The answer has been mistakenly calculated using sin56.3 The correct equation considering ∠JBK should be: ΣFx = 0 +10 + F(BK) + F(BJ) cos56.3 = 0 +10 + F(BK) + 18.02 cos56.3 = 0 F(BK) = -20 kN. In the interest of correctness, you can refer the above mentioned equations. Video editing for the error will be effected before the next course run.
@Rita Singh @AKASH RAUT Your observations are correct. The answer has been mistakenly calculated using sin56.3 The correct equation considering ∠JBK should be: ΣFx = 0 +10 + F(BK) + F(BJ) cos56.3 = 0 +10 + F(BK) + 18.02 cos56.3 = 0 F(BK) = -20 kN. In the interest of correctness, you can refer the above mentioned equations. Video editing for the error will be effected before the next course run.
Your observation is correct. The answer has been mistakenly calculated using sin56.3 The correct equation considering ∠JBK should be: ΣFx = 0 +10 + F(BK) + F(BJ) cos56.3 = 0 +10 + F(BK) + 18.02 cos56.3 = 0 F(BK) = -20 kN. In the interest of correctness, you can refer the above mentioned equations. Video editing for the error will be effected before the next course run.
@@rishabhbiddappa7342 thank you for your reply but i still didn't understand it well.how does the cos component of Fbj acts at a distance 1.5m. can you please help me out ?
A section cutting CD and ED would expose forces F(DC), F(EC), F(FD) and F(EF) in the free body of truss. In case you take moment about C, this would give an equation involving F(FD) and F(EF). Hence, one cannot solve for the force F(FD) at once using this section. Take note that any section selected must cut the truss completely into separate portions. See 4.10
Brilliant teaching methodology!
Confused at 33:09 is there should be sin 56.3 or cos56.3
The answer has been mistakenly calculated using sin56.3
The correct equation considering ∠JBK should be:
ΣFx = 0
+10 + F(BK) + F(BJ) cos56.3 = 0
+10 + F(BK) + 18.02 cos56.3 = 0
F(BK) = -20 kN.
In the interest of correctness, you can refer the above mentioned equations.
Video editing for the error will be effected before the next course run.
at 45:50 in the diagram the force in negative y direction ⬇️ should be FDC not FEF .
Or is it same thing whether we write FDC Or FEF
insane professor ,learning is a joy
33:28 We had to take the cos component of Fbj. Is n't it?
Can some one verify?
i think it is Fbj(cos 56.3)
33:02 I think there is a mistake. It should be Fxcos(56.3) and not Fxsin(56.3)
Right
@@ritasingh7078 exactly
@Rita Singh @AKASH RAUT
Your observations are correct. The answer has been mistakenly calculated using sin56.3
The correct equation considering ∠JBK should be:
ΣFx = 0
+10 + F(BK) + F(BJ) cos56.3 = 0
+10 + F(BK) + 18.02 cos56.3 = 0
F(BK) = -20 kN.
In the interest of correctness, you can refer the above mentioned equations.
Video editing for the error will be effected before the next course run.
Yess i noticed it. 😁
30:40
I didn't get the answer while finding the force Fbj
Can anyone plz tell me
@33.01.... Fbk+Fbj *cos56.31+10=0 gives Fbk=-20 where as the answer given is -25 kN
Your observation is correct. The answer has been mistakenly calculated using sin56.3
The correct equation considering ∠JBK should be:
ΣFx = 0
+10 + F(BK) + F(BJ) cos56.3 = 0
+10 + F(BK) + 18.02 cos56.3 = 0
F(BK) = -20 kN.
In the interest of correctness, you can refer the above mentioned equations.
Video editing for the error will be effected before the next course run.
Iam also got confused bro..thanks for the clarification...
Thanks you sir
Thanks a lot sir !
Hello sir which book is for engineering mechanics.. please tell me..
Very beautifully explained!
Very beauitfully explained!
Thank u sir
Hoping it will helps me sir
Best prof I hv seen yet 🫡
17:49
Sir can u help in solving 2019 gate trusses question in your method
Which method is used to solve gate questions?
@@debjyotidutta8285 generally .. section only..
@@debjyotidutta8285 both depends on question
Thanks sir
30:42 isn't it Fbj cos56.38*1+Fbjsin56.3*1.5?
No u must consider the distance towards the line of action of of the point of moment ur considering
@@rishabhbiddappa7342 thank you for your reply but i still didn't understand it well.how does the cos component of Fbj acts at a distance 1.5m. can you please help me out ?
@@melvindavis3629 cos component should align with the line of action of the point about which u consider the moment
Thanks you. You made my day
@@melvindavis3629 no problem
quite helpful!!
42:38 why can't we solve by section method as CDEF is section &apply moment zero at C to get force FD.
A section cutting CD and ED would expose forces F(DC), F(EC), F(FD) and F(EF) in the free body of truss.
In case you take moment about C, this would give an equation involving F(FD) and F(EF). Hence, one cannot solve for the force F(FD) at once using this section.
Take note that any section selected must cut the truss completely into separate portions. See 4.10
thanks for uploading.