Forces when pulling a cylinder

In summary, the conversation discusses a problem involving determining the force needed to prevent a cylinder from tumbling over. The solution is provided but the individual does not understand it. The provided solution includes an image showing the forces acting on the cylinder, including the pulling force, the force exerted by the ground, and the frictional force. The individual questions why the force exerted by the ground is drawn in a specific location and how it can prevent the cylinder from rotating. The conversation concludes with a clarification that the forces shown are for when the cylinder is about to rise, not when it has already started to rise.
  • #1
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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 [Broken]
[PLAIN]http://www.mediafire.com/imageview.php?quickkey=qv25y3sn7h1cjkc [Broken]

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
 
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  • #2
anyone?
 
  • #3
Hi Tusike! :smile:

(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)
Tusike said:
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) :approve:, the second is dynamic. :yuck:
 
  • #4
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.
 
  • #5


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.
 

1. What is a force?

A force is a push or pull on an object that can cause it to change its motion or shape.

2. How is force related to pulling a cylinder?

When pulling a cylinder, force is applied to the object in order to move it in a certain direction.

3. What factors affect the force needed to pull a cylinder?

The force needed to pull a cylinder is affected by the mass of the cylinder, the friction between the cylinder and its surface, and the angle at which the force is applied.

4. How does the angle of the force affect the pulling of a cylinder?

The angle of the force can impact the direction and magnitude of the force applied to the cylinder. Pulling a cylinder at an angle will result in a smaller force being applied to the object compared to pulling it straight on.

5. What is the formula for calculating force when pulling a cylinder?

The formula for calculating force when pulling a cylinder is F = μmgcosθ, where μ is the coefficient of friction, m is the mass of the cylinder, g is the acceleration due to gravity, and θ is the angle at which the force is applied.

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