Engineering Dynamics question — Belt pulled between two cylindrical rollers

[Pipsqueakalchemist]

Homework Statement

Question below

Relevant Equations

Newton’s 2nd law
Moment equation

For question, in the solution when setting up the moment equation, why does the solution use Fa and Fb? Friction is coefficient times normal force and that can be found with the info given so why does the solution take it to be an variable?

 

[Lnewqban]

Homework Statement:: Question below
Relevant Equations:: Newton’s 2nd law
Moment equation

For question, in the solution when setting up the moment equation, why does the solution use Fa and Fb? Friction is coefficient times normal force and that can be found with the info given so why does the solution take it to be an variable?

The forces that the belt applies onto each cylinder, each of which has a different rotational inertia, can be smaller, equal or greater than the theoretical maximum force that could exist between belt and each of the cylinders surfaces before they start to skid respect to each other.

Static friction force, depending on the normal forces and coefficient of static friction, is the maximum "grip" or "traction" possible between two surfaces having no relative movement.

Once the surfaces break that barrier and skid begins, the force that resists that relative movement is the dynamic friction (normally smaller in magnitude).

 

[Pipsqueakalchemist]

How do you know when to use the max value?

 

[Lnewqban]

How do you know when to use the max value?

The max value is the only one we calculate for friction force.
Let's say you are trying to tow a car with the rear wheels being locked by the parking brake.
The tow truck starts pulling the car from its front, let's say applying 500 Newtons of horizontal force.
At the contact patch of each rear tire, because Newton's second law, a reaction of 250 Newtons develops, but tires do not skid.

Now, we know that the coefficient of static friction between dry asphalt and rubber is 0.8.
We calculate that each rear tire is supporting a partial weight of 500 Newtons; therefore, there will not be any skid until a magnitude of horizontal force of 0.8 x 500 = 400 N is applied: that is the magnitude of your static friction.

Once the tow truck reaches a pulling force of 800 N, it will be able to drag the car with rear blocked tires.
What force will resist the pull after that moment?
The dynamic friction on each of the rear contact patches, which should be around 0.7 x 500 x 2 = 700 Newtons.

 

[Pipsqueakalchemist]

So the reason it’s a variable is bc we don’t know if it’s static or kinetic friction?

 

[vela]

How do you know when to use the max value?

You use it when you have reason to believe the static friction is at its maximum value. Often, there's some wording in the problem like "on the verge of moving" which translates physics-wise into the condition that static friction is as large as it can get.

If you don't have any reason to assume static friction is at its maximum value, then you don't know how big the force is. It will be whatever it needs to be to keep the object from moving relative to the surface it's in contact with.

 

[Lnewqban]

So the reason it’s a variable is bc we don’t know if it’s static or kinetic friction?

Please excuse me the delayed response to your good question.
In this specific problem, they first assume no slipping occurs between belt and cylinders (like if the belt was a string wrapped around each cylinder several times).

By doing so, they calculate the resistive tangential force that the moments of inertia of both cylinders would induce as a function of the angular acceleration of the system (both cylinders acting as one).

Hence, the more force you pull the strings with, the more you overcome the inertial resistance of the system and the greater the angular accelerations of both cylinders get.
But in actuality, you don’t have wrapped strings but only the friction of both surfaces of the belt against the surfaces of the cylinders to overcome that resistance to accelerate.

Then, they calculate the maximum static friction that keeps the cylinders and belt from slipping.
If the belt pulls with a force greater than that value, slipping will occur and that will be the limit to obtain higher values of angular accelerations on both cylinders.

Once slipping begins, the belt pull can only achieve lower angular accelerations, because the dynamic friction will be of a lesser value than the static one.