Lec 06 Equilibrium of Rigid bodies II

[@Shivai1215]

This professor is insane.😄👌

[@Chitrathvidhyarthi]

learning is a joy

[@jeetaaj3438]

I wish all nptel video are like this

[@rajatkumar-tt4dw]

For problem 1 , which force we have to find such that we can analyse that box is in equilibrium? Only three forces will be there as reaction forces. Horizontal force are balanced but what about vertical force balance Fb ?

[@emta1483]

This is an interesting question. Our day-to-day observation suggests that the box will be in equilibrium. However, when we apply the three reaction force-pairs from the discs on the free body of the box, it seems to have an unbalanced vertical force. Therefore, Newton's law dictates it must accelerate which does not happen. This begs the question: is the free body diagram of the box complete then? Consider the following hint and rethink the problem: In reality, the box along with the discs is not suspended midway in the air.

[@Rajendran-k8y]

24:55 Sir, since C is a pin joint reaction forces will be acting on the wheel right? But we haven't considered that in the video. Why sir?

[@learnfast8706]

can we solve problem by individual subsystem directly?

[@harikrishnan3112]

Hi sir. In the first example, there is a reaction force of 200 N at point B and the weight 100 N of the bottom ball is acting along this line of action, but another 100 N force is required to keep the box in vertical equilibrium. At what point of the box this force will be acted upon.

[@saisriramkadiyala754]

Disc 2 weigh 100N which is there to keep the system in equilibrium in vertical position

[@shitizgoel5027]

Balance moment about point B. We get the point on the bottom wall where the normal reaction act on the box.

[@pradhumanchahar]

The best explanation, thanks a lot

[@mohammedmehrajhussaininamd2953]

16:45 how do we got the equation FcD/sqrt(2) -100D/sqrt(2)=0 ?

[@RashiIshaR]

Take moment about E

[@alokmeshram.17]

@@RashiIshaR but in example , moment is taken about b?

[@kidinmyheart8318]

a faculty must be given facility to use a PEN. use of only mouse pointer and slides seems to obstruct freeflow of expression.

[@vinayakhuracan5182]

In crimping tool problem can we consider horizontal forces are zero as no movement in horizontal as it allows to rotate about an axis instead of a three force member ?

[@ShubhamSharma-mq5ko]

These iit folks are awesome

[@shikshaasharma2552]

How did the equation for moment at B come? How did we do F.250.sin(60-theta) ???

[@amitraj6239]

As Radius is given 250mm..First moment of W w.r.t point B will be W.250sin36.87° in clockwise and moment of force F w.r.t point B will be F.250sin(60°-36.87°) in anticlockwise direction..here (60°-36.87°) angle I have taken because angle of F with horizontal is given 30° and angle of W with horizontal is 90° then angle at C become 90-30=60° so angle between Fb and F become 60°-37.87°

[@kritharthbaishnab]

kite ka package laga?

[@dheerajbangar7223]

​@@amitraj6239why you multiplied sin(theta) while calculating moment?

[@shabnamhaque2003]

28:50 crimping problem

[@rakeshkarmakar102]

Thank you very much sir

[@joejose3205]

crimping tool looks like a crocodile's mouth opening.....lol....

[@sumesh6663]

Thank you sir

[@rajkamalaarya75]

THANKUSIR, THANKU VERY MUCH

[@pratikmore6563]

18:43 How we got Fh = 100√2 N?

[@ankuuchauhan7689]

Which book is best for this cource

[@hansdaeducation7982]

f.r steword i think....

[@karanbhanushali6756]

How to find Force on point B along y direction can anyone tell this?

[@shashikanthreddy8526]

How did we get F.250.sin(60-theta) , can anyone explain please ? W.250.sin(theta) is understandable.

[@kiranrocks474]

Same question

[@thinkgrow4296]

since W is acting on the right hand side so W * 250 sin(x) ok !! Now think about Force 'F', which obviously acts on the left hand side of point B... so the transfer F towards that right angle and think about the perpendicular distance it makes with the point B. That will not be due to full 60 degress but only by the angle ( 60- Theta).. so it will be equal to 250*sin(60-Theta)*F ... I have some limitation to explain that in KZbin... since I cant name the points and figure here.. hope you understand it...

[@thinkgrow4296]

If you are not comfortable with my first explanation there is a second method to it .... Let the force F be acting at an angle 30 degree Horizontal line .... its components will be F cos 30 ( horzntl) and F sin 30 downwards along with 'W' take one by one first take the vertical forces and its moment about point B, (Fsin 30 + 200)* 250 sin 36.87 (Clockwise) ---(1) there will be a horizontal component since we resolved F unlike the first method, so F cos 30 * Vertical distance to B now that distance we knew before as 250 mm - 50mm = 200 mm so it will be like F cos 30 * 200 (Anti Clockwise) --- (2) Now equate (1) and (2) (Fsin 30 + 200)*250 sin 36.87 = F cos30*200 by solving you will get the value of F =305.483 N aprx =305.5 N Now I hope you got it clearly.... enjoy learning ..

[@sudhaponram2544]

@@thinkgrow4296 why do you take sin function this is my only doubt

[@thinkgrow4296]

@@sudhaponram2544 That is basic mathematics, it is a prerequisite thing before you join an undergraduate program...I can't be more better than what I have explained above....Sorry friend...

[@Christyv8680]

at 39:37 how is Fbx=0? when we take moment at C=0

[@full-metal_zero0683]

MOMENT AT C = 0 BOTH THE VERTICAL FORCES ( AND THE HORIZONTAL FORCE AT "C" ) PASS THROUGH THE POINT C ONLY SO THEY CAN.T PRODUCE MOMENT, IT THE HORIZONTAL FORCE IS NON ZERO THEN IT WILL MAKE A MOMENT WHICH CANNOT BE CANCELLED BY ANYTHING ELSE

[@pavan4348]

ANOTHER METHOD TO SOLVE THE PROBLEM BY APPLYING "lami's theorem"

[@mdahmed9336]

27:48 how we got w250sin=f250sin(60-teta)

[@abbysaha2091]

Since AB would be 250 sin theta. And if we take CF which is 180 deg. Subtracting 30 deg and 90 deg gives us 60 deg. Thus remaining angle after excluding theta would be 60-theta.

[@ATLmarathon7946]

But ans couldn't come. 300N , what is value of teta?

[@emta1483]

@@ATLmarathon7946 Refer timestamp 27.27 for value of theta

[@Cool_as_Sun]

@@ATLmarathon7946 Theta=36.87°

[@alokmeshram.17]

27:29 if anyone doesnt get it @ reply me W 250 sin theta = F 250 Sin (60 - theta)

[@guesswho-og2wv]

This is by equating the moments of force "F" and weight "W" about point "B"

[@Thakidathom]

Can anyone help me solve the nptel quiz 🤕 questions aren't easy

[@mdahmed9336]

25:55 how we got cos 200/250

[@abbysaha2091]

BC is radius thus 250. The base of triangle is 250-50(height of step)

[@sarthakchoudhury5237]

Complex Analysis

[@salluriganeshsvg]

17:13 how we get FB as 200N

[@arkitkabir6950]

Net vertical forces = 0 (since we assumed body is in equilibrium) Forces acting vertically are weight and normal/reaction at B hence, weight of both disc together = normal force/ reaction force at b

[@SaurabhSingh-xf2pt]

Shoulder is not given by a god

[@karanvirsingh2830]

So doctor welded it on your body🤔

[@ayanokojiuchiha6877]

​@@karanvirsingh2830 wha he mean is It is human evolution from a single cell being

[@create.c0]

19:11

[@AnujKumar-bu9fr]

How Fdx = 0

[@emta1483]

Fdx is the only force acting on member ABD. Now, invoking ΣFx = 0 Fdx = 0

[@AnujKumar-bu9fr]

@@emta1483 thanks

[@hansdaeducation7982]

@@AnujKumar-bu9fr since it is not differentiable and hence not continuous ig

[@santosh8773]

@@emta1483 but why did he take moment about C?

[@nileshrathod3153]

for member ABD, we dont have Sum of forces in Y direction as zero. Should not this be satisfied?

[@muditmishra1811]

did you get the answer to it?

[@muditmishra1811]

I think after doing mathematics, we will get the direction of Fby and Fcy opposite as shown in the example.

[@abbysaha2091]

Exactly. Plus the weight 200N. Do we have to pass it through cg, thats another question.

[@emta1483]

Considering that you are referring to the crimping tool problem: Sum of forces in the y direction must always add to zero as long as the member is in equilibrium. In fact, it is only with this conclusion that we will be able to relate the different forces. Therefore, as @Mudit Mishra pointed out, the different forces acting on the member will mobilize in such a way that the net force is zero. This might even result in the reversal of our initially assumed force direction. Hope this helps.

[@emta1483]

@@abbysaha2091 Which is the problem you are referring to? Can you mention the timestamp for your query?