Pulleys and a Belt, with friction and Torque

[Spencer25]

Homework Statement

Hi Everyone, this question is from a government exam for but I think it is beyond the scope of the course. Any help would be appreciated.

A motor pulley with a radius of 0.2 meters is connected to a drum with a radius of .6 meters. The coefficient of the belt and driver is .35 and the belt and drum .25. The total force is 2300N. What is the Torque on each pulley? Which will slip first? Draw a free body diagram.

Relevant Equations

T=Fxr
Total Force = F1 - F2

Torque on driver
T = 2300N x .2m
= 460Nm

Torque on drum
T=2300N x .6m
T = 1380Nm

Not sure how to calculate friction force on these...Do we have to assume a contact angle of 180 degrees if nothing is given or is that even required?
Not sure what a free body diagram would look like for this.

 

[haruspex]

Homework Statement:: Hi Everyone, this question is from a government exam for but I think it is beyond the scope of the course. Any help would be appreciated.

A motor pulley with a radius of 0.2 meters is connected to a drum with a radius of .6 meters. The coefficient of the belt and driver is .35 and the belt and drum .25. The total force is 2300N. What is the Torque on each pulley? Which will slip first? Draw a free body diagram.
Relevant Equations:: T=Fxr
Total Force = F1 - F2

Torque on driver
T = 2300N x .2m
= 460Nm

Torque on drum
T=2300N x .6m
T = 1380Nm

Not sure how to calculate friction force on these...Do we have to assume a contact angle of 180 degrees if nothing is given or is that even required?
Not sure what a free body diagram would look like for this.

You are right that there is not enough information. We need to know how far apart they are.
If you take them to be so far apart that the contact angle is the same then it is unrealistic and the question becomes trivial.
You could assume they are almost touching.

 

[Spencer25]

You are right that there is not enough information. We need to know how far apart they are.
If you take them to be so far apart that the contact angle is the same then it is unrealistic and the question becomes trivial.
You could assume they are almost touching.

Thank you Haruspex, these exams often have mistakes but since I am not a physics whiz, if we did assume the contact angle was 180 degrees could you tell me how to solve? And a free body diagram?

 

[haruspex]

Thank you Haruspex, these exams often have mistakes but since I am not a physics whiz, if we did assume the contact angle was 180 degrees could you tell me how to solve? And a free body diagram?

There is another difficulty with this question: what is meant by "total force"?
In a belt drive, there are different tensions on the two straight sections of belt. Call these T₁>T₂. On one drum the net torque is (T₁-T₂)r₁, on the other (T₁-T₂)r₂ the other way.
Are you familiar with the capstan equation? If the belt traverses angle $\theta_1$ around drum 1 then to avoid slipping $T_1<T_2e^{\mu_1\theta_1}$.
Clearly $\theta_1+\theta_2=2\pi$.

If we take the drums to be a long way apart then maybe total force means the sum of the tensions, but the second part of the question becomes trivial; obviously the lower coefficient drum will slip first. And, as I wrote, it is wholly unrealistic.

If we take them to be almost touching then we can calculate the angles from the radii, but I am left wondering what the total force refers to.

Does the exam website publish solutions?

 

[Spencer25]

I think what they mean by total force is just tight side minus slack side. I know these questions are well below what people are used to on these forums and is very trivial for your skill level so I really do appreciate you taking the time for this. These questions are from Power Engineering (boiler operation) here in Canada...known as Stationary Engineers in the US. I just learned about Capstan today reading some other threads...it seems to make sense to me. The question as to which one would slip first I was not so sure if it would be related to Mu = Friction/Reaction Normal where maybe Rn would be the total force and which ever one had the higher friction force would slip last?

 

[haruspex]

I think what they mean by total force is just tight side minus slack side. I know these questions are well below what people are used to on these forums and is very trivial for your skill level so I really do appreciate you taking the time for this. These questions are from Power Engineering (boiler operation) here in Canada...known as Stationary Engineers in the US. I just learned about Capstan today reading some other threads...it seems to make sense to me. The question as to which one would slip first I was not so sure if it would be related to Mu = Friction/Reaction Normal where maybe Rn would be the total force and which ever one had the higher friction force would slip last?

Tight minus slack would lead to net torque, as I posted, but "total force"?

The capstan equation is the curvilinear equivalent of, and is derived from, the usual max friction = coefficient x normal force equation.

What hangs on answering the question as posed? Will you get marked on it or is it just practice? I feel the best for purposes of your own education is to ignore the details of the question and just try to develop the general equations, much of which I have already given you.

 

[Spencer25]

The exams are answer 5 of 7 questions so I skipped this one as I had never seen this type. Usually just calculating power and torque with two pulleys and a belt. This was the first time I had seen two coefficients of friction thrown in so thos is more for my own learning. Thank you so much for your help. I understand it better now.