Took me a minute to figure this one out as well. Here's what I came up with. The line showing the 3,4,5 triangle is a reference line and is not actually the force. We know the force is perpendicular to this line, we can rotate this 3,4,5 reference triangle 90°. When you do this simply re-draw the 3,4,5 triangle ON the Force line NA. Now you can do your calc for x-direction and y-direction as normal. Hope this helps!
Where point A is located, how do you determine the direction of the vector components when the resultant line is drawn? does it matter what direction you pick the Ax and Ay to go in if given no guidance or line of reference in a problem? For example, I had a problem similar to this one except that I did not have that line next to A with the 3-4-5 triangle so I had to draw my own x and y component but was not sure if it matters if I draw it to the left or to the right (up or down) and how to know when Ax is left or right etc.
If you know that the moment at A is 0, why do you have to solve for Nb using other equations? Why can't you solve for Nb using the first equation if the moment at A is 0?
@@yousefhammad9858 yeah so I think I know why. It’s the SUM of the moments at A = 0, that doesn’t necessarily mean that the induced moment at A = 0 from the collar.
@europeankid98 You're correct. The sum of the moments at A = 0. M_A is NOT the sum of all moments at A, it's a certain moment around A. That's why you still have to include M_A in the sum of all moments at A.
you can set either clockwise or counterclockwise motion as positive or negative, as long as your convention remains the same throughout the entire problem (i.e. if you decided clockwise is positive and get a positive answer, then that answer is clockwise)
@@farisoqdeh6228 It does kinda. I asked a fellow student today about it as well and it is understandable, yet I am still missing something about it for it to feel intuitave, but I will keep digging and continue to meditate on what you and others have said. Thanks responding
~~And actually the more I think about it, the more it makes sense. I get that M_a overall (net) is zero, but M_a also has a moment. I think I'm understanding. Thanks again.
First of all I appreciate you and thanks you for making these videos.But unfortunately Chapter 5 and after that is not posted . I recommend you to make new videos of chapter 5 full videos and also other chapters Thank you
A question, why if you chose your point of reference in B did not cancel your moment in B? because if you choose your point of reference in A, the moment in A disappears, and so in vice versa
Why is NB in the X direction positive? It's counterclockwise according to the right hand rule
Oh, I didn't realize we still used the moment about A from the collar even when we calculate all the moments relative to A. Thanks!
Taking statics over the summer. This helps a ton!
Thank you.
Thank you! Studying for a test and this helps walk me through it again!
How exactly did you convert the triangular components 4, 3, 5 at A @ 0:39
Took me a minute to figure this one out as well. Here's what I came up with. The line showing the 3,4,5 triangle is a reference line and is not actually the force. We know the force is perpendicular to this line, we can rotate this 3,4,5 reference triangle 90°. When you do this simply re-draw the 3,4,5 triangle ON the Force line NA. Now you can do your calc for x-direction and y-direction as normal. Hope this helps!
Where point A is located, how do you determine the direction of the vector components when the resultant line is drawn? does it matter what direction you pick the Ax and Ay to go in if given no guidance or line of reference in a problem? For example, I had a problem similar to this one except that I did not have that line next to A with the 3-4-5 triangle so I had to draw my own x and y component but was not sure if it matters if I draw it to the left or to the right (up or down) and how to know when Ax is left or right etc.
If you know that the moment at A is 0, why do you have to solve for Nb using other equations? Why can't you solve for Nb using the first equation if the moment at A is 0?
@@yousefhammad9858 yeah so I think I know why. It’s the SUM of the moments at A = 0, that doesn’t necessarily mean that the induced moment at A = 0 from the collar.
@europeankid98 You're correct. The sum of the moments at A = 0. M_A is NOT the sum of all moments at A, it's a certain moment around A. That's why you still have to include M_A in the sum of all moments at A.
at 2:21 why dont we multiply 200 by 6 ft?
It is already a moment, so its distance was already accounted for.
@@stormtroop3016 Thanks bruh i figured it out a long time ago.I am now an Engineer😎
@@brianramz6681 good stuff. Hope I join you one day.
Thank you so so much i finally understand why all my answers were wrong.... i kept forgetting the inversion for the triangle
Thank you very much, you always save my life
since the 200lb is turning clockwise, shouldn't it be a NEGATIVE and not a positive value?
you can set either clockwise or counterclockwise motion as positive or negative, as long as your convention remains the same throughout the entire problem (i.e. if you decided clockwise is positive and get a positive answer, then that answer is clockwise)
Why did u include MA in the equation?
That's my question too.
@@observer4322 Bcz MA is not 0 but the net moment MA is zero so there should be a moment MA to balance the equation.
My comment is 11 months old, hope that helps!
@@farisoqdeh6228 It does kinda. I asked a fellow student today about it as well and it is understandable, yet I am still missing something about it for it to feel intuitave, but I will keep digging and continue to meditate on what you and others have said. Thanks responding
~~And actually the more I think about it, the more it makes sense. I get that M_a overall (net) is zero, but M_a also has a moment. I think I'm understanding. Thanks again.
not the hero we want, but the hero we need!
why dont we take the normal for a like we did for b for the moment?
cant the collar roate about A? why do we moment about A?
First of all I appreciate you and thanks you for making these videos.But unfortunately Chapter 5 and after that is not posted . I recommend you to make new videos of chapter 5 full videos and also other chapters Thank you
why is clockwise positive?
Thank you so much for this!
thanks a lot man
Thank you
A question, why if you chose your point of reference in B did not cancel your moment in B? because if you choose your point of reference in A, the moment in A disappears, and so in vice versa
there is no moment created at b because it is a smooth support. At point A there is a collar support which creates a moment.
he teaches this one but it is so hard to understand. he's good at solving but weak jn teaching
This one:
i.imgur.com/Wb8uB04.png
thank you