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 ?
@emta14834 жыл бұрын
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-k8y9 ай бұрын
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?
@learnfast87063 жыл бұрын
can we solve problem by individual subsystem directly?
@harikrishnan31124 жыл бұрын
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.
@saisriramkadiyala7544 жыл бұрын
Disc 2 weigh 100N which is there to keep the system in equilibrium in vertical position
@shitizgoel50274 жыл бұрын
Balance moment about point B. We get the point on the bottom wall where the normal reaction act on the box.
@pradhumanchahar5 жыл бұрын
The best explanation, thanks a lot
@mohammedmehrajhussaininamd2953 Жыл бұрын
16:45 how do we got the equation FcD/sqrt(2) -100D/sqrt(2)=0 ?
@RashiIshaR Жыл бұрын
Take moment about E
@alokmeshram.17 Жыл бұрын
@@RashiIshaR but in example , moment is taken about b?
@kidinmyheart83183 жыл бұрын
a faculty must be given facility to use a PEN. use of only mouse pointer and slides seems to obstruct freeflow of expression.
@vinayakhuracan51822 жыл бұрын
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-mq5ko5 жыл бұрын
These iit folks are awesome
@shikshaasharma25524 жыл бұрын
How did the equation for moment at B come? How did we do F.250.sin(60-theta) ???
@amitraj62393 жыл бұрын
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 Жыл бұрын
kite ka package laga?
@dheerajbangar7223 Жыл бұрын
@@amitraj6239why you multiplied sin(theta) while calculating moment?
@shabnamhaque20032 жыл бұрын
28:50 crimping problem
@rakeshkarmakar1023 жыл бұрын
Thank you very much sir
@joejose32053 жыл бұрын
crimping tool looks like a crocodile's mouth opening.....lol....
@sumesh66634 жыл бұрын
Thank you sir
@rajkamalaarya752 жыл бұрын
THANKUSIR, THANKU VERY MUCH
@pratikmore65636 ай бұрын
18:43 How we got Fh = 100√2 N?
@ankuuchauhan76895 жыл бұрын
Which book is best for this cource
@hansdaeducation79823 жыл бұрын
f.r steword i think....
@karanbhanushali67563 жыл бұрын
How to find Force on point B along y direction can anyone tell this?
@shashikanthreddy85264 жыл бұрын
How did we get F.250.sin(60-theta) , can anyone explain please ? W.250.sin(theta) is understandable.
@kiranrocks4744 жыл бұрын
Same question
@thinkgrow42963 жыл бұрын
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...
@thinkgrow42963 жыл бұрын
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 ..
@sudhaponram25442 жыл бұрын
@@thinkgrow4296 why do you take sin function this is my only doubt
@thinkgrow42962 жыл бұрын
@@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 Жыл бұрын
at 39:37 how is Fbx=0? when we take moment at C=0
@full-metal_zero0683 Жыл бұрын
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
@pavan43482 жыл бұрын
ANOTHER METHOD TO SOLVE THE PROBLEM BY APPLYING "lami's theorem"
@mdahmed93364 жыл бұрын
27:48 how we got w250sin=f250sin(60-teta)
@abbysaha20914 жыл бұрын
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.
@ATLmarathon79464 жыл бұрын
But ans couldn't come. 300N , what is value of teta?
@emta14834 жыл бұрын
@@ATLmarathon7946 Refer timestamp 27.27 for value of theta
@Cool_as_Sun3 жыл бұрын
@@ATLmarathon7946 Theta=36.87°
@alokmeshram.17 Жыл бұрын
27:29 if anyone doesnt get it @ reply me W 250 sin theta = F 250 Sin (60 - theta)
@guesswho-og2wv Жыл бұрын
This is by equating the moments of force "F" and weight "W" about point "B"
@Thakidathom Жыл бұрын
Can anyone help me solve the nptel quiz 🤕 questions aren't easy
@mdahmed93364 жыл бұрын
25:55 how we got cos 200/250
@abbysaha20914 жыл бұрын
BC is radius thus 250. The base of triangle is 250-50(height of step)
@sarthakchoudhury52373 жыл бұрын
Complex Analysis
@salluriganeshsvg4 жыл бұрын
17:13 how we get FB as 200N
@arkitkabir69504 жыл бұрын
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-xf2pt4 жыл бұрын
Shoulder is not given by a god
@karanvirsingh28303 жыл бұрын
So doctor welded it on your body🤔
@ayanokojiuchiha68775 ай бұрын
@@karanvirsingh2830 wha he mean is It is human evolution from a single cell being
@create.c02 жыл бұрын
19:11
@AnujKumar-bu9fr4 жыл бұрын
How Fdx = 0
@emta14834 жыл бұрын
Fdx is the only force acting on member ABD. Now, invoking ΣFx = 0 Fdx = 0
@AnujKumar-bu9fr4 жыл бұрын
@@emta1483 thanks
@hansdaeducation79823 жыл бұрын
@@AnujKumar-bu9fr since it is not differentiable and hence not continuous ig
@santosh87736 ай бұрын
@@emta1483 but why did he take moment about C?
@nileshrathod31534 жыл бұрын
for member ABD, we dont have Sum of forces in Y direction as zero. Should not this be satisfied?
@muditmishra18114 жыл бұрын
did you get the answer to it?
@muditmishra18114 жыл бұрын
I think after doing mathematics, we will get the direction of Fby and Fcy opposite as shown in the example.
@abbysaha20914 жыл бұрын
Exactly. Plus the weight 200N. Do we have to pass it through cg, thats another question.
@emta14834 жыл бұрын
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.
@emta14834 жыл бұрын
@@abbysaha2091 Which is the problem you are referring to? Can you mention the timestamp for your query?