Forces when pulling a cylinder

[Tusike]

Homework Statement

I have a book with the following problem: What force is needed to pull a cylinder so that it wouldn't tumble over? The frictional constant, and the height and radius of the cylinder are given, and the height of our vertical force exerted on the cylinder.

Now, the problem doesn't really matter, since the book also has the solution, but I don't really understand the solution. They provide the image I have uploaded, which show the forces acting upon the cylinder. F is our pulling force, K is the force exerted by the ground, and S is the frictional force. According to the solution, K=mg. What I don't really understand: why is K drawn where it is? and how should I be able to determine it acts there, by the edge of the cylinder?

url of image:
http://www.mediafire.com/imageview.php?quickkey=qv25y3sn7h1cjkc
[PLAIN]http://www.mediafire.com/imageview.php?quickkey=qv25y3sn7h1cjkc

Homework Equations

none

The Attempt at a Solution

The only reason I could think of putting K there, is that we don't want the cylinder to tumble over, so we are looking at a state in which the cylinder barely started to rise from the ground, so only its right lower edge would be in contact with the ground. However, if this is the case, then why is S still shown on the left? And how could the cylinder start rotating, which would mean it's COM moves up, if K=mg, so the net forces acting vertically are zero?
Thanks!

-Tusike

 

[Tusike]

anyone?

 

[tiny-tim]

Hi Tusike!

(btw, never a good idea to bump your own thread after only a few hours … it removes it from the "no replies" list, and people may not even look at it)

The only reason I could think of putting K there, is that we don't want the cylinder to tumble over, so we are looking at a state in which the cylinder barely started to rise from the ground, so only its right lower edge would be in contact with the ground.

Yes and no.

It shows the forces when the cylinder is about to rise, not when it has already "barely started to rise" …

the first is a static problem (so K = mg) , the second is dynamic. :yuck:

 

[Tusike]

Thanks, it's clear now.

By the way, I bumped it because it was going "down" in the list of posts, and was afraid that people wouldn't see it:) but I'll make sure to bump only in case of having no replies after a few days.

 

[rwisz]

I would like to offer the following response to your question.

Firstly, it is important to note that in order to fully understand the solution provided by the book, we would need to have a clear understanding of the problem statement and all the given information. However, based on the information provided, I can offer the following explanation:

The force exerted by the ground (K) is shown acting at the edge of the cylinder because that is where the cylinder is in contact with the ground. This force is necessary to prevent the cylinder from slipping or tumbling over.

The frictional force (S) is shown acting on the left side because, as the cylinder starts to tilt, the force of friction will act in the direction opposite to the direction of motion. This is to prevent the cylinder from sliding or slipping in the direction of the applied force (F).

As for your question about the cylinder rotating and its center of mass moving up, it is important to remember that the forces shown in the diagram are only the vertical forces acting on the cylinder. There may be other forces acting in the horizontal direction that could cause the cylinder to rotate. However, if we assume that there are no other forces acting on the cylinder, then yes, the net vertical forces would be balanced and the cylinder would not rotate.

I hope this explanation helps to clarify your doubts. It is always important to carefully analyze the given information and fully understand the problem before attempting to solve it. Keep asking questions and seeking understanding, as that is the essence of being a scientist.