Is This Transmission System Exercise Correctly Solved?

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SUMMARY

The discussion focuses on the analysis of a transmission system exercise involving a 38-tooth chainring and a 14-tooth sprocket. The calculations confirm that the input torque (Γ1) is 100.062 N·m, leading to an output torque (Γ2) of 36.86 N·m and an output speed (ω2) of 19.89 rad/s. Participants suggest enhancing the problem statement by including a diagram and additional components, such as the force of the pavement on the tire, to provide a more comprehensive understanding of the system's dynamics.

PREREQUISITES
  • Understanding of torque calculations (Γ = Fd)
  • Knowledge of gear ratios and their implications (i = Z1/Z2)
  • Familiarity with basic physics concepts such as force (P = mg) and angular velocity (ω)
  • Ability to interpret mechanical systems and their relationships
NEXT STEPS
  • Research the principles of mechanical advantage in gear systems
  • Learn about the relationship between torque and power in mechanical systems
  • Explore the effects of friction in transmission systems and how to calculate it
  • Study the dynamics of multi-part mechanical problems and how to structure them
USEFUL FOR

Mechanical engineering students, educators designing physics problems, and professionals involved in mechanical system design and analysis.

Guillem_dlc
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Homework Statement


We have a transmission system (plate-pinion), with a 38-tooth chainring and a 14-tooth sprocket. The distance between the crank (between the chainring and the pedal) is 170 mm and the pedal is overloaded with 60 kg and pedalled at a speed of 70 min-1.

Homework Equations


P = mg; Γ = Fd

The Attempt at a Solution


Data:
Z1=38
Z2=14
m=60 kg
g = 9,81 m/s2
d = 170 mm = 0,17 m
ω1 = 70 min-1 = 7,33 rad/s

The first step is to calculate the force (weight) applied to the pedal:
P = m·g → P = 60·9,81=588,6 N

Once we know the force, we have to calculate the input torque (Γ1):
Γ1 = F·d → Γ1 = 588,6 · 0,17 = 100,062 N·m

The next step is to calculate the gear ratio (i) using the number of teeth, and once calculated we can obtain the output torque (Γ2) and output speed (ω2):
i1→2=(Z1) / (Z2) → i1→2=38/14 = 2,71

i1→2=(Γ1) / (Γ2) → Γ2 = (Γ1) / (i1→2 → Γ2=(100,062) / (2,71) = 36,86 N·m

i1→2=(ω2) / (ω1)→ω2=ω1·i1→2→ω2=7,33·2,71=19,89 rad/s

Is the exercise well done? What could I add because the resolution was more physically correct? Any suggestions or ideas?
 
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In the problem, we have to Find the speed and output torque.
 
I think a more complete problem statement would be a good place to start. It is not clear how the components of the system are related to each other, or what a correct answer would look like. A good diagram of the system would help a lot. Also, you could tell the student what to solve for. Why stop with the output torque and angular rate of the sprocket? Why not make it a two-part problem and ask for the force of the pavement on the tire?
 
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tnich said:
I think a more complete problem statement would be a good place to start. It is not clear how the components of the system are related to each other, or what a correct answer would look like. A good diagram of the system would help a lot. Also, you could tell the student what to solve for. Why stop with the output torque and angular rate of the sprocket? Why not make it a two-part problem and ask for the force of the pavement on the tire?
Thanks! I'm going to add what you are saying!
 
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tnich said:
I think a more complete problem statement would be a good place to start. It is not clear how the components of the system are related to each other, or what a correct answer would look like. A good diagram of the system would help a lot. Also, you could tell the student what to solve for. Why stop with the output torque and angular rate of the sprocket? Why not make it a two-part problem and ask for the force of the pavement on the tire?

The force of the pavement on the tire would be equal to the friction force?
 
Guillem_dlc said:
The force of the pavement on the tire would be equal to the friction force?
The force of the pavement on the tire parallel to the pavement surface would be the frictional force.
 
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The chainring wheel power has to be the same as sprocket power. Then Tq1*V1=Tq2*V2.
One rotation of 38 Teeth wheel1=38/14 rotations of the 14 teeth wheel2.
Then V2/V1=38/14 and and Tq2/Tq1=14/38 as you said.
So your calculation is correct.