Power Transmission with Friction Coefficient: 0.25 to 0.2

AI Thread Summary
The discussion focuses on the impact of reducing the friction coefficient from 0.25 to 0.2 on power transmission in a mechanical system. Calculations show that while power remains constant, adjustments in force and tension are necessary, with a maximum belt tension limit of 1000N. Suggestions include increasing the contact angle of the belt on the pulley to maintain power without exceeding tension limits. Participants discuss the use of Euler's formula to determine the torque at which the belt may slide and how to calculate the lap angles for effective power transmission. Ultimately, the conversation concludes with a clearer understanding of the calculations required to solve the problem.
StevenBennett
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1. Homework Statement

After a certain amount of running, there is likely to be changes in the values of the engineering parameters of the system in section (a). One of these is the friction coefficient, Carry out the activity with calculations to find out the effect on power transmission if the friction coefficient reduces to 0.2 from 0.25: how can the original maximum power value be restored?

Ive worked out the power for both 0.25 and 0.2

The power is the same but i had to change F1 to 1165N, but the maximum tension is the belt can only exceed 1000N how could i get the same powers? would i have to use a belt tension device?

1a - has the calculations for working out 0.25 as the coefficient and the original question
1c - has me working out for 0.2 as the coefficient, also shows how to gain the same power with 1165N as F1
 

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StevenBennett said:
The power is the same but i had to change F1 to 1165N, but the maximum tension is the belt can only exceed 1000N how could i get the same powers? would i have to use a belt tension device?
I haven't seen a sketch of the system, but anyway here is a suggestion:
upload_2015-6-17_22-30-35.jpeg

The purpose is not to increase the tension, but to increase the angle wherein the belt is in connection with the wheel.

There is this "Eulers formula" to calculate at which torque the belt will slide.
 

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The image shows the system...

Small Diameter: 150mm
Large Diameter: 200mm
Apart Length: 600mm

and were working with the large pulley angle when calculating power

how will i know the angle of were the wheel will have to be?

so i guess this will be changing (v) which is 3.14? or the lap angle of the large pulley which is 3.225 Radians?

and will it just be trail and error?
 

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StevenBennett said:
how will i know the angle of were the wheel will have to be?
Say you have calculated the angles θ1 and θ2:
upload_2015-6-17_23-10-55.jpeg

You just draw the tangents ( stipled lines ) and place the small wheel as shown.

StevenBennett said:
and will it just be trail and error?
No, it should be possible to calculate θ1 and θ2. I'm not sure about the name of the law to be used ( I'm not a mechanical engineer ), but there is a law by which max torque can be calculated as a function of tension, angle and friction coefficient.
 
Hesch said:
No, it should be possible to calculate θ1 and θ2. I'm not sure about the name of the law to be used ( I'm not a mechanical engineer ), but there is a law by which max torque can be calculated as a function of tension, angle and friction coefficient.

So once I've found the angles, i just sub them in when working out the lap angles? for example:

Large Pulley Lap Angle

θ = θ2 + 2B =

instead of using 180? or?
 
StevenBennett said:
θ = θ2 + 2B

Sorry, I don't quite understand what you mean. What is θ ?

The lap angle as for the large pulley is θ2, shown in the figure in #4.
 
Hesch said:
Sorry, I don't quite understand what you mean. What is θ ?

The lap angle as for the large pulley is θ2, shown in the figure in #4.

Dont worry but cheers for the help, i fully understand how to work out this question now so cheers for that!
 
Maybe you should make the small wheel larger than drawn by me, otherwise your teacher will get angry with you, because you are bending the belt too much.:smile:
 
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