Traction Force

[studiot]

That last post is college level stuff.

 

So why can't you tell me what the moment of force W is about point C in my diagram?

 

Hint it is exactly the same as the moment of the weight (mg) of your roll in your diagram about the central axis of the roll.

Edited July 17, 2013 by studiot

[bon]

M=m*g*sin(∂)*r

 

???

[studiot]

OK you seem good on the formula side but perhaps lacking in experience in translating this into hardware. This is not a criticism the skill should come with practice.

 

A force cannot exert a moment about any point in its line of action.

 

Or

 

The moment of a force about any point in its line of action is zero.

 

So the moment of W about C is zero - It has no moment about C.

 

Because its line of action passes through C.

 

Now look at your first diagrams.

 

You have shown two forces whose line of action pass through the centreline of the roll - the tension in the support string and the weight of the roll.

 

Therefore these forces have zero moment about the centreline of the roll.

 

That is why I said they are irrelevent.

 

That leaves only two forces that can exert a moment about the centre line of the roll.

 

Can you see this and tell me why we need a moment about the centreline of the roll?

Edited July 17, 2013 by studiot

[bon]

two forces which can exert a moment about the centre line of the roll:

Fz and Fr=Fn*μ

 

 

The tensile force can be split into two components of force in the tangential force and the pressure force is directed to the center of gravity exerts an additional force on the wall.

The frictional force is equal to the pulling force.

[studiot]

 

The frictional force is equal to the pulling force.

 

 

What is your condition for one vector to be equal to another?

 

Now test this against your statement above.

Edited July 17, 2013 by studiot

[bon]

He shall produce the same amount and the same direction.

 

M1=M2

Fr*r=Fz*r

 

r - radius

 

Now test this against your statement above.

mmh, I don´t understand, what I should do???

[studiot]

You said the frictional force is equal to the pulling force.

 

Applying this to test if the frictional force is equal to the pulling force it must have the same amount and direction.

 

Is this true?

[bon]

no it isn`t.

 

I think only the same direction because of the held levers:

F1*a1 = F2*a2

[studiot]

So what do you want to say is true?

[bon]

true:

Fz=Fr (because both have got the same distance from the graviton point)

Fr=Fn*μ=Fz

 

???

[studiot]

 

 

MD65536 post#9

 

Depending on how the roll is oriented, pulling on the paper to unroll it will affect the force with which the roll is pressing against the wall.

 

 

Can you expand on that?

Edited July 18, 2013 by studiot

[md65536]

Now look at your first diagrams.

 

You have shown two forces whose line of action pass through the centreline of the roll - the tension in the support string and the weight of the roll.

 

Therefore these forces have zero moment about the centreline of the roll.

 

That is why I said they are irrelevent.

I don't think this is true. I do think it's incorrect in the diagram, because while gravity pulls on the center of the roll, pulling on the paper doesn't.

 

A downward force through the center of the roll has a moment about the point where the holder is connected to the wall. The greater the downward force through the center of the roll, the harder the roll will press against the wall, making it harder to unravel.

Edited July 18, 2013 by md65536

[studiot]

 

A downward force through the center of the roll has a moment about the point where the holder is connected to the wall. The greater the downward force through the center of the roll, the harder the roll will press against the wall, making it harder to unravel.

 

 

 

Thank you for replying.

 

You haven't got your mechanics quite right. The above is true but the only because it determines the tension in the string, which in turn determines the horizontal reaction from the wall.

Nor does it negate what I said since I did say moment about the centre of the roll, not some other point where there are undoubtedly different moments.The trick in mechanics is to take your moments about points where as many forces as possible pass through and therefore have zero moment.

 

This question is actually quite subtle.

 

There are three situations:

1) With the roll just hanging there and no pull on the paper.

2) With the roll hanging and a pull on the side opposite the wall.

3) With the rolll hanging, and a pull on the side by the wall.

 

Since the tension is not necessarily the same in each case that yields 9 equations in 12 unknowns.

 

There is no deformation so we cannot use compatibility

Stress is not considered so we cannot use elasticity

 

However we can consider the situation at limiting friction which yields a further 3 equations, bringing the total up to 12 - a happy situation since we now have enough to solve the question.

 

Here is the full set of 12.

Note I have replace sin and cos by constants a and b since the geometry does not change so we can more easily see the set of equations is linear and and manipulate them.

What remains is a page of further manipulation, first substitution then setting up and manipulating an inequality to achieve the desired result.

I will leave that part to bon if (s)he is still interested.

Edited July 19, 2013 by studiot

[bon]

I have to say thank you for your great help, I have already solve the problem

[studiot]

I'm glad to learn that you have solved the problem yourself, this is always the best way.

 

 

I did observe that this problem has subleties - you need to invoke most of the principles of mechanics to justify the steps in the solution, so I would be interested to see your solution if you would care to post it.

Edited July 20, 2013 by studiot