Understanding Pulleys, Strings, and Tensions: Common Doubts and Clarifications

[Tanya Sharma]

The frictional force at the axle would be multiplied by the radius of the axle to get the resulting frictional torque.

So you would have:

Iα = (T2-T1)R - fr​

Where T1 is the one tension, T2 is the other tension, R is the pulley radius, r is the axle radius and f is the frictional force at the axle.

Thanks...Things are getting clearer...Now please consider a frictionless pulley with sufficient friction between string and pulley.

Why do we write net torque on the pulley as (T2-T1)R ?

The tangential force acting on the pulley is friction F and not the tension in the string.Tension doesn't act on the pulley.

So,the torque on the pulley is FR.

But,since the friction F is equal to the difference in the tensions at the two ends of the arc we write the torque as (T2-T1)R .

Is my understanding correct ?

 

[jbriggs444]

But,since the friction F is equal to the difference in the tensions at the two ends of the arc we write the torque as (T2-T1)R .

Is my understanding correct ?

Yes.

Note that there is some lack of rigor talking about "friction F" as if it had a well-defined value. Force is a vector. The friction between the string and the pulley is not a single well defined vector whose magnitude is "F". Instead, it is spread out (and may not be spread out evenly) over the surface of the pulley. Fortunately for our purposes, every incremental bit of string that is stretched out over the surface of a circular pulley contributes its frictional force with a moment arm of length R at right angles to the incremental force. So we can safely shrug, treat the net tangential F as if it were a simple number and correctly conclude that its value is T2-T1.

 

[Tanya Sharma]

Thanks jbriggss...The explanation is simply wonderful.

Loosely speaking,this is how I thought that the net friction would be equal to the difference in tensions.

The forces acting on a tiny bit of a string are the net tension difference,say dT and friction f .Now,since the string is massless,the net force on every bit of the string should be zero.So we have dT=f .Summing them along the arc,we get ΔT=F or T2-T1 =F.

Is it correct to think in this manner ?

 

[jbriggs444]

Loosely speaking,this is how I thought that the net friction would be equal to the difference in tensions.

The forces acting on a tiny bit of a string are the net tension difference,say dT and friction f .Now,since the string is massless,the net force on every bit of the string should be zero.So we have dT=f .Summing them along the arc,we get ΔT=F or T2-T1 =F.

Is it correct to think in this manner ?

Yes. This is a good way to come up with a number "F" that is, in some perfectly reasonable sense, the total friction.

The lack of rigor that I had in mind comes up when multiplying this "F" by R to try to compute torque. The result has units of torque and actually gives the right torque in the case at hand, but it would not be a justifiable computation in the case of arbitrary (non-circular) pulleys. There are also complications if the string is wrapped in a slanted or crooked path around the pulley.

 

[jbriggs444]

Loosely speaking,this is how I thought that the net friction would be equal to the difference in tensions.

The forces acting on a tiny bit of a string are the net tension difference,say dT and friction f .Now,since the string is massless,the net force on every bit of the string should be zero.So we have dT=f .Summing them along the arc,we get ΔT=F or T2-T1 =F.

Is it correct to think in this manner ?

Yes. This is a good way to come up with a number "F" that is, in some perfectly reasonable sense, the total friction.

The lack of rigor that I had in mind comes up when multiplying this "F" by R to try to compute torque. The result has units of torque and actually gives the right torque in the case at hand, but it would not be a justifiable computation in the case of arbitrary (non-circular) pulleys. There are also complications if the string is wrapped in a slanted or crooked path around the pulley.

 

[Tanya Sharma]

Thank you jbriggs