Cylinder & Torque: Find Maximum Force w/o Rotation

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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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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!
 
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
 
It's radius, but no radius is given
 
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
 
It's ## Fr \sin \theta ##, but I'm still not sure how to apply it. Can you give a start?
 
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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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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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