# Force required to maintain Equilibrium

member 392791

## The Attempt at a Solution

My problem here is that I keep trying to make substitutions for the normal forces and cannot come up with the number for P. I saw 4.92 and don't understand how there is a range of force for which equilibrium can be maintained. The force I came up with for P is way too big, the answer is 0.955 lb

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• 4.91 attempt 1.pdf
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• 4.91.pdf
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You are assuming that the forces of friction are ## F = \mu N ##. That is not correct for static friction. In statics, the force of friction can be anywhere in the range ##[0, \ \mu N]##.

SteamKing
Staff Emeritus
Homework Helper
Just to save you a little calculation, it's perfectly OK to calculate moments in units of inch-pounds rather than converting all distances to feet.

member 392791
So in this problem, where am I going wrong as far as why the number isn't coming out right? I believe my moment sign convention is consistent, and I am doing the necessary calculations to solve for N_1 and N_2, yet it doesn't come out right. Anyone know where it went wrong?

Voko, if the domain can be between [0,μN], then why doesn't the minimum force P=0?

One problem is that the last term in your moment equation involves ##N_1##. How that could happen for moments about B is a mystery.

Another problem, as I already said, is that you assume the forces of friction are at their maximum.

member 392791
Good catch with the N_1, I see the mistake. I am not sure though what you mean about the assumption that the forces of friction are at their maximum. How do I mathematically express the minimum force if I know the maximum force is F=μN?

You could introduce two independent friction forces, and inequalities for their range. That will give you inequalities for P.

member 392791
What do you mean by that? I don't understand it

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I do not understand how the force of friction at A can be directly replaced with some expression involving the normal force at B - before the rest of the equations are written and analysed at the very least.

Anyway, as I said earlier, you should treat the forces of friction independently of the normal forces, so the whole issue is moot.

member 392791
How do I treat them independently of the normal force? Are you saying the normal force at A doesn't cause a moment at point B? The force of friction is directly proportional to the normal force by the factor of the coefficient of friction. How else can I express the force of friction other than F=μN?

If the range is [0,μN], why isn't the minimum force just zero?

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Just use symbols ##F_A## and ##F_B## for the forces of friction. Do not equate them with ##\mu N##. Instead, use inequalities ## 0 \le F \le \mu N ##.

member 392791
Hmm...I just thought of something, are you meaning to say that I should just consider the forces of friction at points A and B as zero, thus only finding the moments from the factors of W and N? Or are you saying since F_a is zero, then N is zero as well??

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Hmm...I just thought of something, are you meaning to say that I should just consider the forces of friction at points A and B as zero, thus only finding the moments from the factors of W and N? Or are you saying since F_a is zero, then N is zero as well??

I do not mean any of that. Re-read my previous message.

member 392791
I'm looking at it. Using the inequalities, I get that. However, what does that means as far as the minimum force of P required? Does the minimum force of P occur when the frictions are maximums? That should mean that since the friction is a maximum, the force P should not have to exert as much to keep the pole from staying on the wall, since the friction would be doing most of the work.

member 392791
Still not sure how to incorporate the inequalities into my strategy for solving the problem. I have 4 equations and 5 unknowns (N_1, N_2, F_a, F_b, P), so this thing is not solvable with what I have. Is taking a moment around A and B to get 4 equations (2 from the horizontal and vertical forces, and 2 equations from the moments at point A and B) valid, or is there a redundancy and the moments at A and B are not independent?

#### Attachments

• 4.91 attempt 2.pdf
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I said earlier that the forces of friction are in the range ##[0, \ \mu N]##. I now think that was wrong, because that assumes a particular direction. I think the correct range is ##[-\mu N, \ \mu N]##.

As for your question, you have 5 unknowns and 3 equations. That means three unknowns can be expressed as functions of two unknowns. In particular, everything, including ##P##, can be expressed as a function of the friction forces. Then, using the inequalities on the friction forces, you can find the range of ##P##.

member 392791
Problem has been solved, thank you.