86 thoughts on “Homework 25.A”

    1. 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.

  1. 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!

    1. 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.

  2. 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)?

    1. 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.

    1. 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.

  3. 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!

    1. 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.

  4. 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?

    1. 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.

      1. 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.

  5. 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?

  6. 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.

    1. 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.

    1. 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.

    1. 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.

  7. 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?

      1. 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!

    1. 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.

  8. 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?

      1. 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.

    1. 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.

    1. 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.

  9. 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!

    1. 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!

    1. 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.

    1. 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?

  10. 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

    1. 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.

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