Cylinder & Torque: Find Maximum Force w/o Rotation

AI Thread Summary
The discussion focuses on calculating the maximum tangential force that can be applied to a uniform cylinder without causing it to rotate, given a coefficient of static friction. Participants emphasize the importance of using free body diagrams to analyze the forces acting on the cylinder, including weight, normal forces, and friction. A moment balance is necessary to account for torque, with torque defined as the product of force and distance from the axis of rotation. The conversation highlights the need to identify the correct distance for torque calculations and to set up equations based on the forces and moments involved. The final equations derived relate the applied force to the frictional forces and the weight of the cylinder.
minimario
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Homework Statement



Figure P8.74 shows a vertical force applied tangentially to a uniform cylinder of weight w. The coefficient of static friction between the cylinder and all surfaces is 0.500. Find, in terms of w, the maximum force F that can be applied without causing the cylinder to rotate. [Hint: When the cylinder is on the verge of slipping, both friction forces are at their maximum values. Why?]

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Homework Equations



Torque = F * d

F = Mg

Friction = coefficient * normal force

The Attempt at a Solution



There are a few force: weight force, F, <right wall>: normal force to the left, friction force, <lower wall>: friction force to the right, normal force upward. However, I don't know how to combine these forces to solve the problem.
 
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Have you started properly by drawing a free body diagram showing all the forces acting on the cylinder, or do you feel that you have progressed beyond the point where you need to use free body diagrams?

Chet
 
You must have heard about Newton's laws. There is one that says ##\ \vec F = m\; \vec a\ ## (*) .
If the thing don't move, the forces must add up to zero. That's your combination !

(*) Wrote it as vectors, so you may use it for horizontal non-movement and for vertical idem as well.

And if you're still short on equations, something similar for non-rotation also holds. You know what I mean ? :wink:

[edit] bedtime for me; leave you to Chet !
 
@chet, Yes, I have drawn free body diagrams, but I cannot upload here. The forces on the FBD were written in the "attempt at a solution" section.

From the FBD, I have n1 = coefficient * n2, coefficient * n1 + F + n2 = w. Is that correct?
 
minimario said:
@chet, Yes, I have drawn free body diagrams, but I cannot upload here. The forces on the FBD were written in the "attempt at a solution" section.

From the FBD, I have n1 = coefficient * n2, coefficient * n1 + F + n2 = w. Is that correct?
Yes. Sorry. I didn't read your post carefully enough.

You have 2 equations here and 3 unknowns. The third equation would be a moment balance.

Chet
 
Can you explain more about how to set up the moment balance?

Thanks!
 
minimario said:
Can you explain more about how to set up the moment balance?

Thanks!
Well, first of all, do you know what a moment (torque) is? Do you know how to calculate the moment of a force about an axis of rotation?

Chet
 
Yes, torque = F * d, to calculate moment, is mr^2.
 
minimario said:
Yes, torque = F * d, to calculate moment, is mr^2.
Not exactly. mr^2 is what you use to calculate the so called "moment of inertia." That's different from the moment of a force around an axis. A moment is the same thing as a torque. They're different words for the same entity. Your equation torque = F*d needs a little more explanation. How do you determine what value of d to use?

Chet
 
  • #10
It's radius, but no radius is given
 
  • #11
minimario said:
It's radius, but no radius is given
There's a rule for determining the correct distance d to use. You drop a normal from the axis of rotation to the line of action of the force. d is the distance from the axis to the line of action of the force. Does this ring a bell? If not, go back and review your textbook or notes.

Chet
 
  • #12
It's ## Fr \sin \theta ##, but I'm still not sure how to apply it. Can you give a start?
 
  • #13
minimario said:
It's ## Fr \sin \theta ##, but I'm still not sure how to apply it. Can you give a start?
OK. Make a list of the moments of each of your forces around the axis of the cylinder. Let's see what you get.

Chet
 
  • #14
KUgUi6l.png


Torque caused by f1: ## f_1 \cdot R##
Torque caused by n1, W, n2: 0
Torque caused by f2: ##f_2 \cdot R##
Torque caused by F: ##F \cdot -R##

Therefore, ##F = f_1 + f_2##?

Now we have that equation and ## n_1 + f_2 +F= W## and ##f_1 = n_2##?
 
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  • #15
minimario said:
Torque caused by f1: ## f_1 \cdot R##
Torque caused by n1, W, n2: 0
Torque caused by f2: ##f_2 \cdot R##
Torque caused by F: ##F \cdot -R##

Therefore, ##F = f_1 + f_2##?

Now we have that equation and ## n_1 + f_2 +F= W## and ##f_1 = n_2##?
That all looks right.
 
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