59 thoughts on “HOMEWORK 22 - Fall 24”

    1. Yes, you should consider the unlabeled point because it anchors the pulley system that supports the weight W. A downward force of P is applied at this point due to the tension in the cable from the pulley system. While this point does not directly influence the internal forces, the force applied here is important for the overall equilibrium of the structure. Hope this helps!

  1. Could someone help explain why the forces in truss members change or remain the same as you analyze different joints? Specifically, when moving from one joint to another, how does the tension or compression in the connected members affect the equilibrium at each joint?

    1. from what i understand the forces in each member, whether tension or compression, remain constant along the length of the member but affect equilibrium differently at each joint. At every joint, the sum of forces must equal zero in both horizontal and vertical directions. As you move from one joint to another, the force in a member remains the same but is applied in the opposite direction at the new joint. This force then influences the equilibrium at the next joint, helping determine the unknown forces in the other connected members. Forces change between joints based on the overall structure and load distribution.

    2. The forces should stay the same no matter where you look at the members from. The only reason they may seem different is when you are finding the moment at different points. This can also be seen if you find a member through the sum of forces in a direction and how you can only get one number (if you do it right) no matter what order or method you use to get the answer.

    3. Each joint in a truss must be in equilibrium, meaning the sum of all forces acting on the joint must be zero. So, when you analyze a joint, the forces in the connected members, either tension or compression, are determined by the requirement that the joint remains in equilibrium. As you move from one joint to another, the forces in the members connected to the joint you just analyzed become known quantities. These forces are then used to determine the forces in the other members connected to the next joint. Hope this helps!

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    4. For example, say you determine a member is connecting two joints in tension. When analyzing one joint, the tension will pull outword, where as at the other joint it will be pulling inward, keeping the members at tension. Hope that helps!

    5. The forces in truss members stay the same (either tension or compression) no matter which joint you analyze because they are internal forces consistent throughout the member. However, how these forces affect equilibrium changes as you move from one joint to another due to the different angles, loads, and connections at each joint.

      When you solve one joint, the force you find in a shared member carries over to the next joint it connects to, ensuring consistency. The equilibrium conditions (ΣFx=0 and ΣFy =0) at each joint account for the geometry and loads, so the way forces balance changes at each joint, but the member forces themselves don’t.

    1. This would be the easiest spot to find the force at member 1. I would make a separate cut to find the forces on members 2 and 3. The strategy for deciding on where to section the structure is to find where the cut goes through the members you are trying to solve for. So in this case, cutting laterally between points I and E allows you to cut through member 1 so you can solve for that force.

    2. I split the truss between I and E, E and H, and H and C so that I could have all 3 of the members as forces. By doing this you also have to calculate the force of member EH, but that is not too difficult. You can split it that way too, it really depends on how you plan to attack the problem.

    1. Yeah I think so, as it acts as an anchor for the pulley system. Therefore, a downward force of P would be acting at this point. You will need this for the overall equilibrium of the structure.

  2. How do we solve for the reaction forces Ay and By? I am considering the full truss as a rigid body, but if I were to do a moment, r would be used for some terms but not all and therefore not cancel. Any suggestions?

    1. I'd assume that you would have to look at the location the reaction force of the pully acts on the truss itself and not the ground. So, the reaction force is acting on joint K. And I'm pretty sure you can have the reaction force acting directly on the joint, so you do not need to account for the R of the pully for the perpendicular distance.

    2. To solve for Ay and By, using the entire truss as a rigid body is the right approach. When taking moments, be sure to consistently include the perpendicular distances from the forces to the point of rotation. If you’re using the pulley radius r for some terms, you need to ensure it is properly incorporated wherever applicable, such as for the tension forces from the cable system. Double-check that all distances are measured relative to the chosen pivot point to avoid inconsistencies.

    3. You can take the moment about one of the points and solve for the reaction forces. But depending on where you make your cut, you might not have to solve for them.

    1. You have to look at the sum of the forces around that point. When you set the sum of forces in the x or y direction to zero, it is usually apparent which forces are equal to zero. You can also use methods from the zero force member lecuture to quickly find which ones are zero force.

    2. You can look at the types of joints. A T-joint will have a zero member if no external forces are acting on it. This is the same for Y-joints and elbow joints.

    3. To determine if a member has zero load on it, look at the FBD for that joint. Because the truss is in static equilibrium, the sum of the forces in the x and y directions add up to zero. So for the FBD, if there is only one member acting in one direction, then that member will have zero force on it. The easiest example is a T pr Y joint, where there is one member pointing at an angle compared to the other two. The member in that direction will have zero force, unless there is an external load on it. The other two members will have forces with equal magnitude and act in opposite directions.

  3. Is it ok to still use the method of joints to find the loads in the members in this problem or do is it the method of sections make the problem more concise and faster?

    1. Using the method of joints is possible and would ultimately result in the same answer, but I believe it would be a lot more work. Instead, I first solved for the moment about point A and used equilibrium equations to find the reaction forces Ax, Ay, and By and then made two splits (horizontally through 1, and horizontally through 2 and 3) to solve for the loads using the method of sections. The method of joints may work, you would just need to start from the outside of the truss and work your way inside to these three points, which would require solving many more systems of equations.

    1. There are multiple ways to solve the problem, but I don't think it can happen with just one break since 1, 2, and 3 don't all share one line where there could be a break. Try using two separate breaks, such as one for 1 and another break for 2 & 3.

    1. For member 1, I think you can ignore the weight from the pulley/cables since it should factor into your calculations for the reaction forces at A and B.

    1. If you are trying to find the total moment about point A, then you should find the moments from the forces in DH, CH, and DE (although I think that the moment from DE will be equal to 0 since the force intersects point A). You can use this to solve for the force in CH.

  4. Can we essentially ignore everything happening with the pulleys since we can select a smaller section of the entire system to find the load carried by members 1, 2, and 3?

    1. It depends on what part of the system (the smaller section) that you work on. If you are focusing on the top part of the system after you divide it, then no, you can not ignore everything that is happening with the pulley cause there is a force tension acting on it... but you can neglect R.

  5. Is it possible that the forces FDE and FIE are associated such that FDE = -FIE. This would prevent having to do multiple cuts and such and I figured it would be okay because any forces acting on those two compents dont act in the vertical direction

    1. That would not be necessary for this homework as we don't care about the loads in the cables of the pulleys. Just the truss components and the load from the pulleys onto the frame.

  6. How are we supposed to determine the forces caused by the pulley's, and would those effect the way that we are solving this question if we are just going to use sections?

    1. You can solve for the forces in the y direction for the pulley with the block, then you'll find that Tension = P. This will then help you when solving for the forces with the different sections.

  7. I am not sure if the equilibrium equations should be taken with the entire truss system in mind, or as we go member by member/section by section. The approaches seem to lead to different answers.

    1. Initially, I would consider all of the forces in the overall free body diagram with the entire truss. This would include the reaction forces at A and B as well as the forces in the pulley system. By doing that, you can calculate your reaction forces. Then, you can split up the truss into just the section that you desire, and those would include the forces in the overall free body diagram as well as the forces of the members in the sections that you cut. From there, you can solve for the forces of the members.

    1. Yes, you should take into account the unlabeled points when solving. Specifically the point directly below K. This is because when you solve for the moment at H (or whatever point you choose) and plug in to find the equilibrium equations, you will need to include this unlabeled point as there is a tension force acting on it.

    1. So for the pulley system, there is the block of mass W which is 2P, and it has two strings attached to it, so each one of them has a weight of P. For the long string that goes across both pulleys, that one has a weight of P across the whole thing, resulting in a P force going down on the left side of the pulley at K. So there is not a force at K, but r distance away from K, and the same for the right pulley. But there is a force P at M because it is directly connected to that point.

  8. I’m wondering how to incorporate the K and M pulley radius in our equation. Does it have to do with our moment or reaction forces for just the pulley system? The Truss system? Or both?

    1. You could solve it using method of joints or method of sections, but sections would simplify the solution. Using method of sections, you can cut between 1, and make another cut between 2 & 3

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