# Solving Part (b): Find F3, F4, F1 and F2

• gnits
In summary: F4 * 4a * sin(Ө) + F * 2a * cos(Ө) = F1 * 2a * cos(Ө) + F2 * 2a * sin(Ө) + w*a*sin(Ө) + w1 * 2a * sin(Ө)which simplifies to:4*F4*tan(Ө) + 2F = 2*F1 + 2 * F2 * tan(Ө) + w * tan(Ө) + 2 * w1 * tan(Ө)(Call this "equation 2")Now, from part (a) I know that:
gnits
Homework Statement
To find the internal forces acting in a certain system
Relevant Equations
moments and equating of forces

I have done part (a) and agree with the answer given in the book of 2a (2w + w1) sin(Ө)

It is part (b) where I am stuck. Here is my diagram:

(In green are show the forces of the couple - but they are not needed in my following working)

In grey, I have labelled the internal forces F1, F2, F3 and F4 on the system when considered as being made up of the two parts ABC and AC.

I need to find F3, F4, F1 and F2.

I have these four answers from the book, they are:

(1/2)*(w + w1) tan(Ө)
(1/2)*w1
(1/2)*(w + w1) tan(Ө)
(1/2)*(2w + w1)

Considering rod AC only and resolving horizontally I have:

F3 = F1

and resolving vertically I have:

F2 = F4 + w

These equations are consistent with the given answers.

2*F3 = w *tan(Ө) + 2 * F4 * tan(Ө)

(again, this is satisfied by the given answers)

2*F1 + w * tan(Ө) = 2 * F2 * tan(Ө)

(again, this is satisfied by the given answers)

So I have these four equations, which the required answers satisfy, but I can't see how to determine the values of F1, F2, F3 or F4 from these. I need another equation.

I thought to consider the part ABC and resolve vertically, but this leads to:

F4 = w + w1 + F2

Which is NOT satisfied by the required answers.

How can I proceed? (Have I labelled the internal forces with the correct directions?).

Thank for any help,
Mitch.

Last edited:
gnits said:
I thought to consider the part ABC and resolve vertically, but this leads to:

F4 = w + w1 + F2

Which is NOT satisfied by the required answers.
There is an external force acting on this part of the system which you have not included.

TSny said:
There is an external force acting on this part of the system which you have not included.

It is not the couple forces (F) as I am resolving vertically, and they both act horizontally.

Ahh, would it be the force at the pivot B?

If so then I don't want to introduce this unknown into my equations. So I could consider ACB and take moments about B. This would lead to:

F4 * 4a *sin(Ө) + F * 2a * cos(Ө) = F1 * 2a * cos(Ө) + F2 * 2a * sin(Ө) + wa * sin(Ө)

(Call this "equation1")

But of course now I have F involved.

Can I use the result from the first part of the question to say the the moment of F about B is:

2a (2w + w1) sin(Ө) ?

So that would give:

F * 2a * cos(Ө) = 4aw * sin(Ө) + 2a * w1 * sin(Ө)

I would then substitute this into "equation 1" to give a new equation relating F1, F4 and F2.

From this I could substitute in from my earlier equations for F1 and F4 both in terms of F2 and so solve for F2.

Would that be the way to go, or is it wrong to use the answer to the first part of the question in this way? Or indeed, if it is ok, then it there a simpler approach?

gnits said:
Would that be the way to go, or is it wrong to use the answer to the first part of the question in this way?
That's a reasonable approach. Why should it be wrong to do that?
I doubt there's an easier way.
You got four equations for AC using forces in two directions and moments about two points. For 2D statics, you can only ever get three independent equations for a rigid component. You could deduce the fourth from the other three.

Unfortunately, this approach hasn't worked.

So, if I consider part ABC and take moments about B (so as to exclude the reaction force on the pivot at B) I get:

F4 * 4a * sin(Ө) + F * 2a * cos(Ө) = F1 * 2a * cos(Ө) + F2 * 2a * sin(Ө) + w*a*sin(Ө) + w1 * 2a * sin(Ө)

which simplifies to:

4*F4*tan(Ө) + 2F = 2*F1 + 2 * F2 * tan(Ө) + w * tan(Ө) + 2 * w1 * tan(Ө)

(Call this "equation 2")

Now, from part (a) I know that:

F * 2a * cos(Ө) = 4*a*w*sin(Ө) + 2 * a * w1 * sin(Ө)

this simplifies to:

2F = 4*w*tan(Ө) + 2 * w1 * tan(Ө)

and I also know from my earlier working when considering AC only that:

F4 = F2 - W

and

2*F1 = 2 * F2 * tan(Ө) - w * tan(Ө)

So I can substitute for F, F4 and F1 into "equation 2"

and when I do this, everything cancels leading to 1 = 1.

Can anyone see where I have gone wrong?

Thanks.

gnits said:
Unfortunately, this approach hasn't worked.

So, if I consider part ABC and take moments about B (so as to exclude the reaction force on the pivot at B) I get:

F4 * 4a * sin(Ө) + F * 2a * cos(Ө) = F1 * 2a * cos(Ө) + F2 * 2a * sin(Ө) + w*a*sin(Ө) + w1 * 2a * sin(Ө)

which simplifies to:

4*F4*tan(Ө) + 2F = 2*F1 + 2 * F2 * tan(Ө) + w * tan(Ө) + 2 * w1 * tan(Ө)

(Call this "equation 2")

Now, from part (a) I know that:

F * 2a * cos(Ө) = 4*a*w*sin(Ө) + 2 * a * w1 * sin(Ө)

this simplifies to:

2F = 4*w*tan(Ө) + 2 * w1 * tan(Ө)

and I also know from my earlier working when considering AC only that:

F4 = F2 - W

and

2*F1 = 2 * F2 * tan(Ө) - w * tan(Ө)

So I can substitute for F, F4 and F1 into "equation 2"

and when I do this, everything cancels leading to 1 = 1.

Can anyone see where I have gone wrong?

Thanks.
Note that the question depends on the fact that all joints are freely hinged. If B were fixed the set up would be statically indeterminate.
So you need to consider the balance on e.g. AB.

I see, yes, all joints are hinged. So I must consider a rigid part. ABC is not such a part. AB is.

If I take moments for AB about B I get:

2a*F*cos(Ө) = 2a * F1 * cos(Ө) + 2a*F2*sin(Ө) + wa * sin(Ө)

which simplifies to:

2F = 2 * F1 + 2* F2 * tan(Ө) + w * tan(Ө)

Now I substitute in for F and F1 as I did previously, and this gives:

4*w*tan(Ө) + 2 * w1 * tan(Ө) = 2 * F2 * tan(Ө) - w*tan(Ө) + 2*F2*tan(Ө) + w * tan(Ө)

and this gives:

F2 = 2w + (w1)/4

whereas I had hoped for F2 = w + (w1) / 2

Have I just made some stupid arithmetical error?

Last edited:
gnits said:
I see, yes, all joints are hinged. So I must consider a rigid part. ABC is not such a part. AB is.

If I take moments for AB about B I get:

2a*F*cos(Ө) = 2a * F1 * cos(Ө) + 2a*F2*sin(Ө) + wa * sin(Ө)

which simplifies to:

2F = 2 * F1 + 2* F2 * tan(Ө) + w * tan(Ө)

Now I substitute in for F and F1 as I did previously, and this gives:

4*w*tan(Ө) + 2 * w1 * tan(Ө) = 2 * F2 * tan(Ө) - w*tan(Ө) + 2*F2*tan(Ө) + w * tan(Ө)

and this gives:

F2 = 2w + (w1)/4

whereas I had hoped for F2 = w + (w1) / 2

Have I just made some stupid arithmetical error?
OK, found it. Was indeed an arithmetical slip. Thanks so much for all of the help. My main lesson here has been to be mindful of which parts of the system are rigid. Thanks again, Mitch.

haruspex

## 1. What is the purpose of solving for F3, F4, F1, and F2?

The purpose of solving for F3, F4, F1, and F2 is to determine the values of these variables in a given equation or problem. This allows us to find a complete solution and fully understand the problem at hand.

## 2. How do you solve for F3, F4, F1, and F2?

To solve for these variables, you need to use the given equations or problem to set up a system of equations. Then, you can use algebraic techniques such as substitution or elimination to solve for the unknown variables.

## 3. Why is it important to find all four variables, F3, F4, F1, and F2?

It is important to find all four variables because they are all interconnected and necessary for a complete solution. Each variable affects the others and finding all of them allows for a comprehensive understanding of the problem.

## 4. Can you use a calculator to solve for F3, F4, F1, and F2?

Yes, you can use a calculator to solve for these variables. However, it is important to understand the underlying mathematical concepts and not solely rely on the calculator's results.

## 5. What should I do if I am stuck on solving for F3, F4, F1, and F2?

If you are stuck, it is helpful to go back and review the given information and equations. You can also try breaking down the problem into smaller parts and solving for each variable separately. It may also be beneficial to seek help from a teacher or tutor.

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