Need help solving this friction problem

[dlacombe13]

Homework Statement

A gate is supported by two rings at A and B that fit over a vertical rod as shown. If the coefficient of friction between the rings and the rod is 0.2, find the maximum distance between the rings for which the gate is in equilibrium. The reaction at the upper ring is directed away from the gate and at the lower ring toward the gate.

Homework Equations

∑Fx = 0
∑Fy = 0
∑M = 0
F_m = (f_s)(N)

The Attempt at a Solution

Well I am having trouble even figuring out a free-body diagram for this one. I am also having trouble with the equations of equilibrium, since I appear to have too many unknowns. I know the normal force is opposite of W. I am very unclear about the reaction forces at A and B. The problems suggests that the force at A would be pointing horizontally away from the gate, and horizontal towards it at B. So then is the friction vertical (I assume the gate is trying to slide down the rod, so friction is acting against it)? If so, wouldn't the normal force be perpendicular to the friction, and so horizontal? Furthermore, the moment equation is throwing me way off since I have to include the distance h, which just adds another unknown and makes it seem unsolvable. Any help please? I feel like I am missing some concept here :/

 

[Simon Bridge]

If there were only one ring, what would the weight make the gate do?
You can draw the force on the picture that stops it doing that... these are the horizontal forces.

You can see what the gate would do left to itself... you can draw friction force arrows at the point the act.

Once you have the forces in their right places, you can work the moments etc as needed.

H is the distance you are asked to find.
I only see 2 unknowns, H and W.

 

[dlacombe13]

If there were only one ring, the gate would try to fall counter-clockwise. The force that stops it from doing that would be pushing or pulling upwards. So would that be the friction? Since the gate is contact with the rod, the normal force would be horizontal and perpendicular to the rod. Can there be more than one normal force? Like, would each ring have a normal force? Or just the gate as a whole?

 

[haruspex]

I know the normal force is opposite of W.

How do you know that?
How would you define "normal force"? If unsure, check out https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/

wouldn't the normal force be perpendicular to the friction, and so horizontal?

Quite so.

 

[Mister T]

I know the normal force is opposite of W. I am very unclear about the reaction forces at A and B.

Consider the contact force exerted on one of the rings by the pole. Resolve that force into two components so that one component is parallel to the pole, and the other perpendicular. The parallel component is called the friction force. The perpendicular component is called the normal force, or in this problem statement the reaction force. The coefficient of friction is the ratio of the friction force to the normal force.

In a situation where you have an object on a horizontal surface, the normal force is upward and the weight force is downward. They are opposite in direction in that case, but that is not the general case.

 

[dlacombe13]

Okay so I am now aware that the normal force is perpendicular to the contact plane. In this case, the contact plane is the vertical plane where the gate and the rod meet, so the normal force is indeed horizontal. Now for each of the rings, A and B, I had a strong feeling that I need components of each, so let's call them Ax, Ay, Bx, and By. I also am aware now that the frictional force is what is preventing relative motion of the two forces, so that would be upwards, preventing the gate from falling. Do I at least have this information correct? If so, I am a little fuzzy as to where exactly this friction force would be drawn. Is it Ay and By? Or must I draw some other force vector to depict it? I am also a little fuzzy as to whether where the normal force would be drawn. Is in Ax and Bx? Or is it only once force drawn elsewhere?

 

[Mister T]

The vertical component of the contact force is the friction force.

 

[dlacombe13]

Okay I am starting to understand it more. I am confused about points A and B, which are each points of contact. Each has a vertical and a horizontal component right? Let's say Ay and By are the vertical components. Is Ay or By the friction force? I mean, which would I use in the equation?

 

[Mister T]

Both! Each is a friction force.

 

[dlacombe13]

Alright, so I have this as my free-body diagram, but the amount of unknowns is just throwing me off, and I can't seem to solve for anything :/

I have:
ΣFx=Ax-Bx=0
ΣFy=Ay+By=0
ΣM_b= -(Ax)(h)+2.5w=0
Fm=0.2Bx
Fm=0.2Ax

But I just can't seem to solve for any of the unknowns. I feel like I have too many unknowns, and perhaps that my FBD is still drawn wrong.

 

[Mister T]

What's Fm?

Why is there no w in the equation for the sum of the vertical forces?

 

[dlacombe13]

Ah okay, that is a mistake, the W should certainly be in there. Fm is the maximum friction, which equals the coefficient of friction X normal force. So the the sum of the y-components should be By + Ay - W = 0. I still am lost as to how to solve for the values, there seems to be too many unknowns. I mean, I have Ax, Bx, Ay, By, W, h. Are Ay and By equal to Fm (maximum friction)?

 

[Mister T]

Are Ay and By equal to Fm (maximum friction)?

Ay and By are the friction forces, yes, but they are not equal to each other.

 

[dlacombe13]

Okay so the friction force is the vertical component of the contact force. So is the friction force = Fm (maximum friction) = Ay = By? Even if this is the case, I am just having trouble bringing these equations together. My biggest fall back is the h (height), which just won't allow me to eliminate any variables. I have tried and tried, but I keep ending up with equations with two unknowns. I am sorry to drag this on, I just am really frustrated and feel like I am totally missing something obvious.

 

[haruspex]

Okay so the friction force is the vertical component of the contact force. So is the friction force = Fm (maximum friction) = Ay = By? Even if this is the case, I am just having trouble bringing these equations together. My biggest fall back is the h (height), which just won't allow me to eliminate any variables. I have tried and tried, but I keep ending up with equations with two unknowns. I am sorry to drag this on, I just am really frustrated and feel like I am totally missing something obvious.

You have 5 unknowns, Ax, Bx, Ay, By, H.
You have 5 equations: vertical force sum, horizontal force sum, moment sum, and the relationship between normal forces and frictional forces.
Just go through the usual process of eliminating one unwanted unknown at a time.
If still stuck, please post all your working.

 

[Mister T]

Okay so the friction force is the vertical component of the contact force. So is the friction force = Fm (maximum friction) = Ay = By?

A_y is the friction force exerted on Ring A by the pole. B_y is the friction force exerted on Ring B by the pole. They are not equal to each other.

 

[haruspex]

A_y is the friction force exerted on Ring A by the pole. B_y is the friction force exerted on Ring B by the pole. They are not equal to each other.

Why not? Aren't the normal forces equal? The coefficients equal? Can it slip at one and not the other?

 

[Mister T]

Why not? Aren't the normal forces equal? The coefficients equal? Can it slip at one and not the other?

Oh, yes, they are. But I reached that as a conclusion, it wasn't premise, so I should have said they are not necessarily equal to each other. If, for example, there was a third horizontal force, then the normal forces might not be equal in magnitude and then the two friction forces wouldn't be equal. The fact that one ring can't slip without the other slipping doesn't necessarily mean that the two friction forces have to be the same.

 

[haruspex]

I should have said they are not necessarily equal to each other.

Yes, that would have been better.

 

[dlacombe13]

I'm back today, since it was like 1AM. I am about to solve for the unknowns, however I just realized I have 6 unknowns, not 5. I have Ax, Ay, Bx, By, H, W. I don't think I have. I am not really sure what the 6th equation would be. I also have never solved for a system with more than 3 unknowns, and I am completely drawing a blank. Don't all the equations have to contain all the variables? What I have so far is:
ΣFx=Ax-Bx=0
ΣFy=Ay+By-W=0
ΣMb= -(Ax)(h)+2.5w=0
By=0.2Bx
Ay=0.2Ax

 

[haruspex]

I'm back today, since it was like 1AM. I am about to solve for the unknowns, however I just realized I have 6 unknowns, not 5. I have Ax, Ay, Bx, By, H, W. I don't think I have. I am not really sure what the 6th equation would be. I also have never solved for a system with more than 3 unknowns, and I am completely drawing a blank. Don't all the equations have to contain all the variables? What I have so far is:
ΣFx=Ax-Bx=0
ΣFy=Ay+By-W=0
ΣMb= -(Ax)(h)+2.5w=0
By=0.2Bx
Ay=0.2Ax

When all variables involving a particular dimension are unknown, and not sought, there's one less unknown than you think. That's because you only care about the ratios between them. In the present case, you could divide through everywhere by W, leaving the four unknown force ratios Ax/W, Ay/W, etc.

 

[dlacombe13]

Well I got the answer of 1, which is correct. It just seems like I did it by accident. I will show my work here, just to verify that my process was indeed correct. I admit though, I don't understand the whole "divide through everywhere by W". If you could just perhaps elaborate a little and perhaps shed some light on the concept behind it. I never had to solve a problem this complex before, and in this manner (using ratios). Anyways, here's my work:

My Equations:

(1) Ax/W - Bx/w = 0 --> Ax/W = Bx/W
(2) By/W=0.2Bx/W
(3) Ay/W=0.2Ax/W --> Ax/W = Bx/W therefore By/W = Ay/W
(4) By/W + Ay/W - W/W = 0 --> By/W + Ay/W = 1
(5) -AxH/W + 2.5W/W = 0 --> -AxH/W + 2.5 = 0

So:
Substituting Ax/W for Bx/W in equation (2), I get By/W = 0.2Ax/W
Plugging in By/W and Ay/W into equation (4):
0.2Ax/W + 0.2Ax/W = 1 --> Ax = 2.5w
Plugging in Ax into equation (5):
2.5WH/W + 2.5 = 0 --> H = 1

 

[haruspex]

2.5WH/W + 2.5 = 0 --> H = 1

Typo, missing minus: -2.5WH/W + 2.5 = 0 --> H = 1

I don't understand the whole "divide through everywhere by W".

In the equations, if you were to replace every occurrence of W with 2W, of Ax with 2Ax, etc. for each force, the equations would not change. The extra factors of 2 would cancel out. So the value of W does not affect the answer for H.
It was not really necessary to do the dividing by W. You could have just proceeded with elimination of unknowns in the usual way for simultaneous equations and you would have ended up with an equation like WH=W. But I mentioned dividing all the forces by W in order to demonstrate that there were really only 5 unknowns (insofar as finding H was concerned).
A simpler and very common example of the process arises in SUVAT equations. Suppose you know an initial velocity, a constant acceleration and a final velocity and you want to find distance. You could apply the SUVAT equation v²=u²+2as; or you could use energy: 1/2 mv²=1/2 m u²+Fs, F=ma. Some students make the mistake of thinking they cannot use the second if they do not know the mass, but as you can see, the mass just cancels out.

 

[Mister T]

ΣFx=Ax-Bx=0
ΣFy=Ay+By-W=0
ΣMb= -(Ax)(h)+2.5w=0
By=0.2Bx
Ay=0.2Ax

1st equation gives you $A_x=B_x$
Substitution of 4th and 5th equations into 2nd equation gives you $0.2A_x+0.2B_x=w$
Thus $0.4A_x=w$

3rd equation gives you $2.5w=A_xH$
Thus $2.5(0.4A_x)=A_xH$ or $H=1 \ \mathrm{ft}$.

Why do you think you just got lucky? The thing that made the algebra easy was the fact that there were only two horizontal forces, so they had to be equal in magnitude.