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Hello, I love your videos they make this topic so clear! However I have a simple question. When you write the equation w(x) for the triangular distributed load, when is it -300, not 300? I am talking about the intercept or whatever you are going to call it. How do you decide on the sign?
@@studentengineering Thank you for the response. Usually all distributed loads go downwards, does that mean that I will always have the w(x) intercept as negative?
Okay, so it definitely works, but what I don't understand is the reason for flipping the curve, can someone explain the reasoning? Thank you, excellent video
I am confused on why 3 is being plugged in instead of 6. I am getting different boundary conditions but ending up with the same max moment of -2250. I see you are defining the part where the triangular distributed load starts as 0 on the x axis? Why isn't this defined as 3 on the x axis?
Hi, thanks for the great video, I am trying to find the volume of a an entire tree, picture a silhouette of a tree, let's say a pine tree, I am thinking of a 21m high tree, the trunk is 0.4m diameter with a taper of the trunk at 0.8% on each side and the branches start at 1.8m off the ground and then there is a parabolic taper to the top of the tree from that point on to zero. The largest width of the tree, where the branches first start is 12m wide. I am thinking to revolve half the silhouette and so half of trapezoid of the trunk at 0.2m tapering to 1.8m high and then 6m branch start going parabolic to the top. Could you perhaps kindly outline how because I want to get this exact? Can this be done with an equation of the edge of the silhouette, or is it better to spin the area of half the silhouette, I'm not sure. Thank you.
Sounds like an interesting problem. I think that your problem would be solved the easiest by just finding the cones and partial cones that make up the tree and add up those volumes. I don’t think many of the equations in the video need to be used because you are revolving the area in a full circle and the lines you are using are all straight.
@@studentengineering I've done Pappus Guldinus theorem volume calculations in Calculus in high school, and wanted to do it once again, as the tree volumes, need to do a bulk density check for the entire enclosed volume of the trees, quite important because I have to analyze a parking garage roof slab with trees on top acting as additional loads on top of a thick soil overburden. FYI, the tree outline is sometimes parabolic and not always straight because there is a mix of maple/pine etc. Yes, I do understand that I could do cones and cylinders but preferred to do it in one shot. Not a problem but tks for the quick reply.
Great question! They actually are connected but in order to find the centroid of all of them we break them up into more manageable parts. Does that make sense?
Fantastic explanation. You saved me hours of time from reading hundreds of notes when you summarised the information i need in just a few minutes. Truly a life saver!
Great question! I go over that in this video: Frictional Forces on Flat Belts Explained // Equations and Explanation ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-AB-Oxslw9mI.html
Great question! This video explains what C is and the other variables in more detail: CATENARY LOADED CABLES (Free-Hanging Cables) // How To Solve For The Unknowns // Equations Included ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-xqWa5In5Jlw.html
Great question! Here is a link to another one of my videos that explains that: SHEAR and MOMENT DIAGRAMS EXPLAINED (calculus especially) ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-bj8jvZ6C8l4.html
Would you please show us how to find the centroid of the irregular solid according to the following descriptions. Rectangle: Along X direction = 3.8 Along Y direction = 2.5 At the corners of that rectangle along Z direction: corner A = -95 corner B = -150 corner C = -290 corner D = -230. Corner A is at the bottom left of the rectangle. Corners B to D is counter clockwise from corner A. Thanks in advance.
Using the instructions in this video you can find the centroid of that irregular solid. I’m not going to make a another video specifically for this problem
Great question! It’s because those forces pushing against the frame cause it to rotate about point A and the result is an uplift force at B. Since the reaction is in the opposite direction of the force in the member, the reaction is downwards.
Tbh this method was the only method that really helped me understand problems with negative slope. I tried doing the section method calculating the moment and vertical force with respect to x and couldn’t find a solution. With this method it was so straight forward. Thanks for the help.
you are gorgeus sir. my old man professor know anything or he is very selfish and greedy. whatever, It helps when information is shared. No one is trying to reinvent gravity to understand quantum physics. But university professors sometimes have the opposite idea. Thank you for your very clear and unambiguous explanation.
What if the pole can rotate with some friction, how does that change the dynamics? If the friction of the rope is greater than the pole does only the pole friction matter after that?
Great question! I’m not exactly sure but it would definitely matter how easily the pole could rotate. I would think that if the pole could rotate freely then there would be no slipping between the pole and the rope. If the friction preventing the pole from freely rotating was greater than that of the friction between pole and the rope then the pole would probably rotate while the rope was also slipping around it. You would probably need to find the friction force from the rope and see when it matched the friction force of the pole rotating.
Why multiply 450 with 1? I understood the centre of gravity for the right angled is 1/3 its length, but since you took moment at point A, shouldnt it be 450x1/3? Because its 1/3 away from point A?
