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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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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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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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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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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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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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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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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