Engineering Mechanics_Forces on a Plane_Level 2_Problem 1
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- เผยแพร่เมื่อ 26 ก.ย. 2024
- Download the Manas Patnaik app now: cwcll.on-app.i... Problem Description: A prismatic bar AB of length l = 5m and of negligible weight is hinged at A and supported at B by a string that passes over a pulley C. A vertical load of 60 N applied at the end B of the bar is supported by a force P applied to the string. Find the axial forces in the bar and the limiting value of tension T when the bar approaches vertical position. Take distance between hinge and pulley h = 6m.
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Sir why didn't you resolve the force P acting at point C?
The same force T is transferred into point P by the pulley, since it is an ideal pulley. So no need to consider the force P. Thank me later
@@muhzinfirozam559 great justification bro
@@muhzinfirozam559 thnks
sir i think that the solution is wrong ucannot directly predict that the net hinge force is directly acting along
ab pls reply
yes I also have the same question. you cannot take SAB as theta inclined to the horizontal.
@@janakamohotti sab is not a resultant its force in bar
Dear sir why didn't consider Ha & Va at hinged support.
yes I also have the same question. you cannot take SAB as theta inclined to the horizontal.
Why the hinge force doesn't change in vertical condition ?
Sir how did you take hinge force constant
You are doing very good work, may you reach 1 million soon
Sir what about the force P at Point C? U didn't resolved it?
Kindly reply
That is a ideal pulley a T is same in all string
10:15 Can't we also consider that limiting tension case if the bar rotates anticlockwise till it's vertical?
Sir ,is it necessary that at the hinge we take only single force along AB axis, can we take individual horizontal and vertical forces?
Just try to free the bar AB from the support at A. It will experience a compressive force at that same angle theta. Just think about it...
@@ManasPatnaikofficial But can we either take the individual horizontal and vertical forces in the place you took that Sab force sir?
@@ManasPatnaikofficial then why in previous questions u took as ha and Va ??
@@ManasPatnaikofficial I could think that if we remove support from A.. the resultant of P and W would be somewhat directed towards A but can't figure out that it will align with AB.
yes I also have the same question. you cannot take SAB as theta inclined to the horizontal.
sir plz gv me some solution in related to slender in rigd body..i dnt knw u teach mechanics also..i see frm coment section..i rly so hpy for u ......sir u god gifted..tq
Splendid teaching Manas bhai
Sir why P force was not considered anywhere in the soln?
It is similar.as T
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Sir solution of limiting value of tension , I don't understand plz explain this
gaurav...watch it again brother...
How can i explain that in words....
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Sir why didn't u resolve tension at b????
Thankyou 🤗
All the best Sitaram......
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ty sir
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The force p =100 N