Statics problem

Hey guys,

I have a question about a Statics problem I ran into the other day. Here's a link to the basics of the problem:

http://physicsforums.com/showthread.php?p=3777621

I was helping a friend with this problem, and I found that if I solved for T using a moment about the point of reaction, my answer was HALF compared to doing the entire problem with summing forces only.

I have showed this to a few colleagues, and they all get the same answer as me. Some amount with forces, half starting with moments.

Any insight?

If R is an actual piviot point, why do you have it running straight up the arm? It's not a two force member, since there's a load in the middle, and it's not a pin at the end.

If T is a cable, it's the only two force member you have, and solve T by the moments around R, then solve R by what T is.

Basically, I don't see how R's vector direction is defined as you have drawn.

alfadriver wrote: If R is an actual piviot point, why do you have it running straight up the arm? It's not a two force member, since there's a load in the middle, and it's not a pin at the end. If T is a cable, it's the only two force member you have, and solve T by the moments around R, then solve R by what T is. Basically, I don't see how R's vector direction is defined as you have drawn.

Am I making an incorrect assumption that the reaction force will be along the length of the member? The more I think about it, the more wrong this assumption seems.

It's always something really simple that you overlook. And I think you may have just knocked it out of the park.

Yeah, you need to be more clear with how the end is constrained.

Is it a pin joint?

The problem is extremely unclear it's from an MET statics textbook. They do not define it as a pin joint. It's literally a pole butted against a wall and a floor. When I took ME statics we never ran into anything arbitrary like that.

Yes, that is an incorrect assumption. It's not a two force member, so you don't know what direction the force vector goes.

A pole that is butted against a wall and floor have independant X and Y forces, which don't have to go up the arm as if it were a two force member. That point does provide a pivot point, thought. IF (and being that's it's an MET question, I know it's not) there were some constraint that the arm fit into the corner, then you would have to account for a possible moment. But being as I think it's being written, the point acts as if it's a pin- normal force X and normal force Y, and no moment.

Solve T by the sum of moments around R = 0 (one equation, one unkown T).

Solve R by sum of forces in X and Y = 0, and you are done (two equations, two unknowns, Rx and Ry).

That gives you 3 equations and 3 unkowns.

OK so a 2 force member, the forces at each end will always be coaxial with the length of the member. Tension or compression.

3 force members, the direction of the support reactions can be any direction.

4+ force members dont really exist because all the forces that don't act on the ends can be summed making it a 3 force member.

Is that correct?

These are the times when I regret opening a thread. HUHHHHUHH!!???

In reply to Taiden:

Close enough. But it's pretty easy to have more than 3 forces on a beam.

Since we are at GRM, picture a span that has 4 cars sitting on it. That's a 6 force member- two on each end, and the 4 cars. Although classic statics would have it at 5, since you would not pin both ends....

The main distiction is that there are 2 force members, and there's everything else.

Good enough. Thanks.

I'm also noticing something... very few of my classmates are willing to explore these problems. I feel like I learn the most important stuff by never giving up on these things. I wonder how many engineering students miss out on important tidbits with the apathetic approach to problems presented outside of classes.

I feel smarter just having read this thread.

alfadriver wrote: Since we are at GRM, picture a span that has 4 cars sitting on it. That's a 6 force member- two on each end, and the 4 cars. Although classic statics would have it at 5, since you would not pin both ends....

Rereading this, did you make a typo by chance?

I agree that in this example it would be a six force member. Two support reactions and four loads on the span.

I agree that in statics the problem would have two supports. A pin type and likely a roller with only a Y component.

But this would still be a "six force member." Wouldn't it?

And a different question... in a classic statics problem, you could find the magnitude of all the four loads, find the centroid of the load distribution, and equate it to a single force in a specific location making it effectively a three force member.

Would you agree?

EastCoastMojo wrote: I feel smarter just having read this thread.

I had the exact opposite reaction

Taiden wrote:

alfadriver wrote: Since we are at GRM, picture a span that has 4 cars sitting on it. That's a 6 force member- two on each end, and the 4 cars. Although classic statics would have it at 5, since you would not pin both ends....

Rereading this, did you make a typo by chance? I agree that in this example it would be a six force member. Two support reactions and four loads on the span. I agree that in statics the problem would have two supports. A pin type and likely a roller with only a Y component. But this would still be a "six force member." Wouldn't it? And a different question... in a classic statics problem, you could find the magnitude of all the four loads, find the centroid of the load distribution, and equate it to a single force in a specific location making it effectively a three force member. Would you agree?

I should have posted 8 down to 7 (two at each end + 4 cars to one at the ball, 2 at the pin and 4 cars). but the point is really that it doesn't matter if it's more than two forces.

The problem with the 3 force member idea is that the moment of each is not really like putting them all in one spot. When you get into dynamics, where the forces are applied matters.

monark192 wrote:

EastCoastMojo wrote: I feel smarter just having read this thread.

I had the exact opposite reaction

Statics = the analysis of stuff that doesn't move.

Remember F=ma? this is F=0.

∑F=0 is surprisingly useful

Taiden wrote: Good enough. Thanks. I'm also noticing something... very few of my classmates are willing to explore these problems. I feel like I learn the most important stuff by never giving up on these things. I wonder how many engineering students miss out on important tidbits with the apathetic approach to problems presented outside of classes.

I'm just happy knowing there are engineering students that do what you're doing with problems like that. I haven't studied enough physics to understand WTB "statics" are, but when I see Alfadriver start talking about infrastructure, I sure as hell want somebody with your tenacity working on it..

I am taking dynamics currently, took statics last semester,, and my statics professor told our class of a classmate he had decades ago. This classmate of his would not worry about any zeros or ten to the power of whatever and just do the calculations then add kN or MPa depending on the type of problem. Scary stuff. That stuff ends up killing people, no one cares if you get the equation right but your units are wrong.

but i think it has been covered, its not a two force member so reaction at base is not coaxial. I wasnt able to tell from your picture where the angle of beam made a 3,4,5 triangle. its a pretty simple problem though. force and moment balance, then just put them together. angle beam makes with horizontal is necessary though.

Sometimes if my answers are screwed up, I do the problem with only units and no magnitudes. If my units are correct at the end, I inspect for arithmetic errors

Units are god like.

Also, statics is a neat class because you don't really need calculus for it. Anyone who knows SOH CAH TOA and algebra can learn most of statics. It's pretty dern useful too.

fritzsch wrote: I am taking dynamics currently, took statics last semester,, and my statics professor told our class of a classmate he had decades ago. This classmate of his would not worry about any zeros or ten to the power of whatever and just do the calculations then add kN or MPa depending on the type of problem. Scary stuff. That stuff ends up killing people, no one cares if you get the equation right but your units are wrong.

I was a TA teaching statics 20 years ago, I to make the students think a little, I told them that you get X off a problem if the number is wrong, but everything else is right, but if you KNEW that something was wrong, and said pointed it out even though that you had no time to correct the math, I would credit you with x points. Basically, if the process was right, but the answer was wrong AND the student knew it, I would give full credit. For the time I taught, I think I only got to do that once or twice.

Considering the reliance of CAD, even back then, I thought it was very, very important that students had an idea of what the answer should be so that they can filter bad CAD data.

Getting statics right is very important. For basically every single other class that is taken.

Being honest, though, I haven't used statics for 20 years. Just not the nature of what I do.