Calculate the coefficient of friction between crates and a conveyor belt

[sperrya]

Homework Statement

Hi can anyone help me calculate the coefficient of friction between crates and a conveyor belt. The crate is moving up a 25^O angle at a constant speed to a height of 12m.

Homework Equations

CoF=F/R

The Attempt at a Solution

I don't know if this is a trick question. My immediate response would be that he coefficient of friction is an empirical measurement and it has to be measured experimentally, it cannot be found through calculations. Therefore referencing the standard coefficients in a table which have already been published, such as 0.2 for wood on wood?

Have I got this right or should i be looking at another calculation such as angle of repose to answer my query?

Thanks in advance

 

[PeterO]

Homework Statement

Hi can anyone help me calculate the coefficient of friction between crates and a conveyor belt. The crate is moving up a 25^O angle at a constant speed to a height of 12m.

Homework Equations

CoF=F/R

The Attempt at a Solution

I don't know if this is a trick question. My immediate response would be that he coefficient of friction is an empirical measurement and it has to be measured experimentally, it cannot be found through calculations. Therefore referencing the standard coefficients in a table which have already been published, such as 0.2 for wood on wood?

Have I got this right or should i be looking at another calculation such as angle of repose to answer my query?

Thanks in advance

The tangent of the angle of a slope will give the co-efficient of friction if the body [crate] is on the point of slipping. The co-efficient could be anything greater than that and it still won't slip.

 

[sperrya]

Hi there, thanks for your help, so can I confirm:

at an angle of 30^o the crate starts to slip, therefore the set angle is 25^o. So to work out the coefficient of friction is:

tan25 = 0.466

Is this the correct coefficient of friction between the crates and the conveyor belt?

Thanks

 

[Tomer]

The tangent of the angle of a slope will give the co-efficient of friction if the body [crate] is on the point of slipping. The co-efficient could be anything greater than that and it still won't slip.

True, and yet I assume you're more interested in the solution itself.

As always with bodies on slopes, draw a x-y system, so that the x-axis is parallel to the surface of the slope and in the direction of the crates movement.
When you analyze forces, you couldn't care less about velocities, with the exception that the fact that the velocity is constant (v = 12) implies that the sum of forces working on the x-axis (the axis of movement for the body) is 0!
Of course the sum of the forces on the y-axis is also 0 (otherwise the crate would start to float :-) )
There are three forces in act: Gravity, pointing downwards, Friction, pointing in the direction of the x-axis (it resists the potential fall-back, that's the nature of friction, and therefore points to the way of movement, as always), and the normal force, point like always to the positive direction of the y axis.
Gravity has two components: mgsin$\alpha$ on the x-axis and mgcos$\alpha$ on the y axis. Make sure you understand why.
By calculating the sums of forces at each axis you get:

x: F_fric - mgsin$\alpha$ = 0
y: N - mgcos$\alpha$ = 0

F_fric = $\mu$ * N

(N is the normal force)
Isolate N from the first equation, put it in the second, and you'll get your $\mu$...

 

[PeterO]

Hi there, thanks for your help, so can I confirm:

at an angle of 30^o the crate starts to slip, therefore the set angle is 25^o. So to work out the coefficient of friction is:

tan25 = 0.466

Is this the correct coefficient of friction between the crates and the conveyor belt?

Thanks

No. It is set at 25^o for safety. You are saying that slipping begins at 30^o so that is the angle used to calculate the co-efficient of friction

 

[PeterO]

True, and yet I assume you're more interested in the solution itself.

As always with bodies on slopes, draw a x-y system, so that the x-axis is parallel to the surface of the slope and in the direction of the crates movement.
When you analyze forces, you couldn't care less about velocities, with the exception that the fact that the velocity is constant (v = 12) implies that the sum of forces working on the x-axis (the axis of movement for the body) is 0!
Of course the sum of the forces on the y-axis is also 0 (otherwise the crate would start to float :-) )
There are three forces in act: Gravity, pointing downwards, Friction, pointing in the direction of the x-axis (it resists the potential fall-back, that's the nature of friction, and therefore points to the way of movement, as always), and the normal force, point like always to the positive direction of the y axis.
Gravity has two components: mgsin$\alpha$ on the x-axis and mgcos$\alpha$ on the y axis. Make sure you understand why.
By calculating the sums of forces at each axis you get:

x: F_fric - mgsin$\alpha$ = 0
y: N - mgcos$\alpha$ = 0

F_fric = $\mu$ * N

(N is the normal force)
Isolate N from the first equation, put it in the second, and you'll get your $\mu$...

The maximum possible friction is given by F_fric = $\mu$ * N
The actual friction is the classic "just strong enough" force. Only when the angle is at the maximum before slipping does this formula work to give you the actual co-efficient of friction.

 

[sperrya]

Hi again, thank you for your replies.

So is it as easy as 'tan30=0.577' which is the coefficient of friction in this example.

Thanks

 

[PeterO]

Hi again, thank you for your replies.

So is it as easy as 'tan30=0.577' which is the coefficient of friction in this example.

Thanks

It is, but you need to be able to understand the the reasoning/formulas in the other explanation posted in this thread.

 

[Tomer]

Does it say in the question that it starts slipping on 30?

PeterO, you're right - but still, that's the way to show why miu equals the tan of the critical angle, no? I don't recall being able to use this formula automatically...

 

[PeterO]

Does it say in the question that it starts slipping on 30?

PeterO, you're right - but still, that's the way to show why miu equals the tan of the critical angle, no? I don't recall being able to use this formula automatically...

I post #3 sperrya added that key bit of information.

I agree with the limit on formula use.

If the question was a 1 mark "what is the coefficient of friction" you just use the tangent formula.
If the question was a multi-mark "Show that the coefficient of friction is ..." you would certainly want to see the expressions you presented.

 

[sperrya]

The question states that at an angle of 30^o the crates start to slip down the slope, so they set it at 25^o for safety. It asks, calculate the coefficient of friction between the crates and the belt?

The learning material I have been reading through is full of errors and mistakes, and basically only gives 2 sentances on this subject and then moves through the angle of repose to calculate force required to prevent slippage or to start it moving. In all the worked examples it refers to a table of pre-determined coefficient factors to use. But now it is asking me to solve it but there is no easy reference as to how! It gives the CoF=tan'lambda'

Thanks

 

[Tomer]

Sounds like terrible learning material. I guess I agree with PeterO then - tan30 is the answer - it should always be the critical angle, just before it starts slipping. Or just after. :-)

 

[sperrya]

It is terrible, I am disgusted with ICS for the shoddy material. Some material is wrong, some answers to the progress check sections are wrong, and most of the questions in my assessment have not even been covered in a way of supporting my learning. Terrible!

Anyway, thanks for your help...

I so far have this:

F=mgSin30
N=mgCos30

Therefore F/N = mgSin30/mgCos30

Therefore F/N = Sin30/Cos30

Therefore F/N = Tan30 (=0.577)

Which is the short answer I began with, but is the methodology correct in my calculations?

Thanks

 

[Tomer]

Glad to hear you agree it's terrible :)

When you write F = mgsin30, what do you mean by F? Is that supposed to be the friction force? It's true, but make sure you understand why. It's derived from F - mgsin30 = 0.
The rest seems good, but I would only add that, like PeterO wrote before, F = miu * N at the critical angle (when the crate just begins to slip), and therefore:

miu = F / N = tan30 = ...

Looks good other than that!

 

[sperrya]

F = Friction
N = Pressing Force

At least that's how it is described in the learning material.

Thanks v.much for your help both of you..