Dynamics: Lesson 15 - Drawing Kinetic Diagrams, The Quintessential Dynamics Problem
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- Опубліковано 13 лют 2021
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DR. Hanson ,thank for another great lecture on Drawing Kinetic Diagrams. These problems are not too difficult to solve in Mechanics II.
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The Return of the King
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Prof you forget the negative sign -1.504 T
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You have noticed it well! Nevertheless, the solution got worked out correctly when the Proff shifted the negative number to the left: Over all, Dr Hanson is a great teacher and I've been benefited a lot through his video lectures.😇
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Three years up.... But question, If a=a and there's no such thing as negative mass, How come MAsub(b) is negative? You mentioned it was due to the sign convention, but does it really count there overall?
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Let’s go ‘freedom units’!!!
To save time, from two equations T - f = mA a and T - wB = - mB a, isolate a => (T - f ) / mA = ( - T + wB )/ mB => ( T - 3 ) / ( - T + 8 ) = m A / m B = wA / w B = 3 / 2 => 2 ( T - 3 ) = 3 ( - T + 8 ) ==> T = 6 lbs
Thank youu
why is the weight not multiplied by the acceleration due to gravity?
3 months late here, but the weight is equal to mass times gravity (W=m*g). Acceleration and gravity are the same thing so when you have a weight, it already accounts for the gravity/acceleration. Be careful not to mix up weight and mass. :)
Why isnt it -m(b)a(b)?
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