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## 86 thoughts on “Homework 25.A”

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Do we need to know the length of DC to solve this question?

I was able to get an answer without knowing the length of DC.

Does this mean that we have to avoid taking moments? And do we need to know the reaction forces at A and B to solve?

I was able to get the answer by trying to get reaction at A but not B. For B, it may be possible to use momentum.

Can we treat member DC as a two-force member?

Yes!

Member DC can be treated as a two force member because only two forces act upon this point.

yes

Does the moment M acting on BC not allow us to treat BC as a two-force member or can we treat BC as a two-force member?

Hi Jack. If only the loads applied to an element are acting at the extremes we would be dealing with a two-force member. Any additional load or moment would make the element a non-two-force member. So you're right, the moment M acting on BC not allow us to treat BC as a two-force member.

Thank you so much for the knowledge you have shared. You have changed my life immensely.

When starting this problem, should I make two FBDs or three, one for each bar?

Hi! I made three FBDs, one for each member and that made the problem very easy to solve! Keep in mind that DC is a two force member!

Do we not need to consider the weight of the entire structure or even of each beam?

Since the problem does not provide any value for the weights of the beams, it is safe to assume that the weight of the structure and beams can be neglected.

I am confused how DC is a two force member when on of its forces acts in line with arm AD and the other with CB. Since they are not equal and opposite i'm confused how to simplify it to a two force member. If someone could explain it to me that would be greatly appreciated!

Yes definitely, DC is a two force member since it has a set of reaction forces at both its ends with no additional forces acting on DC.

If we consider that the machine is not currently in motion, the segment DC is additionally not moving. Since DC has only two points where force is applied, the total forces on each end must be equal and opposite, otherwise the segment would move.

Just to double check, we cannot treat BC as a two force member due to the additional moment?

Yes that is correct! We cannot treat BC as a two force member. There are two forces at b, two forces at C and a moment.

I wrote out the FBDs for the three links and found that there were 8 unknown forces (Ax, Ay, Dx, Dy, Cx, Cy, Bx, By), on top of the unknown force M. Are we supposed to use the sum of forces for the entire machine, since without this we would only have seven equations (three for AD, three for BC, and one for the two-force member DC)? Or is there something that I am not noticing (i.e. BC is a two-force member, Cy has to be 0 due to the right angle, etc)?

Yeah I used the equations of the total forces on the machines to solve. (I don't think any of them are two-force elements or Cy is 0)

I'm pretty sure that DC is a two force member since there are no other forces acting on that element.

Do we not need to know the lengths of any of the members AD, DC, or CB to solve for M?

The vertical length of both BC and AD are given in terms of d, while the slope of AD allows for its horizontal to be determined (as well as its total length, but that isn't necessary). The length of DC is not needed.

Is it recommended to make 3 FBD's to solve this question? Or can we make 4 by splitting length AD at point E?

I am trying to write the moment equation about A for member AD. How can I find the horizontal distance from A to E?

I think you can use the given vertical distance and the given ratio to find out the horizontal distance with some trig.

Should the force at C be shown vertically while the force at D shown on an angle?

Do we need to consider a FBD for the pin at E that forces 2P and P act on? Something like this was done in example 8.C.8 for reference.

Is the moment M acting directly on the joint B or is it just acting on the member BC?

Do we assume that the length of CD is d?

I don't think that we can assume that, but the length of CD should not be needed to solve this problem.

To find the moment, do we add it to the moment equation about B?

Hey Elena,

If I understand your question correctly, I think you're asking about the moment M that you have to find. If you're taking the moment of the system about B, yes, you include the moment M in that equation. You wouldn't include forces acting at B in an equation for the moment about B, but a moment acting at B should be included.

If DC is a 2 force member do we ignore forces at those points? So would we just have 2 unknowns and the moment we are trying to find? Thank you for your time and help!

I believe that you are right. Of course there are the unknowns at the reactions, but I do not think that they are necessary to solve this

Are we allowed to assume that the moments at A and B are equal?

I don't think so because if you're talking about the system as a whole, the external forces would be at A and B, but the forces at E are also external, and those would impact the moment about A differently than they would impact the moment about B.

I'm afraid that I may have done this problem wrong. It seems to be too simple and I didn't have to use any of the reaction forces at A or B in my calculations. Am I going about this right?

Hi Clark,

That is how I solved the problem, you don't need to use the reactions because everything in the machine is moment based.

Does M act as the moment about point A for point B. Like instead of Bx and By we use M in a moment equation?

M is a applied load. in this problem, you can consider it as acting anywhere. It does not affect the way you draw the constraint forces at B.

In moment equation, moment is calculated by either moment caused by force or applied moment.

Revised:

M is a applied load. It does not affect the way you draw the constraint forces at B.

In moment equation, moment is calculated by either moment caused by force or applied moment.

Does this problem require us to draw 3 FBDs or 4 (by dividing AD into AE and ED)?

My suggestion:

1st FBD: CD, 2 force member;

2nd FBD: AD, calculate force at D;

3rd FBD: CB, calculate M.

How are we supposed to write the value of M for equilibrium as a vector? Isn't it just a value in terms of P and d that will be negative or positive depending on rotational direction?

in k_hat direction

Since the moment M is clockwise, the answer should be negative right? And in my equilibrium moment equation about B should I use M as positive or negative? If I use it as negative in the equilibrium equation I keep getting a positive value.

In FBD, assume CCW; in equation use plus (+); when replacing the variables by the numbers, take whatever positive or negative you get.

I got that M is positive. This is because we assume the clockwise direction and get a positive value with that assumption (provided in the problem diagram). Although the direction of M in the diagram is negative, but we calculate it to be positive in that direction.

How does knowing DC is a 2 force member help us in this problem? I'm struggling to figure out how to tie that member into equilibrium equations.

It helps since there would then be only one unknown force, Fcd, at D and C instead of two unknowns, Dx Dy or Cx Cy. So you'd need less equations to solve for the unknowns.

Is there any differences between a frame and a machine? If so, what are they? Thank you in advance.

Hey Matthew, both frames and machines are types of structures with at least one multi-force member. However, the key distinction is that frames are stationary whereas machines are designed to do mechanical work and contain moving parts which alter forces.

Since CD is a 2 force member will the D force in AD also be associated with the angle calculated from the 345 triangle? and allowing us to calculate the D force easier?

Also since A is a pin there is no moment at A right?

Yes, only if you set the moment equation at point A.

Hi Everest! There is no moment at point A, because joint A is grounded using a pin joint which allows for rotation. This is true of both joints A and B. The moment shown at joint B is an additional moment (not caused by the joint itself). You were correct!

Yes, in AD, if you set up a moment equation at point A, you have to use the 345 triangle to calculate the torque from point D.

As we supposed to have M in the negative direction, because that is the direction of the arrow? Or should we assume it is positive.

I believe that the moment is meant to be negative. We take into account the directions that forces are drawn in, so I assume the same principle applies to moments as well.

Yes, M at point B is negative since it is clockwise.

I'm trying to solve this problem by first analyzing AD and solving for the Force D but whenever I do the moment about A I end up getting 0 every time because something cancels out. Is my approach wrong?

Also is there a Bx reaction or just By?

There is a Bx reaction as well as a By reaction. If B was a roller joint, then there would only be a By reaction. Additionally, there is a moment at B.

Since the force at D has a line of action through point A, wouldn't its moment about A be zero?

At D, there is only a Dx reaction because member DC is a two force member. Therefore, when solving the moment at A, you would need to cross Rad X . From here, you should be able to solve for Dx.

If you take the moment about A, do you have to find the distance between points A and B? Then multiply that by the moment about B?

no. the moment at B is applied locally.

Do we need to use the length of Cd, or can we assume it is a two force member?

Hi Isabelle, the length of CD is not required and as outlined by Luke above, it is considered a two-force member which can be seen by evaluating the forces on the member.

When taking the moment about A, would you multiply P by (d/2) as its moment arm and would you need to multiply P by its 3-4-5 triangle relation as well? Also, what would the moment arm be for 2P? Thank you!

Hi Gabriella! When taking the moment about A, for P the relevant distance would be (d/2) as you thought. This is since you take the distance between the point of reference and the closest point of the line of action of force P. However, you will need to consider the 3-4-5 triangle relation when accounting for 2P. Since you can consider the distance (d/2)*(4/4), you can consider the distance when considering 2P in the moment calculation as (d/2)*(3/4), or (d/8). Hope this helps!

Can we assume that DC is a two force member? I don't see how we could solve the equation with multiple forces on each joint

Hi Hadassah. Any rigid body with applied forces only in its extremes is considered a two-force member which is the case of DC.

Can someone explain why DC is a two-force member? Wouldn't there be two reaction forces at each end?

Hi Marina! You can determine that DC is a two force member by evaluating the local free-body diagram of the member. The y force equilibrium equation reveals the relationship between Dy and Cy (Cy = -Dy). You can easily verify this. The moment equation can be taken about either point D or point C to reveal that Cy = 0 or Dy = 0. You do not need to know the distance to determine this. So now we know that the forces at points D and C do not have any vertical component: they are co-linear with member DC itself. Another requirement for DC to be a two-force member is that its x-components are opposite in magnitude. You can verify this by solving for the x equilibrium equation in the same local free body diagram. You will find that they are equal and opposite (Cx = -Dx). For these reasons, member DC is a two-force member.

So, in AD is there a Dx and Dy or just Dx. I'm confused from something mentioned above.

Since DC is a 2-force member, it only has internal forces (i.e. compression or tension forces). With that, the force at D acting on AD will be equal and opposite to the internal force of DC, which means there will be only one reaction force at D. Does that help clarify?

DC is a two force member. Imagine if you have Dy, will DC start turning around no matter what force you apply at C.

Kind of late but does DC have any effect on the final solution aside from sorting out whether or not the reaction force at C is positive or negative. Just wondering because I found the reaction Dx from the Ad bar so I wasn't sure if the CD bar had any further effect

Yes Dc has a final effect on the solution, it can be used to find M if you take the moment about B.

Why do we need to solve for F at C and F at D since these are internally balanced could we not just ignore them in our moment calculation

You can ignore them when you are doing calculations involving the whole frame since, as you said, the forces are internally balanced. That being said, when we do calculations involving the separate sections those forces need to be taken into account.

Is point E a joint or just a position where the forces are applied?