Problems about statics. Force and Moment calculations.

[jonjacson]

Homework Statement

I´ll use an image to show the problem that you can see as attached file. I need to calculate a force-couple system applied to the point C in red, equal to the force f0 applied to the point E.

Homework Equations

M=fd

The Attempt at a Solution

Well in these kind of problems you just need to add two equal and oppossite forces f1 and f2 at the point C.

Now f1 is the equivalent force, and f0-f2 is the couple.

f1 is simply a 250 N applied at -45º to the point C.

My doubt is about the couple, it´s a bit confusing to me to guess what is the moment arm of the forces.

If the bar BD weren´t in the system it would be easy:

M=f2*0 (because it´s a force applied to the point and the moment arm is 0) + f0*0.72 (0.4+0.2+0.12).

But now I think that BD it´s like the basis of a seesaw. B is the point were the system would be turning around so if I project the forces f0 and f2 to the line CA, the components lying down on the line CA don´t have moment arm, only the orthogonal ones do.

So, the angle u must be 45, that means that v is 75 degrees.

f2 orthogonal = 250*sin(75)=241.48 N

The moment arm of this force is 0.4 meters, the distance CB.

f0 orthogonal = f2 orthogonal

The moment arm of this force now is the distance BA plus the radius of the disc 0.12, finally:

M total of the couple as function of f2 and f0= 250*sin(75)*0.4 + 250* sin(75)*0.32=173.86 Nm

The problem is that the book gives the answers and it´s 174.9 Nm, I know 1 Nm it´s a small difference but maybe I didn´t use a correct reasoning, so I´d like to check if the moment arms are right or not.

And it´s not too clear to me why the problem suggests you to start calculating the force couple system at the point A, I can´t see any benefit in doing that.

 

[haruspex]

If the bar BD weren´t in the system it would be easy:

As I read the question, you do not have to worry about forces in that arm. It is merely asking what force+couple would be equivalent to F0.

And it´s not too clear to me why the problem suggests you to start calculating the force couple system at the point A, I can´t see any benefit in doing that.

The principal difficulty is computing the distance from the line of action of F0 to the point C. The point A is a useful go-between.

 

[jonjacson]

Thank you for your answer :)

Well I get the same result if I do that, if I join the line of the force f0 with a perpendicular line to the point C, there is a right triangle, and one of his angles is 15º (60-45), so:

Cos(15)*0.72*250= 173.9 Nm

----

You are right, the problem doesn´t ask anything about the BD arm, it´s just that I am a bit confused when it´s needed to calculate the arm, because I try to think about the axis of rotation of the system, and that is not necessary.

I mean, in the case I show in this picture:

What is the moment arm of the 1N force about the green point u?

If I follow the rule which says you need to know the minimum distance between the line of application and the point, the arm would be the hole length of the seesaw.

If I consider that there is a fulcrum in the midle point, the arm would be only half the length of the seesaw, and that is what I think it´s the correct answer, Isn´t it?

If that is the case, WHy shouldn´t we consider the bar BD in the original problem?

 

[haruspex]

Cos(15)*0.72*250= 173.9 Nm

Your error is here in the OP:

force f0 applied to the point E.

It isn't applied at E (which is in the straight line of CBA), but 15 degrees around the curve.
Correcting for that gives me 174.89.

What is the moment arm of the 1N force about the green point u?

If I follow the rule which says you need to know the minimum distance between the line of application and the point, the arm would be the hole length of the seesaw.

If I consider that there is a fulcrum in the midle point, the arm would be only half the length of the seesaw, and that is what I think it´s the correct answer, Isn´t it?

The moment of the 1N force about u involves the whole length. The fulcrum provides a moment in the other direction.

 

[jonjacson]

Your error is here in the OP:

It isn't applied at E (which is in the straight line of CBA), but 15 degrees around the curve.
Correcting for that gives me 174.89.

The moment of the 1N force about u involves the whole length. The fulcrum provides a moment in the other direction.

I see, I wasn´t using the correct application point, I recalculated everything and I found the same as you.

Very interesting, so only to check if now it´s clear. In this case:

IF the problem asks you to find the equivalent force-couple of the 1N force, you can forget the fulcrum, the problem is not asking the real couple of the system, only the equivalent to the 1N force at point u.

IF the problem were asking us to calculate the total forces and moments on the system we should include everything in that calculation the 1N force and the fulcrum.

The total force would be 0, since the normal of the fulcrum equals the 1N force.

The system only rotates around the fulcrum with a total moment= 1N*L - 1N*L/2=L/2 N*m is the answer, counter clock wise. L is the length of the seesaw.

So if in the original problem the question had been "What is the total force-couple around the point C?"

We should have include in our calculations not only the f0 force but the force made by the arm BD too.

Is this right?

Thank you very much!

 

[haruspex]

... since the normal of the fulcrum equals the 1N force.

Not in general. Are there no other forces in this picture? No acceleration? What here has mass: the plank? u?

 

[jonjacson]

I went too fast with that. You are right, it´s not general, if the bar is massless and the point u weights 1N, the fulcrum must react with 2N to get equilibrium.

I didn´t think this correctly.