58 thoughts on “HOMEWORK 20 - Fall 24”

  1. What should the answer be in terms of? The best I can get is in terms of μ and W, and I'm not sure I can simplify it further. Do I need more equations to solve for μ and W?

    1. I would say you also need to assume friction between the wedge and the block since there is a μs pointing to the intersection of the ground and wedge in the lower left corner.

    1. Due to B being a pin joint there are no moment reactions at B. There are X and Y reactions at B, but if you define the moment equilibrium equation at B, you need not consider them as their lines of action pass through B.

    2. Yup, if you calculate the moment around point B that will help eliminate the reaction forces and some of the component forces from force normal/friction from the wedge in your overall calculation

    3. You need to put them in your X and Y equations, but you dont really use those on to solve bc you don know the reactions, and then when you take the moment at b they don't matter.

    4. You might have to consider the x and y reactions as a pin joint, but given the method you might use to solve the problem you can definitely cancel out those forces. At the same time, you can't initially, without defining how you plan on solving the problem, ignore those reactionary forces.

    5. Yes, you need to consider the reaction forces at B because the pin joint can provide both horizontal and vertical reactions. These reactions are necessary to maintain equilibrium, especially since the wedge force P and the friction at A will introduce forces in both directions. Make sure to include them when writing the equations for equilibrium to solve for P.

    1. My answer contains both the coefficient and the Weight as we are not given a value for the coefficient of static friction, so there is not a way for us to cancel out the value for our answer. When solving for the normal forces you're going to have to equate the friction forces to the coefficient times the normal forces. Therefore, your answer could be in terms of both coefficient and the Weight of the block.

    2. My answer contains both the coefficient and the Weight as we are not given a value for the coefficient of static friction, so there is not a way for us to cancel out the value for our answer. When solving for the normal forces you're going to have to equate the friction forces to the coefficient times the normal forces. Therefore, your answer could be in terms of both coefficient and the weight.

    1. That should be fine, I don't think its possible to be able to answer it without both (unless their given). The way I thought about it is that P is dependent on the friction caused by the object on the ramp.

  2. Do, we need to factor in the weight of the wedge? I don't know if factoring it in would have any impact, but often the question specifies not to, such as in the lecture examples.

    1. In my lecture and the lecture book, we were told that the wedge will often be of negligible weight and dimensions in comparison to the block. Considering that the problem does not specify a weight for the wedge, I think it's valid to assume that the weight of the wedge is negligible in this problem too.

    2. According to my professor, we can pretty much always assume that the weight of the wedge is negligible. They'll either tell us, or we can assume that it is not necessary within the problem.

  3. Since the point of contact with wedge of the block at A is round, are the friction force and normal force aligned with the x and y axes respectively (straight up and horizontal), or are they at the angle of the wedge and normal to the wedge respectively?

    1. The friction force is aligned with the angle of the wedge and the normal is perpendicular to that. They are not aligned with the axes even though it is a round surface contact.

  4. I assumed that the friction force would be along the slant of the wedge since the corner of the wedge would move along the slant as the wedge is moved. For the normal I set it orthogonal to the wedge.

  5. When looking at the sum of the moments about point B, wouldn't the friction and normal force from the friction have to be horizontal and vertical because you don't have any known height of the block?

    1. I don't think so, because if you look at how the friction and the normal force are on the bottom left corner, the normal force would have to be diagonally up and to the right and the frictional force would have to be diagonally downward and to the right to keep the block stationary on the wedge. When you do the sum of moments you will have to use cos/sin of the angle for each force and then multiply 3d, but an angle will be necessary because having to keep the block stationary on the wedge means the forces will have to be at an angle to cancel out the angle from the wedge.

    2. You have to remember that because this is a pin joint. There is no moment at B. What is happening at B are X and Y reactions at. So, when solving this problem you need not consider the reactions, Fx, and Fy, as passing through B. After that solve for their respective forces and then after finding the required normal force equations for the normal force between the floor and ramp, and then in between the ramp and the block, after this you can solve for P.

  6. In the given statement it mentions that the block is inhomogeneous, in examples I have seen strictly homogenous blocks. Does this effect our approach to this style of question at all?

    1. Inhomogeneous means that the density of the block is not uniform, meaning the block's center of mass is not acting directly in the middle like we usually see. In this case, note that the block's center of mass is shifted slightly to the left. This difference shouldn't change your overall approach to the question, just be sure to account for this new position of the center of mass when summing the moments.

    2. This simply means the density of the block isn't uniform and therefore its center of gravity is not at the center of the block. The approach is the same though the force W doesn't act in the center as is shown in the image.

    1. There is friction between the wedge and ground because of the force P acting on the wedge to the right. You can imagine pushing the wedge in the direction given and since the surface is rough, there will be friction on the wedge and ground as well as the block in the same direction.

    1. Great question, I believe it's generally best to treat them separately. Different external forces are acting on each of the objects, for example, the wedge has applied force P and friction with the ground, and the block has friction between itself and the wedge and weight - from an example we saw in lecture, there could also be a wall restricting either object from moving horizontally. To generalize the process, I would analyze the forces for each system individually before assuming that the objects move together in the same direction because that might not always be the case.

  7. In the first step, I set the sum of vertical forces to zero to find the conditions where the block starts lifting at A. For the second step, how do you correctly set up the moment equation about point B while accounting for the friction force at A?

    1. I believe the surface between the triangle and the block you are referencing would only have the normal force and the frictional force. Keep in mind that the friction force would act in opposite directions as they block moves. But yeah, you are correct that it has a perpendicular normal force at that rounded surface point.

    1. This is not a slipping or tipping problem, it is a wedge problem. It says in the problem statement that you should find the minimum value of the wedge force in order to raise the block. Thus, to raise the block, you would need to push in the wedge. That means that your impending motion would be the wedge moving in the positive direction to raise the block. From there, you can determine the direction of your friction forces and solve for the correct forces.

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