Static Equilibrium - Tension, Torque, Lever, Beam, & Ladder Problem - Physics

[@TheOrganicChemistryTutor]

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As an engineering student, subjects like Physics, Calculus, and Chemistry are some of the important subjects in this course. But not all students can immediately understand what the professor teaches in only one sitting. This video gives a great introduction to static equilibrium as this topic is very important for engineers. It is essential since it is the engineer's responsibility to develop and construct a structure that is stable and does not fall, slide, or tilt. Static equilibrium is not an easy lesson and there are many formulas that need to be remembered but the lecturer of this KZbin channel gave an easy-to-understand explanation of the topic so that students could understand it right away. He also gave real-world practice problems to further deepen the students' understanding of static equilibrium. Furthermore, the ability to rewind a tutorial video on social media will be very helpful to learners if they are having trouble understanding a part of the lesson they are studying. The video is a guide for students to clearly understand those complicated topics. However, it is crucial that students have their notes from discussions with their professors that could be a big additional help upon reviewing the topics that had been tackled. #Surviving #TheFutureEngineers #NextGenerationEngineers

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55:14 could be my calc setting but when you simply use the triangle to determine the angle it does not match the value you got using the force component - sin 0 = 8/10 which is 53.13🤔

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24:33 We know that: Sum of forces = 0 Sum of clockwise tension= sum of anti-clockwise tension T sin 30 * 2m = 200kg(9.8)(2m) + 30kg(9.8)(1m) Tension = 4214 N

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Thank you so much. But, at 44:47 The force exerted by the wall would be horizontal only if the wall was perfectly smooth, right? If the wall had a friction like the ground, then that force would also have a vertical component, as I see. Sorry if I'm being too nerdy.🤓☺

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I want to clarify to the problem of a Beam and 20kg block. Isn't Normal Force supposed to be added to the sum of the Forces in the Y Direction?

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Hey, Hey, Chief! 1 Question: If you say that the ground force is exerting a force along the ladder's length, and decomposes into FGy and FGx, it's also true that the wall ALSO exerts a force onto the ladder and should be decomposed into FWx and FWy!!!

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I think so too :/ I was literally scrolling through the comments to search if someone had addressed this note and if they got a reply explaining why this isn't the case.

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At 55:15, is the angle measure between the forces really supposed to be different to the angle measure between the ladder and the ground? Because if you take the length of the ladder (hyp = 10) and the height (opp = 8), then you'd get theta=arcsin(8/10)=53.13deg

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For the question that has the uniform beam supported by two legs with a beam mass of 100kg and a 20kg box resting on top of it, Why is it that the normal force acting on the 20kg box that is created by the surface of the beam and the box not considered when equating forces? Video- 19mins- 24mins.

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At 16:35 sin 50 = 0.766 by adding sin 40 and 50 we get value of 1.409 and if we divide 784/1.409=556

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Hi! This is in relation to the last problem. Should we really use half of 12 (the horizontal distance) as the lever arm of the beam? What I initially thought was to use half of the length of the beam itself, using the center of gravity of the beam, which in this case is 15/2=7.5. Anyways, this is just to clarify everything! :) Thank you so much for your help! You definitely help me survive all my college classes Y_Y

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On the Problem starting at 18:10 why isn't the Force Normal F(n) for the 20kg accounted for. Wouldn't that affect the answer when putting Net Fy=0. Doesn't the 100kg block exert a force normal (Fn) on the 20kg Block?

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