Determine the position x and the tension developed in ABC
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- čas přidán 20. 12. 2016
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Cable ABC has a length of 5 m. Determine the
position x and the tension developed in ABC required for
equilibrium of the 100-kg sack. Neglect the size of the
pulley at B.
This method was better than the ones provided by Chegg. Thanks a lot
your solutions is a million times clearer than the one in the solutions manual. Thank you for making us understand
the way you solved this was so clear and easy, thank you so much!!
Well detailed outline of the solution. Thanks. You made it easy to understand.
Great job . Thank you since Colombia.
you are my hero
GODBLESS YOU ...
Thank you so much 😊
You explained everything beautifully except the ending. Never assume we know where you get values from. Where did you get 3.574 from? Where did you get 2.04 from? Where did you get 2.824 from? Gave me a headache trying to find the origin of those values
Could you do 3-45 it is very similar to this problem but the cord is elastic.
THANKS
thank you !!
You rock
Thnx
Why is alpha the same theta?
which page number is this question from?
please tell me how did you get tan
tan of theta would equal (BD) divided by x, so BD equals xtan(theta) just multiply both sides by x.
or...[ABcos(theta)]+[BCcos(theta)] = 5cos(theta)m => 3.5m = 5cos(theta)
why is tension AB the same as tension BC? I know that its the same string but why ?
just a rule for frictionless pullys.
plz can u tell me how is theta equal to alfa
?????
the cable is not split at the pulley. It is one cable so the tension is the same throughout.
A body with a mass of 250 kg is supported by the flexible
cable system shown in Figure below. Determine the
tensions in cables A, B, C, and D.
I want solution plz 🙏 😩
ily
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