It certainly can – a planet tidally locked with an equal-sized planet in the same solar orbit can still have wicked tides due to its sun.
But “tidal force” refers to something more specific: it is the force left over by differences in gravitational force, not the gravitational force itself. The sun exerts a far stronger gravitational pull on the Earth than the moon, but the tidal effects are smaller because of the distance to the sun.
So while “tides are just gravity” is correct in a sense, it omits the important nature of tides.
For a second, I was mildly concerned that your post may have been a whoosh. Clearly I was worried over nothing.
Your attempt at insulting me not withstanding (and I shall inform you good sir that for me stupid would be a vast upgrade. There are days when I look up in awe at the vast intellectual heights reached by an average garden squash.), I’d simply ask you to reread the original post from Bricker that you have quoted. I feel you are misunderstanding it greatly. Perhaps based upon some prior experience with him. Personally I find your interpretation of the quote to be laughably inaccurate. To me it seemed he was offering a possible defense of the O’Reilly event but deemed it to be inaccurate and seemed to agree with the views of the prior posters. But then I’ve the reading ability and mental faculties of a radish burp.
I love a miracle.
I have a question: would you say that, relatively speaking, the strength of the gravitational pull that the sun exerts on the earth os equal to the strength of the gravitational pull that the earth exerts on the moon.
I’m not sure how to even go about measuring that. You have mass and distance as variables. And the distance of orbit may come into play. But can it be said that any object held in orbit by another object share, relatively speaking, the same gravitational pull?
I’m not quite sure what you’re asking here, but I can run some numbers, and maybe it’ll either answer your question or you can clarify based on that.
Assuming we ignore General Relativity (this is safe, as long as we don’t require our results to be too precise), the equation for gravitational force is just this:
F = Gm1m2/r^2
That is to say, it’s the constant G, times the masses of each object, and scaled by the inverse square of the distance between them.
Our inputs are then the following:
G = 6.674e-11 N(m/kg)^2
Dse = 1.496e11 m (distance between sun and earth)
Dem = 3.844e8 m (distance between earth and moon)
Ms = 1.9891e30 kg (mass of sun)
Me = 5.9736e24 kg (mass of earth)
Mm = 7.3477e22 kg (mass of moon)
Plugging these numbers in, we get:
Fse = 3.54e22 N (force between sun and earth)
Fem = 1.98e20 N (force between earth and moon)
As you can see, the sun pulls on the earth almost 200 times more than the moon. Note that Newton’s laws mean that the earth pulls on the moon with the same force as the moon pulls on earth.
But perhaps you mean the force relative to the mass of the object. Well, that’s not true in general: Fse/Me=0.0059, while Fem/Mm=.00269.
However, we can also look at the force between the sun and moon:
Fsm = 4.36e20
Note that the sun has a stronger pull on the moon than the earth does, even though the moon orbits the earth! In any case, Fsm/Fm = 0.0059, just like the ratio of Fse/Me. This is of course expected since the only variable term in our original equation gets canceled out.
Some further information. We might ask how to calculate the relative strength of tidal forces. We can look at it this way: what is the tension on a string holding two 1 kg masses together, 1 m apart, and such that the string is aligned with the radius toward the gravitational objects?
So we do this by calculating the difference between the forces on the two test masses.
Fsm1 = 5.931706207641053e-3 N
Fsm2 = 5.931706207561752e-3 N
Fsd = 7.9301e-14 N (the difference between Fsm1 and Fsm2)
I may get some flak at using such overly-precise looking numbers in my calculations. While it’s true that these numbers are imprecise in an absolute sense, they are precise in a relative sense. Their difference, then, is fairly accurate.
Moving on, we can do the same calculation for our test objects at earth’s distance from the moon.
Fmm1 = 3.318721894250374e-5 N
Fmm2 = 3.318721876983351e-5 N
Fmd = 1.7267023e-13 N
And as you can see, Fsd/Fmd = 0.46, which is right around half–just what other posters have mentioned as the strength of the sun’s tidal forces compared to the moon’s.
Hopefully this more or less answers your questions?
You killed it!
From the slightly privileged perspective of the guy who write the post:
Yes, you’re right on the money. I observed it would be possible to mount some sort of defense, but that I wouldn’t, because it was clear to me that O’Reilly wasn’t being quoted out of context or otherwise misunderstood.
It’s very amusing that Hentor, whose only criticism of liberals is for being too conservative, accuses me of partisanship when I’m the one demonstrating my willingness to criticize “my side” when deserved.
Huh? I just wanted to know why we sometimes have a full moon, and not all the time.
Only kidding. That’s close to what I was trying to get at. Mainly the 0.0059 and 0.00269 numbers. More specifically (or should I say, less specifically?) I was trying to get a feel for the relative importance of size and distance in determining gravitational pull. Is one a more dominant factor than the other? If so, how much more?
Tell ya, there are times you gotta love The Dope. and this is one of them. Thanks for taking the time you did, Dr. Strangelove. Awesome answer. Wow.
It would not be difficult, Mein Führer!!.. Uh, Excuse me, Mr. Martiju. 
Moon waxes. Moon wanes. Never a miscommunication.
Next question.
Sure you guys are all smart when it comes to tides, but what about rainbows?? Huh, and what about DOUBLE rainbows!? Don’t try and tell me that shit ain’t real miraculous. I’d like to see “science” explain that shit!
Which is also not to be confused with God of GAAP, who rules over the forces of accounting principles and balances Jesus’s checkbook.
This should help:
bad post
Wow! Thanks.
I will be bookmarking this and freaking my amateur astronomer friends tonight with it.
heh heh heh
But only for the backward accounting fundies in the US. The rest of the world believes in the more enlightened Deity (or not) of IFRS! 
The pull goes down with the square of distance–in other words, if something is twice the distance away, it will exert 1/4 of the force.
Whether that’s more or less important than size depends on how we define size. If we use “diameter”, then size is more important, because mass goes up with the cube of diameter. Twice the diameter, eight times the mass.
So if we double all the dimensions in our system, keeping everything else the same, the total pull will go up: the extra distance decreases the force, but the extra mass more than makes up for it.
An extra wrinkle is density. The sun is very big, but not very dense. If it were the same size as it is but the density of earth, it would have 4x the gravitational pull. So while size is important, it’s not everything.
Glad to. To be honest, I had a selfish interest here. I’d never done the calcs before, and I wanted to see if they actually matched what I had separately heard (particularly the 1/2 ratio for the tides). It’s cool to see theory match reality.
It’ll be a cold day in hell before I let you write your inventory back up.
eta: My god can beat up your god.
Actually, there are at least four other objects in similar orbits besides Cruithne: