Effect of Increase in Load Torque on Belt Tension & Power

AI Thread Summary
Increasing load torque while maintaining constant speed results in higher tension on the tight side of the belt and lower tension on the slack side. The power transmitted also increases as torque rises, given that angular velocity remains constant. The relationship between the forces on the tight and slack sides is governed by the capstan equation, which maintains a constant tension ratio. As torque increases, both tensions will adjust until they reach their maximum and minimum values, potentially leading to belt slippage. Understanding these dynamics is essential for effective belt and pulley system design.
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Homework Statement


What would be the effect on the following if the load torque is increased and the speed maintains constant:

a) the tension in the tight side of the belt

b) the tension in the slack side of the belt

c) the power transmitted

Homework Equations

The Attempt at a Solution


I believe I need to find a relationship between the torque T and the F1, F2 and power P.

I will start with the easiest one which I believe is c).
P=T*w
If w is constant, then when the torque will increase, power will also increase.

I have a problem with F1 and F2. Can anyone point me in the right direction on how to create a relationship between the P and F1 and P and F2?
 
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Power = Force * Velocity

In the case of a belt, the force applied to the pulley is the belts tight side minus the slack side so:
P = (F1 - F2)v

For a given belt & pulley arrangement, the ratio of F1 to F2 is constant.

For a flat belt the ratio is:
F1/F2 = eμθ
where μ is the coefficient of friction and θ is the angle (in radians) of the arc of the contact surface.

If the pulleys are of different size then the smallest θ should be used, that is, the smallest pulley will always slip first.
The equation for a V belt is a bit more complicated as you need to account for the angle of the v but you don't need it to solve your problem.
 
billy_joule said:
Power = Force * Velocity

In the case of a belt, the force applied to the pulley is the belts tight side minus the slack side so:
P = (F1 - F2)v

For a given belt & pulley arrangement, the ratio of F1 to F2 is constant.

For a flat belt the ratio is:
F1/F2 = eμθ
where μ is the coefficient of friction and θ is the angle (in radians) of the arc of the contact surface.

If the pulleys are of different size then the smallest θ should be used, that is, the smallest pulley will always slip first.
The equation for a V belt is a bit more complicated as you need to account for the angle of the v but you don't need it to solve your problem.

Question is about the v-belt pulley - see topic :)

I've actually done the question before I've seen your reply. My answer is:

a) & b)
According to the equation T=(f1 - F2)*r, assuming r will not change, when torque is increased, the effective tangential force (F1 - F2) is increased. With no torque applied when the belt is in equilibrium with F1 and F2 is equal, so tension force F = F1 + F2. Once torque is applied and the pulley is turned clockwise F1 (tight side) will increase and F2 (slack side) will decrease. When we continue to increase the torque the F1 and F2 will continue to increase and decrease until they will reach their max and min values (due to limiting ratio of tensions) and the belt will start to slip.

c)
According to equation power P = T * w. If the angular velocity w is constant and the torque is increased, the power will also increase.

Do they make sense, what you think?
 
Can I suggest going for a ride on a bicycle? What happens to the tension in the chain when you come to a hill and try to ride up at the same speed? The answers should really be intuitive for an engineering student.
 
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CWatters said:
Can I suggest going for a ride on a bicycle? What happens to the tension in the chain when you come to a hill and try to ride up at the same speed? The answers should really be intuitive for an engineering student.

Is that not what I've said? Is my thinking incorrect? I believe, using your analogy, when going up hill with same speed the tension on the chain will increase as more torque needs to be applied. As with a bikes chain and gears there is no slip, if you apply to much torque the chain will break.
 
sponsoraw said:
Question is about the v-belt pulley - see topic :)
Yes. Like I said, the flat belt equations are adequate here, There's is no need to overcomplicate things for this introductory problem. In fact I think it's borderline misleading, some students may get stuck attempting to use the V belt equations without the V angle.

Do they make sense, what you think?

C is correct.

A & B are not.

"According to the equation T=(f1 - F2)*r, assuming r will not change,when torque is increased, the effective tangential force (F1 - F2) is increased. With no torque applied when the belt is in equilibrium with F1 and F2 is equal, so tension force F = F1 + F2. "

F = F1 +F2 is not correct. That implies both forces are in the same direction, which we know they are not either from logic, a FBD or from T=(f1 - F2)*r

When no torque is applied to the pulley:
T= 0 = (f1 - F2)*r ,

r is a non zero constant, therefore
F1-F2 = 0 = F

"Once torque is applied and the pulley is turned clockwise F1 (tight side) will increase and F2 (slack side) will decrease. When we continue to increase the torque the F1 and F2 will continue to increase and decrease until they will reach their max and min values (due to limiting ratio of tensions) and the belt will start to slip."
That disagrees with the tension ratio equation in post #2. the ratio of F1 and F2 is constant so if F1 increases by a factor of, say, 5, F2 also increases by a factor of 5.

It's an application of the capstan equation, you can see a derivation here:
https://en.wikipedia.org/wiki/Capstan_equation#Proof_of_the_capstan_equation
 
billy_joule said:
Yes. Like I said, the flat belt equations are adequate here, There's is no need to overcomplicate things for this introductory problem. In fact I think it's borderline misleading, some students may get stuck attempting to use the V belt equations without the V angle.
C is correct.

A & B are not.
F = F1 +F2 is not correct. That implies both forces are in the same direction, which we know they are not either from logic, a FBD or from T=(f1 - F2)*r

When no torque is applied to the pulley:
T= 0 = (f1 - F2)*r ,

r is a non zero constant, therefore
F1-F2 = 0 = F

That disagrees with the tension ratio equation in post #2. the ratio of F1 and F2 is constant so if F1 increases by a factor of, say, 5, F2 also increases by a factor of 5.

It's an application of the capstan equation, you can see a derivation here:
https://en.wikipedia.org/wiki/Capstan_equation#Proof_of_the_capstan_equation

I think it's getting complicated a bit now, where it should be straight forward. Let me start again, maybe I don't explain good enough.

When the belt is in equilibrium and a force F is applied to pre-tension the belt, that force is F=F1+F2, where F1=F2=0.5F. Both tight and slack side are same at that point. I believe, correct me if I'm wrong, they are of the same magnitude but opposite direction (sign). Once you apply torque to rotate the pulley the F1 will start to increase and F2 decrease. They will increase and decrease proportionally to a ratio described in the limiting ration of tensions equation. That ratio, when discussing v-belt, depends on coefficient of friction, angle of lap and the angle of pulley groove. When we continue to increase the torque, the F1 and F2 will eventually reach their max and min values and the belt will start to slip.

How this sound now? I hope it's more clear. If it's still wrong then I might be to thick or missing something.
 
sponsoraw said:
I think it's getting complicated a bit now, where it should be straight forward. Let me start again, maybe I don't explain good enough.

When the belt is in equilibrium and a force F is applied to pre-tension the belt, that force is F=F1+F2, where F1=F2=0.5F. Both tight and slack side are same at that point. I believe, correct me if I'm wrong, they are of the same magnitude but opposite direction (sign). Once you apply torque to rotate the pulley the F1 will start to increase and F2 decrease. They will increase and decrease proportionally to a ratio described in the limiting ration of tensions equation. That ratio, when discussing v-belt, depends on coefficient of friction, angle of lap and the angle of pulley groove. When we continue to increase the torque, the F1 and F2 will eventually reach their max and min values and the belt will start to slip.

How this sound now? I hope it's more clear. If it's still wrong then I might be to thick or missing something.

The tight and slack side tensions are equal only when T = 0, when that occurs, both tensions are zero; F = F1 = F2 = 0.

Regarding the bolding: what you describe is not a constant ratio.

F1/F2 = e^μθ
Lets say e^μθ is 5, which is the sort of value you could expect. So:
F1/F2 = 5
then
F1 = 5*F2

So if the tight side increases (F1), then the slack side (F2) increases also.
eg if F1 is 10 Newtons then F2 is 2 Newtons
If F1 doubles to 20 N, as torque increases, then F2 also doubles to 4 N - the ratio remains constant.
 
Thanks for explaining. I went back to my books and I know where I got confused, not a very good wording. I finally got it now.

The only question I've got is how do I calculate the maximum torque allowed before the belt start to slip?
 
  • #10
You'll need a lot more information to answer that question.
It depends on the speed - as rpm increases centrifugal forces increase so the normal force and friction force decreases. (linear mass of the belt is relevant here)
Or maybe it'll never slip - the tensile strength of the belt could be exceeded before slipping ie the belt breaks.
Work through your textbook, it'll probably have a similar example (with enough info given to answer)
 
  • #11
I don want to be a Necto poster but If anyone can just confirm it, Is this answer correct for parts a+b?
sponsoraw said:
Question is about the v-belt pulley - see topic :)

I've actually done the question before I've seen your reply. My answer is:

a) & b)
According to the equation T=(f1 - F2)*r, assuming r will not change, when torque is increased, the effective tangential force (F1 - F2) is increased. With no torque applied when the belt is in equilibrium with F1 and F2 is equal, so tension force F = F1 + F2. Once torque is applied and the pulley is turned clockwise F1 (tight side) will increase and F2 (slack side) will decrease. When we continue to increase the torque the F1 and F2 will continue to increase and decrease until they will reach their max and min values (due to limiting ratio of tensions) and the belt will start to slip.

c)
According to equation power P = T * w. If the angular velocity w is constant and the torque is increased, the power will also increase.

Do they make sense, what you think?

I am currently studying the same topic with the same assignment question. I am not sure my answer is entirly correct as I have used the ulitmate belt strength to calculate my answer, which eliminates the chance of F_1 getting tighter. although it does not become slacker.
 
  • #12
Al_Pa_Cone said:
I am currently studying the same topic with the same assignment question. I am not sure my answer is entirly correct as I have used the ulitmate belt strength to calculate my answer, which eliminates the chance of F_1 getting tighter. although it does not become slacker.

It sounds like your problem statement is slightly different. I suggest starting a new thread.
 
  • #13
It is the exact same question from tees university mechanical principals assignment.

My reference in post 11 was subject to the measurements taken in the previous question. However I cannot use those measurements as the question required me to give the maximum power transmitted.

I have started a new thread with all of question 1 parts labelled, calculating power/ torque transmitted between two pulleys. Thanks
 
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