# Force required to maintain Equilibrium

1. Sep 23, 2013

### Woopydalan

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

2. Relevant equations

3. 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.pdf
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2. Sep 23, 2013

### voko

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 $[0, \ \mu N]$.

3. Sep 23, 2013

### SteamKing

Staff Emeritus
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.

4. Sep 23, 2013

### Woopydalan

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?

5. Sep 23, 2013

### voko

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.

6. Sep 23, 2013

### Woopydalan

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?

7. Sep 23, 2013

### voko

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

8. Sep 23, 2013

### Woopydalan

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

Last edited: Sep 23, 2013
9. Sep 23, 2013

### voko

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.

10. Sep 23, 2013

### Woopydalan

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?

Last edited: Sep 23, 2013
11. Sep 23, 2013

### voko

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

12. Sep 23, 2013

### Woopydalan

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??

Last edited: Sep 23, 2013
13. Sep 23, 2013

### voko

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

14. Sep 23, 2013

### Woopydalan

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.

15. Sep 24, 2013

### Woopydalan

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?

#### Attached Files:

• ###### 4.91 attempt 2.pdf
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16. Sep 24, 2013

### voko

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 $P$.

17. Sep 24, 2013

### Woopydalan

Problem has been solved, thank you.