Forces when pulling a cylinder

Tusike

1. The problem statement, all variables and given/known data
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.p...v25y3sn7h1cjkc


2. Relevant equations
none

3. 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)
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.

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.