = PCD + 2 ( m)                             = m ( T + 2)
                          = 156 + 2 ( 3 ) ✓         OR             = 3 ( 52 + 2 ) ✓
                          = 162 mm ✓                               = 162 mm ✓                             (2)

          6.1.4      Circular pitch:
                     CP = m π
                          = 3π ✓
                          = 9, 42mm ✓                                                                     (2)

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Mechanical Technology: Fitting and Machining     9                            DBE/May/June 2024
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6.2       Dove tail calculations:

          W = 145 + 2(DE)

          m = W - 2 (AC) - 2 (R) or m = W - 2 ( AC +R)

          6.2.1      Maximum width of dove tail (W):

                     Calculate DE:

                                                                             AD
                                                                    tanθ =        ✓
                             DE ✓                                            DE
                      tan =
                             AD                                              AD
                                                                   tan60  =     ✓
                        DE = AD tan ✓               OR                      DE
                           = 32 tan30                                         32
                                                                      DE =
                           = 18, 48 mm ✓                                     tan60 
                                                                           = 18,48 mm ✓

                      W = 145 + 2 ( DE ) ✓
                         = 145 + 2 (18, 48 ) ✓
                         = 145 + 36, 96
                         = 181,96 mm ✓                                                                (6)

          6.2.2      Distance between the rollers (m):

                     Calculate AC:

                             BC
                      tan α =      ✓
                             AC                                    AC
                              BC                             tan θ =    ✓
                        AC =        ✓                              BC
                             tan                   OR        AC = BC tan60  ✓
                               12                                = 12 tan60 
                           =
                             tan30                              = 20,78 mm ✓
                           = 20, 78mm ✓

                      m = W − 2 ( AC ) − 2 (R ) ✓               m = W − 2 (AC + R ) ✓
                        = 181, 96 − 2 ( 20, 78 ) − 2 (12 ) ✓      = 181,96 − 2 (20,78 + 12) ✓
                                                             OR
                        = 181, 96 − 41, 56 − 24                   = 181,96 − 65,56
                        = 116, 40 mm ✓                            = 116,40 mm ✓                       (6)

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Mechanical Technology: Fitting and Machining    10                          DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

6.3       Milling of spur gear:

          6.3.1      Indexing:
                                 40
                      Indexing =
                                  n
                                 40
                      Indexing =
                                  A
                                  40
                               =
                                 160 ✓
                                 1 6
                               = 
                                 4 6
                                 6
                               =
                                 24 ✓
                     Approximate indexing:
                                     No full turns and 6 holes on a 24-hole circle ✓
                                             OR
                                      No full turns and 7 holes on a 28-hole circle ✓               (3)

          6.3.2      Change gears:
                      D DR             40
                           = (A − n) 
                      D DN             A
                      D DR                   40
                           = (160 − 163 )        ✓
                      D DN                  160
                                   40
                           =−3 
                                  160
                             − 120
                           =
                              160
                             3 8
                           =       ✓
                             4 8
                      D DR   24 ✓
                           =
                      D DN 32 ✓                                                                  (4)
                                                                                                [28]

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Mechanical Technology: Fitting and Machining    11                      DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 7: TOOLS AND EQUIPMENT (SPECIFIC)

7.1       Define hardness:
          Hardness is a material’s ability to resist deformation, ✓ usually by
          indentation/penetration/scratching. ✓                                                 (2)

7.2       Hardness testers:

          •   Brinell hardness tester ✓
          •   Rockwell hardness tester ✓
          •   Vickers hardness tester ✓
                                                                         (Any 2 x 1)            (2)

7.3       Microscope:

          To measure ✓ the diameter/depth of the indentation ✓ left in the test
          material.                                                                             (2)

7.4       Tester:

          7.4.1      Tensile tester ✓                                                           (1)

          7.4.2      Hand wheel ✓                                                               (1)

7.5       Hardness tester:

          Rockwell hardness tester ✓                                                            (1)

7.6       Moment and Force:

          Test the reaction ✓ on either side of simply loaded beam. ✓                           (2)

7.7       Reading:
                              ✓ ✓
          2,00 + 0,40 + 0,5 = 2, 90 mm
                                                                                             (2)
                                                                                            [13]

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Mechanical Technology: Fitting and Machining    12                         DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 8: FORCES (SPECIFIC)

8.1       Forces:

          8.1.1      Horizontal components:

                       HC = 75cos35 − 15cos45 − 5cos0 − 45cos60
                               ✓       ✓     ✓ ✓
                       HC = 61,44 − 10,61 − 5 − 22,5
                       HC = 23,33 N ✓                                                            (5)

          8.1.2      Vertical components:

                       VC = 75sin35 + 15sin45 − 5sin0 − 45sin60
                                ✓      ✓           ✓
                       VC = 43,02 + 10,61 − 0 − 38,97
                       VC = 14,66 N ✓                                                            (4)
                                                       OR
       Force   θ  8.1.1  ∑HC/x = Fcosθ                     8.1.2    ∑VC/y = Fsinθ
       25 N   90º HC = 75cos35°    61,44 N ✓               VC = 75sin35°    43,02 N ✓
       40 N    0º HC = 15cos135°   10,61 N ✓               VC = 15sin135°   10,61 N ✓
       55 N  290º HC = 5cos180°    -5 N    ✓               VC = 5sin180°    0N
      120 N 210º HC = 45cos240°    -22,5 N ✓               VC = 45sin240°   -38,97 N ✓
                             Total 23,33 N ✓                                14,66 N ✓             (9)

          8.1.3      Resultant:

                      R 2 = VC 2 + HC 2
                       R=    (14,66 )2 + (23,33 )2 ✓
                       R = 759,20
                       R = 27,55N ✓                                                               (2)

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Mechanical Technology: Fitting and Machining    13                                  DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

          8.1.4      Angle and direction of resultant:
                     Angle:
                                   VC                                          HC
                            tanθ =                                    tanθ =
                                   HC                                          VC
                                             14,66  ✓                             23,33  ✓
                                  θ = tan−1               OR           θ = tan−1        
                                             23,33                                14,66 
                                  θ = tan−1 (0,63 )                      θ = tan−1 (1,59 )
                                  θ = 32,14 ✓                           θ = 57,86 ✓
                     Direction:
                                                                                N      R
                      R = 27,55 N 32,14  North of East
                                                            ✓                         θ E
                      OR
                      R = 27,55 N 57,86  East from North                                                  (3)

8.2       UDL Beam:

          8.2.1      Distributed load:
                     Uniform distributed load:

                      UDL = 15  6 ✓
                          = 90 N ✓                                                                         (2)

          8.2.2      Reaction in support A:
                     Take moments about B:
                          ✓             ✓          ✓
                     (160  2,5 ) + (90  7) + (55  14) = (A  14)
                                     400 + 630 + 770 = 14A
                                                           1800
                                                      A=
                                                            14
                                                      A = 128,57N ✓

                     Reaction in support B:
                     Take moments about A:
                                               ✓           ✓
                     (B  14) = (55  0) + (90  7) + (160  11,5 )
                         14B = 0 + 630 + 1840
                                 2470
                            B=
                                  14
                            B = 176,43 N ✓                                                              (7)

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Mechanical Technology: Fitting and Machining    14                              DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

          8.3.1      Diameter:
                            F                                 π  d2
                         σ=                                A=
                            A                                   4

                                 F                          ✓
                           A =     ✓                               4A
                                 σ                          d=        ✓
                                                                    π

                           A = 3,19  10 −4         OR
                                                                   4 ( 3,19  10 −4 ) ✓
                                                           d=
                                                                           π
                        πd2
                            = 3,19  10 −4 ✓
                         4
                                                             = 0, 02015 m

                      π  d2 = 1,28  10 −3
                                                             = 20,15 mm ✓

                          d2 =   4,06  10 −4 ✓

                             d = 0,02015 m

                             d = 20,15 mm ✓                                                             (4)

          8.3.2      Change in length:
                         σ
                     E =
                         ε
                         σ
                     ε=
                         E ✓
                         56,5  10 6
                     ε=                ✓
                          90  10 9
                     ε = 6,28  10 − 4 ✓

                           ΔL
                        ε=
                           oL
                      ΔL = ε  oL ✓
                      ΔL = 6,28  10 −4  0,275 ✓
                      ΔL = 1,73  10 −4 m
                             (          )
                      ΔL = 1,73  10 −4  1000
                      ΔL = 0,17 mm ✓                                                                 (6)
                                                                                                    [33]

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Mechanical Technology: Fitting and Machining    15                 DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 9: MAINTENANCE (SPECIFIC)

9.1       Preventative maintenance:
          • Checking for wear and tear on belt. ✓
          • Checking belt alignment. ✓
          • Checking the tensioning devices. ✓
          • Checking the tensioning setting. ✓
          • Make sure all guards are in place. ✓
          • Checking for dirt on belt and pulleys. ✓
                                                                   (Any 4 x 1)            (4)

9.2       High power drives:
          • Gear drive ✓
          • Multiple belt drives ✓
          • Tooth belt drives ✓
          • Chain drives ✓
                                                                   (Any 2 x 1)            (2)

9.3       Bonding methods:
          • Plastic welding (heat) ✓
          • Adhesive/PVC-Weld ✓                                                           (2)

9.4       Uses of the materials:

          9.4.1      Nylon:
                     • Pulleys ✓
                     • Ropes ✓
                     • Bushes ✓
                     • Gears ✓
                     • Wear pads ✓
                     • Wheels ✓
                     • Rollers ✓
                     • Gaskets ✓
                     • Seals ✓
                     • Machinery parts ✓
                                                                   (Any 2 x 1)            (2)

          9.4.2      Fibreglass:
                     • Machine covers ✓
                     • Roof covering ✓
                     • Woven cloth ✓
                                                                   (Any 2 x 1)            (2)

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          9.4.3      Bakelite uses:
                     • Circuit boards ✓
                     • Electrical components ✓
                     • Electrical insulators ✓
                     • Kitchenware ✓
                     • Jewellery ✓
                     • Pipe stems ✓
                     • Toys ✓
                     • Distributor rotor ✓
                     • Distributor cap ✓
                     • Aircraft components ✓
                     • Bearings ✓
                     • Clutch linings ✓
                     • Brake linings ✓
                     • Laminated materials ✓
                                                                   (Any 2 x 1)            (2)

9.5       Thermo-hardened/Thermosetting or Thermoplastic composite:

          9.5.1       Thermoplastic ✓                                                     (1)

          9.5.2       Thermo-hardened / Thermosetting ✓                                   (1)

          9.5.3       Thermo-hardened / Thermosetting ✓                                   (1)

          9.5.4       Thermoplastic ✓                                                   (1)
                                                                                       [18]

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Mechanical Technology: Fitting and Machining    17                 DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 10: JOINING METHODS (SPECIFIC)

10.1      Uses of multiple threads:
          • Fire hydrants ✓
          • Valves ✓
          • Aircraft landing gear ✓
          • Industrial machines ✓
          • Lids of containers/jars ✓
          • Fly press ✓
                                                                   (Any 3 x 1)            (3)

10.2      Multiple screw threads:
          • They provide more bearing surface. ✓
          • Do not strip easily. ✓
          • To provide faster linear movement. ✓
          • They are more efficient. ✓
          • They lose less power to friction. ✓
                                                                   (Any 3 x 1)            (3)

10.3      Square Thread:

          10.3.1     Pitch:
                                   Lead
                      Pitch =
                              Number of starts
                              46
                            =
                               2 ✓
                            = 23 mm ✓                                                     (2)

          10.3.2     Pitch diameter:
                                  P
                     Dm = OD −
                                  2
                                 23 ✓
                         = 80 −
                                  2
                         = 68, 50 mm                                                     (2)

          10.3.3     Helix angle of the thread:
                              Lead
                     tan θ =
                             π  DM
                                    46
                            =             ✓
                                π  68,50
                                             ✓
                         θ =✓tan −1 (0,213755544 )
                           = 12,07 or 124' ✓                                            (4)

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Mechanical Technology: Fitting and Machining    18                         DBE/May/June 2024
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          10.3.4     Leading tool angle:

                      Leading tool angle = 90 − (helix + clearance angle )
                                         = 90 − (12,07 + 3) ✓
                                         = 74,93 or 7455' ✓                                      (2)

          10.3.5     Following tool angle:

                      Following tool angle = 90 + (helix − clearance angle )
                                           = 90 + (12,07 − 3) ✓
                                           = 99,07 or 994' ✓                                  (2)
                                                                                               [18]

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Mechanical Technology: Fitting and Machining    19                   DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

QUESTION 11: SYSTEMS AND CONTROL (DRIVE SYSTEMS) (SPECIFIC)

11.1      Hydraulic calculations:

          11.1.1     The fluid pressure in the hydraulic system in MPa:

                                          π d2
                      A (Plunger ) =
                                           4
                                          π(0,03) 2
                                    A =             ✓
                                               4
                                    A = 0,71  10 −3 m2 ✓

                           F
                      P=
                           A

                              900      ✓
                      P=
                           0,71  10 ✓
                                    -3

                      P = 1,27  106 Pa

                                          ✓
                      P = 1,27 MPa                                                           (5)

          11.1.2     Mass in kg:

                        FA  F
                           = B
                        A A AB
                                  FA  A B
                           FB =            ✓
                                     AA
                                900  31, 42  10 -3 ✓
                           FB =
                                   0, 71  10-3      ✓
                           FB = 39828,17 N
                      Mass = 3982, 82 kg OR 4059, 96 kg ✓                                    (4)

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11.2      Function of hydraulic components:

          11.2.1     Motor
                     Drives the hydraulic pump. ✓                                                     (1)

          11.2.2     One-way-valve
                     To prevent backflow of hydraulic fluid. ✓                                        (1)

          11.2.3     Reservoir
                     Contains the hydraulic fluid. ✓                                                  (1)

11.3      Belt drive:

          11.3.1        The rotational frequency in r/sec:

                      NDR  DDR = NDN  DDN                NDR  DDR = NDN  DDN
                              NDN  DDN                            NDN  DDN
                      NDR =             ✓                  NDR =             ✓
                                DDR                                   DDR
                            5,83  0,5           OR              350  500
                      NDR =            ✓                   NDR =            ✓
                               0,09                                 90
                            2,92                                 1944,44 r/min
                      NDR =                                NDR =
                            0,09                                       60
                      NDR = 32,39 r/sec ✓                  NDR = 32,41 r/sec ✓
                                                                                                      (3)

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          11.3.2     Power transmitted in Watt:

                      Ρ=
                           ( T1 − T2 ) π D N
                               60
                                ✓            ✓        ✓
                      Ρ = (1900 − 450 ) π  0, 09  32, 39
                      P = 13279,18 W ✓
                                 OR

                      Ρ=
                           ( T1 − T2 ) π D N
                               60
                                ✓            ✓      ✓
                      Ρ = (1900 − 450 ) π  0, 5  5, 83
                      P = 13278, 73 W ✓
                                 OR

                      Ρ=
                           ( T1 − T2 ) π D N
                               60
                                ✓            ✓      ✓
                      Ρ = (1900 − 450 ) π  0, 5  350
                      P = 13286, 32 W ✓                                                   (4)

11.4      Avoid slipping:
          • Adding a belt tensioning device. ✓
          • Not subjected to sudden loads. ✓
          • Do not overload the drive. ✓
          • Cover the drives to guard against dust and fluids. ✓
          • Increase the contact area of the belts. ✓
          • Use toothed belts and pulleys. ✓                                              (2)
                                                                   (Any 2 x 1)

Copyright reserved                                                     Please turn over

### Page 22

Mechanical Technology: Fitting and Machining    22                          DBE/May/June 2024
                                     SC/NSC – Marking Guidelines

11.5      Gear drive:

          11.5.1     Number of teeth:

                      Ninput  Product of teethon driven gears
                               =
                     Noutput Product of teethon driver gears
                      NA      T  TD
                            = B
                      ND      T A  TC
                              N  T A  TC
                         TD = A             ✓
                                 ND  TB
                              3500  33  25
                         TD =                 ✓
                                 1050  55
                         TD = 50 teeth ✓                                                         (3)

          11.5.2     Torque:

                        2 π NT                                 P = 2 π NT
                     P=
                          60                                           P
                                                               T=
                     T=
                         P  60                                    2π N ✓
                        2π N ✓                    OR            (737,4  103 )
                                                               T =
                        (737,4  10 )  60                          2  π  17,5 ✓
                                       3
                     T =                   ✓
                            2  π  1050                       T = 6706,33 Nm ✓
                     T = 6706,33 Nm ✓                                                            (3)
                                                                                                [28]

                                                                                TOTAL:          200

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