Is This Transmission System Exercise Correctly Solved?

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Homework Help Overview

The discussion revolves around a transmission system involving a plate-pinion setup with specific parameters such as a 38-tooth chainring and a 14-tooth sprocket. The problem includes calculating forces, torques, and speeds related to the system while considering the physical relationships between the components.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations of force, input torque, gear ratio, output torque, and output speed. There are suggestions to clarify the problem statement and to consider additional aspects such as the force of the pavement on the tire. Some participants question the completeness of the original problem and propose the inclusion of diagrams.

Discussion Status

The discussion is ongoing, with participants providing feedback on the clarity of the problem statement and suggesting improvements. Some guidance has been offered regarding the relationships between the components, but no consensus has been reached on the completeness of the exercise.

Contextual Notes

There are indications that the problem may lack sufficient detail regarding the relationships between components and expected outcomes. Participants are exploring the implications of the calculations and questioning assumptions about the system's setup.

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.