# Having a difficult time in statics. (EM 221)

• celtics777
In summary, the question is about determining the moment of a 21-pound force about point B by resolving the force into horizontal and vertical components. The concept of moments and resolution of forces is discussed, with an explanation of how to calculate a resultant force and the effect of a single force in a given direction. The formula for finding the horizontal and vertical components of a force is also provided. However, the student is still struggling to understand how to apply this information to the specific problem at hand.
celtics777
A 21-pound force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α=25°, determine the moment of the force about point B by resolving the force into horizontal and vertical components.

I wish I could say I have some work to show progress, but I don't. Worst professor ever.

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Well what don't you understand?

Have you resolved the force into horizontal and vertical components as the question asked?

I honestly don't know how to do that.

I may be over thinking everything in this class so far, but this is my first "problem solving" class I've had in my engineering program.

Well what do you know about moments?

Perhaps a definition?

Also do you know what is meant by 'resolve the force into horizontal and vertical components'?

I know moment is the physical effect of a force to cause an object to spin about a point. Same as torque, as far as our class is concerned. (professor said so)

I know that M=Fd, where F ~ magnitude of force. and d~ orthogonal distance from point to line of action of force.

I know moment is always cw or ccw in 2d.

I am not sure what it means by resolving the force..

My professor is only giving me definitions, and not doing a very good job of showing me how to use what I know.

OK so let's do the resolution first.

Do you know how to calculate the resultant of two forces (which act together at the same point)

In other words do you know what a resultant is?

Don't worry if you don't know, but I do need to find out.

Sorry, I don't. He's assuming we know more than we do.

A Kentucky lad or lass huh?

Well I'm going to use a Kentucky catapault as as example.

My first sketch shows a view looking down on the catapault with a stone loaded into the V and the elastic drawn back.

So the elastic is stretched and exerting force on the stone.

If we let go the stone will be slung forwards along the dashed line with a force I have marked as R.

Symmetry tells us that each of the forces in the elastic are equal and I have marked them as F.
We note that they are at an angle alpha to the eventual trajectory of the stone.

Now this is very important.

R is the resultant of the two forces F acting on the stone.
That is we could achieve the same effect if we slung the stone forwards with a single force R

This combining of two forces to make a resultant can always be done.

********************

Resolution of forces is the exact opposite process.

That is we split a single force into two other forces or if you like replace it with two other forces.
Usually is makes sense to replace with two forces at right angles to each other.

I have done this in the second sketch.

Here I have replaced each of the forces F with two forces at right angles to each other. One direction is along the dashed line of R.

I have used the formula that the resolution of F along R is Fcosα and at right angles to R is Fsinα.

Since the effect is the same if I replace the forces F with the resolved forces Fcosα and Fsinα, I can pick two and recombine in a different order.

So If I calculate the resultant of two forces Fcosα I get R = 2Fcosα acting along the line of R

Interestingly if I take the two forces at right angles to R I get them opposing each other and cancelling out ie their resultant is zero.
This, of course is what we expect if the stone is to be slung along the dashed line - there is zero force pushing it sideways.

Does this help so far?

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Lad. haha. :) I think we just got smashed by Louisville though.

So a resultant is the the "result" of two (or more) forces acting on the same object in different directions? So it's like adding forces? Is there a formula for this so I will know for the future? Is it always R=2Fcos(α)?

I think I understand how a resolution is like replacing a single force with two other forces, but I'm still not sure how to do that.

Thanks for helping me by the way.

Looks like we're doing swell.

Yes a resultant is exactly the result of combining forces. √

As to formulae, look carefully at my sketch.

There are two forces F acting on the stone (one for each side of the elastic) so we double the effect of a single F acting (hence the 2 in the formula).

The effect of a single force F in a direction alpha to its line of action is Fcosα

By effect I mean the force acting in the direction alpha I would have to replace it with.

The effect of a force acting in a direction at right angles to alpha is Fsinα

Now there is more to it than this but this is enough to complete your question and to take forwards to the next stage.

Can you now use this information to see that by resolving your force horizontally and vertically we mean replace it by a pair of forces, one acting horizontally and one vertically.

Can you post the formulae for these two forces now say Fh and Fv?

I'm trying really hard over here, but I'm having trouble knowing what I should be looking at on my image. I know both AB and the force already displayed is 20-lb, but which do I make the right angles to? Or am I not supposed to do that here? The one with alpha makes most sense, but then why does the questions say determine the moment of force about point B? I seem to understand everything until I actually go to do the calculations.

I tried what you said and got Fh=40cos(25) and Fv=40sin(25). Those are almost guesses, but at least I'm starting to understand more.

Alpha is the angle the 20lb force makes to the horizontal.

And we are told the value of alpha.

So the 20lb force has the same effect as a horizontal force of 20cosα

(There is only one 20lb force so it is not 40cosα)

and the same effect as a vertical force of ____? Since the vertical is at right angles to the horizontal unless you are drunk.

So we can replace the 20lb force by 2 forces, one horizontal and one vertical.

Fill in the blank and we can proceed.

haha. I'm not drunk. I'm trying.

20sin(25)? Approx. 8.452 lb?

Great stuff

So you can now draw a diagram showing the horizontal and vertical forces instead of the 20lbs.

Since you know the length of the lever (9") and the angle it makes to the horizontal and you already know the definition of moment

You can now calculate separately the moments of the horizontal and vertical forces.

Can you see the required orthognal distances in your new diagram?

All you then have to do is decide whether the meoments are ccw or cw and total them up

M=Fd

Vertical: M=(20 lb)(9")(sin(25)) approx. 76.07 lb-in

and

Horizontal: M=(20 lb)(9")(cos(25)) approx. 163.14 lb-in

Are those the two separate moments? Do I just add them?

And what should the diagram look like? Just a right angle with 20cos(25) horizontal and 20sin(25) vertical?

Thanks for being so patient with me.

Did you draw the diagrams?
Civil engineers draw lots of diagrams.

It should look something like my top sketch with the horizontal and vertical forces acting at A.

Now I have shown the beginning of the calculation of the moment for the horizontal force in the bottom diagram.

Yes we have the forces.

But the lever is leaning over at 65°.
So we must do some geometry to find the perpendicular (orthogonal) distance between the line of action of the horizontal force and B.
This is again simple trigonometry, you should be able to do.

Knowing that AB = 9 inches, can you state what the perpendicular distance is and calculate it?

When you know the force and the perpendicular distance you can calculate the moment.
Is this moment clockwise or counterclockwise?

Can you do a similar diagram for the vertical force? (there is no need to post it)

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## 1. Why is statics considered a difficult subject?

Statics is considered a difficult subject because it involves complex mathematical calculations and requires a strong understanding of physics concepts. It also requires a lot of practice and critical thinking to solve problems.

## 2. How can I improve my understanding of statics?

To improve your understanding of statics, it is important to attend lectures regularly, take notes, and actively participate in class discussions. You can also practice solving problems and seek help from your professor or peers when needed.

## 3. What are some common mistakes made in statics?

Some common mistakes made in statics include incorrect application of equations, forgetting to consider all forces acting on an object, and neglecting to draw accurate free body diagrams. It is important to pay attention to details and double check your work to avoid these mistakes.

## 4. How can I prepare for statics exams?

To prepare for statics exams, create a study schedule and review your notes and class materials regularly. Practice solving problems and seek help from your professor or classmates if you are struggling with any concepts. It is also important to get enough rest and stay calm during the exam.

## 5. What are some real-world applications of statics?

Statics has many real-world applications, including structural engineering, architecture, and mechanical design. It is used to analyze and design structures that can withstand various forces, such as bridges, buildings, and machines. Statics is also used in the study of fluid mechanics and electromechanical systems.

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