Problem statements
Solution video - H25.A
Solution video - H25.B
DISCUSSION THREAD
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HOMEWORK 25 - Fall 24
47 thoughts on “HOMEWORK 25 - Fall 24”
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It says to give the value of M as a vector, in this case, does it mean that you would write the magnitude as a vector as the k component?
Yes
Yes, since the rest of the system exists in the xy-plane, any moments should be perpendicular and therefore in the z-direction!
Yes, as it is in the xy-plane. This means the moment will be in the k direction.
yes!
Yes, it is meaning in the moment will be in the k component
In the image, M is clockwise meaning that it's -z using RHR, so when we write the answer, (if the answer is clockwise), should we write it as positive or negative?
since M is clockwise, my moment was negative, since it would be opposing the normal counter-clockwise rotation of M so instead you would -M in moment of b equation.
My moment was negative because it opposes the counterclockwise direction of positive moments. Therefore, you should subtract M in the moment equation for point B.
The sign of your answer would be whichever direction the z component of the moment points towards. If you found a clockwise moment, the z component would be negative according to the right hand rule, and you would get a vector that looks something like . Hope this helps!
Is the right angle at point C relevant to helping me take the moment about point B?
The right angle at C just proves that the member BC is vertical, as we know that DC is horizontal. This allows us to get the vertical length of the bar BC with the given geometries, which helps us solve for the moment.
Is there a way to determine the length between A and B for the moment about A or am I overcomplicating things?
I dont think that is necessary for this problem. In order to visualize the distances useful in this problem it helps to break up the overall system into smaller systems(do this by drawing FBD of the different bars). From there, think about what forces you would need in order to set up a moment equation to find the moment M and you might be able to narrow down what force balances or a moment balances are necessary to compute.
When calculating the moment about A, it is best to only look at the FBD containing member AED. When you calculate the moment in this FBD, you do not need to include the reactions at B since they are not part of this FBD diagram. In this moment equation, you should only have one unknown which will then help you solve for other components in another FBD.
For this problem, the length between in A and B is not needed. This is because it is best to take the moment about A when the free body diagram is just member AD, and not the entire machine. This eliminates the forces around point B.
You shouldn't need to calculate that length. If you break each segment of the free body diagram into separate joints, you should be able to solve for everything asked for and your moment about a shouldn't require any information from joint B.
Since there are many ways of solving this problem, which is the easiest way to go about it? And is there a strategy we could use to figure out which way to go about similar problems to save time?
The one I found was the easiest was solving for the minimum moment required to keep the frame in equilibrium was to first start with the first free body diagram A E D, then moved to the next once I found force D then to the second free body and then once I got to the third there was only one unknown which was the moment at B. When approaching these problems, I typically go straight to the free body diagram that has the term i'm solving for and if I cant solve then I go to another one and find that extra unknown.
Im having a hard time understanding how to draw the direction for the x and y components of each force from diagram to diagram. Anyone have any helpful pointers there?
It does not matter which direction you sketch the arrows; it will not change the magnitude of the components, only their signs.
The way I approach it is drawing them in the positive direction for the first instance you see them and then in the negative direction for the second instance since they will be equal and opposite.
Can I do the force on each of the members and 1 total force ( F ad) or do I have to do Dx and Dy and then solve?
The way I solved it, I did not need to do Dx or Dy to solve it.
You should not have to solve for Dx or Dy to solve this problem. In fact, if you look at the far-right member and apply Newton's second law in the y-direction, you may find that you don't need to solve for any component other than By...
How do we write out the final vector for M? Is it only in the z direction since it's a couple and it doesn't really have an x and y direction?
Yeah, you got it. The moment due to M only has a non-zero component in the z direction.
Are there zero-force members at point C because it's a corner?
I don't think so because that is only for trusses and two force members
No, it isn't a zero force member, but instead, you can treat it as a two force member because there are only two forces acting on it, in the x direction at D and C, which cancel out and you can separate the system into the components and slowly solve for each variable to find the overall value of M.
I’m wondering where to start in order to be able to find the distance from point A and E in order to use the Moment at A equation.
You should use the length D on the vertical side and that its a 3-4-5 triangle and set it up like a ratio in order to find the horizontal length.
How exactly are we supposed to express a moment force in terms of a vector?
I believe what is being asked for is that when you get the final value for M, they want it in standard vector notation, meaning i hat for x, j hat for y, and k hat for z. Hope this helps!
I would just solve it like normal and then at the end you can I could better see that there is only a z component.
How does solving for the moment about A help to find the moment about B? Is MA = -MB?
Solving for the Moment at point A, will solve for the X component of D, which than you can use to solve for the X component of C, which than can be used to solve for the Moment about point B, Hope this helps.
When I solved for M, I got a positive number, but I have reason to believe that it should be a negative number, I double checked and all my signs are correct, so I'm not sure where I went wrong.
I think in this case, you may want to check your internal forces at joint D and C, because they are internal forces when you draw FBD respect to different members it will change direction. By doing this I think you will get the correct answer.
You likely did not make any error. You probably assumed that M was negative to start in your calculations because that is how it was drawn. Then, when you get a positive number, it confirms that you were correct and M is indeed negative. It can be confusing, but essentially a positive number means that your initial assumption was correct.
It may be the case that you found the magnitude for the moment, but the problem asks for the moment as a vector, so I would refer to the direction of the moment from the diagram.
I had the same issue. It makes more sense for the moment to be negative (meaning that it is opposite of the way it is draw in the diagram). I went back and reversed the direction of some of the reaction arrows I drew initially. When I went back through to correct the signs, I found the final moment to be negative. You can try this and see if it helps.
When drawing the individual free body diagrams of each of the links is there some convention as to the order to go to make sure your forces all face the correct direction? For example, to get the correct direction for the moment the forces must be going into DC in the FBD so that they face the correct directions, is this because these are internal forces through that member?
When putting the answer in terms of a vector am I putting (0i,0j, answer) ?
Since the problem is asking for a moment, yes, you would write the answer in that form with only the z-component having a value
Conceptually, does anyone know why the moment (M) on member CB doesnt translate to the other members when you make isolated FBD's? I would think an external moment being applied would have an effect on the whole system, no matter where its being applied.
The isolated FBDs create isolated systems, so within the system for segment AD you will not consider any forces from the segment CB because you have broken them into two different systems. So for segment AD, you don't consider M or the forces at B, and for CB, you don't consider the forces at A.