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Calculate: 3.1 The belt speed in m/s 3.2 Maximum and minimum tension in the belt

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4 parts
3.1

2.09 m/s\text{2.09 m/s}

3.2

15335.01 N\text{15335.01 N}

28737.97 N\text{28737.97 N}

4.1

204.29 r/min (clockwise)\text{204.29 r/min (clockwise)}

4.2

51.33 r/min (clockwise)\text{51.33 r/min (clockwise)}

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Part 3

Part 3.1

2.09 m/s\text{2.09 m/s}

The belt speed in m/s

  1. Convert the driven pulley diameter to meters and its speed to revolutions per second.

    Diameter of driven pulley Dd=210 mm=0.210 mD_d = 210 \text{ mm} = 0.210 \text{ m}.
    Speed of driven pulley Nd=190 r/min=19060 r/sN_d = 190 \text{ r/min} = \frac{190}{60} \text{ r/s}.

  2. Calculate the belt speed (vv).

    v=πDdNdv = \pi D_d N_d v=π×0.210 m×19060 r/sv = \pi \times 0.210 \text{ m} \times \frac{190}{60} \text{ r/s} v=π×0.210×3.1666... m/sv = \pi \times 0.210 \times 3.1666... \text{ m/s} v≈2.09 m/sv \approx \boxed{\text{2.09 m/s}}

Part 3.2

15335.01 N\text{15335.01 N}

Maximum and minimum tension in the belt

  1. Convert the angle of contact to radians.

    Angle of contact θ=180∘=π rad\theta = 180^\circ = \pi \text{ rad}.
    Coefficient of friction μ=0.2\mu = 0.2.

  2. Calculate the tension ratio eμθe^{\mu \theta}.

    eμθ=e0.2×π=e0.6283185≈1.874e^{\mu \theta} = e^{0.2 \times \pi} = e^{0.6283185} \approx 1.874

  3. Use the power transmitted formula to find the difference between the tight-side tension (T1T_1) and slack-side tension (T2T_2).

    Power transmitted P=28 kW=28000 WP = 28 \text{ kW} = 28000 \text{ W}.

    P=(T1−T2)vP = (T_1 - T_2)v T1−T2=Pv=28000 W2.089115 m/s≈13402.80 NT_1 - T_2 = \frac{P}{v} = \frac{28000 \text{ W}}{2.089115 \text{ m/s}} \approx 13402.80 \text{ N}

  4. Set up a system of equations using the tension ratio and the tension difference.

    1. T1T2=1.874  ⟹  T1=1.874T2\frac{T_1}{T_2} = 1.874 \implies T_1 = 1.874 T_2
    2. T1−T2=13402.80 NT_1 - T_2 = 13402.80 \text{ N}

  5. Solve for T2T_2 (minimum tension).

    Substitute (1) into (2):

    1.874T2−T2=13402.801.874 T_2 - T_2 = 13402.80 0.874T2=13402.800.874 T_2 = 13402.80 T2=13402.800.874≈15335.01 NT_2 = \frac{13402.80}{0.874} \approx \boxed{\text{15335.01 N}}

  6. Solve for T1T_1 (maximum tension).

    T1=1.874×T2=1.874×15335.01 N≈28737.97 NT_1 = 1.874 \times T_2 = 1.874 \times 15335.01 \text{ N} \approx \boxed{\text{28737.97 N}}

Part 4

Given:
Annulus (ring gear) TA=75T_A = 75 teeth.
Sun gear TS=35T_S = 35 teeth.
Planetary gear TP=15T_P = 15 teeth.
Arm D.

We use the tabular method for epicyclic gear trains. The fundamental ratio for this configuration (Sun driving Planet, Planet driving Annulus, Arm carrying Planet) is NA−NDNS−ND=−TSTA\frac{N_A - N_D}{N_S - N_D} = -\frac{T_S}{T_A}. The negative sign indicates that if the arm is fixed, and the sun rotates clockwise, the annulus rotates counter-clockwise.

Part 4.1

204.29 r/min (clockwise)\text{204.29 r/min (clockwise)}

Determine the speed and direction of the sun gear if the annulus is fixed and the arm (D) rotates at +65 r/min+65 \text{ r/min}.

Given: NA=0N_A = 0 (annulus fixed), ND=+65 r/minN_D = +65 \text{ r/min}. Find NSN_S.
Using the formula:

NA−NDNS−ND=−TSTA\frac{N_A - N_D}{N_S - N_D} = -\frac{T_S}{T_A} 0−65NS−65=−3575\frac{0 - 65}{N_S - 65} = -\frac{35}{75} −65NS−65=−715\frac{-65}{N_S - 65} = -\frac{7}{15} 65NS−65=715\frac{65}{N_S - 65} = \frac{7}{15} 65×15=7×(NS−65)65 \times 15 = 7 \times (N_S - 65) 975=7NS−455975 = 7N_S - 455 7NS=975+4557N_S = 975 + 455 7NS=14307N_S = 1430 NS=14307≈204.29 r/min (clockwise)N_S = \frac{1430}{7} \approx \boxed{\text{204.29 r/min (clockwise)}}

Part 4.2

51.33 r/min (clockwise)\text{51.33 r/min (clockwise)}

Determine the speed and direction of the annulus if the sun gear is fixed and the arm (D) rotates at +35 r/min+35 \text{ r/min}.

Given: NS=0N_S = 0 (sun fixed), ND=+35 r/minN_D = +35 \text{ r/min}. Find NAN_A.
Using the formula:

NA−NDNS−ND=−TSTA\frac{N_A - N_D}{N_S - N_D} = -\frac{T_S}{T_A} NA−350−35=−3575\frac{N_A - 35}{0 - 35} = -\frac{35}{75} NA−35−35=−715\frac{N_A - 35}{-35} = -\frac{7}{15} NA−35=(−35)×(−715)N_A - 35 = (-35) \times \left(-\frac{7}{15}\right) NA−35=35×715=7×73=493N_A - 35 = \frac{35 \times 7}{15} = \frac{7 \times 7}{3} = \frac{49}{3} NA=35+493N_A = 35 + \frac{49}{3} NA=1053+493=1543≈51.33 r/min (clockwise)N_A = \frac{105}{3} + \frac{49}{3} = \frac{154}{3} \approx \boxed{\text{51.33 r/min (clockwise)}}
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