Calculate the force on the tight side of a belt drive

[Girn261]

Homework Statement

Belt coefficient of friction is .35, pulley diameter is 100cm and maximum torque is 200nm. Calculate the force on the tight side

Homework Equations

T2=T1/e^friction*angle

The Attempt at a Solution

T=fr, 200nm=f x .5m , f=400N

T2=T1/e^friction*angle

400N=T1/e^.35*pie

T1=1201.134N answer
I know this is a fairly simple calculation but I'm just wondering if what I did was right. Thanks

 

[TSny]

Does "maximum torque" refer to the net torque due to both T1 and T2? If so, then I don't think T2 = 400 N.

 

[Girn261]

Does "maximum torque" refer to the net torque due to both T1 and T2? If so, then I don't think T2 = 400 N.

Thank you for the reply, I think you are correct. Thanks for pointing out my mistake , now I need to find T2 & T1.. hmm

 

[TSny]

now I need to find T2 & T1.. hmm

Think about how T1 and T2 are related to the 400 N force that you calculated from the max torque.

 

[Girn261]

Think about how T1 and T2 are related to the 400 N force that you calculated from the max torque.

T1-T2=400N Max torque.. hmm, now two unknowns

 

[TSny]

..and two equations

 

[Girn261]

..and two equations

Sorry bear with me, I'm not that good at this. So,

F2^e*friction*angle - F1/e*friction*angle = 400N. I think that's right so far, now need to figure out next step

 

[TSny]

F2^e*friction*angle - F1/e*friction*angle = 400N.

I don't understand this. What's wrong with your original equation T₂=T₁/e^μθ ? (The tool bar has editing tools for superscripts, etc. Σ on the tool bar will bring up math symbols.)

This, along with T₁ - T₂ = 400 N gives you two equations to work with.

 

[haruspex]

Calculate the force on the tight side

I would just like to point out that there is only enough information to find the minimum tensions. Any equal increase on that to both tensions will still satisfy the conditions.

 

[TSny]

I would just like to point out that there is only enough information to find the minimum tensions. Any equal increase on that to both tensions will still satisfy the conditions.

Would the condition T₁ = e^μθ T₂ still be satisfied?

 

[haruspex]

Would the condition T₁ = e^μθ T₂ still be satisfied?

That is not a given condition. It is just like linear friction: if slipping, T₁ = e^μ_kθ T₂; if not slipping T₁ ≤ e^μ_sθ T₂.

 

[TSny]

That is not a given condition. It is just like linear friction: if slipping, T₁ = e^μ_kθ T₂; if not slipping T₁ ≤ e^μ_sθ T₂.

Yes, I see that now. Thanks, haruspex.