Something that I have come to think of when looking for the direction of moment in each component is thinking of each force as it moves along its own line of action, and then considering how I would have to 'steer' the arrow head in order to reach the point that I are trying to find the moment around. This has worked well for me in 2D problems, I am not sure how it would hold up in 3D problems.
+Rozalind Aria Since the horizontal component of the first two line up perfectly with point A... as in they pass through it, there is no moment created by that component of the force. So you could break them up into components but you'd find the horizontal components come out to 0. The second question: The triangle is used to resolve that force into components. If you're trying to find the Y component, the leg of the triangle parallel to the Y axis is used. Sin = y/r cos = x/r Same thing with the X component of that force using the triangle. The part that is parallel is 3, so the X component of that force would be 3/5 x 500.
Did you understand the answer by Rusty. Another smart guy who is thinking 2 steps ahead of us. 1.) Your question is valid. You can break the first forces(F1 and F2) into it's component. When you do that you will get F1(horiz) x 0 meters = 0 and F2(horiz) x 0 meters = 0. So basically, do the extra work to write it out completely and you can't. Later it becomes instinctive. 2.) Work out the trigonometry of F3 directions. In fact, find the angle between the 4 and the 5 legs. What do you get? Then review your trigonometry again. You know opp/hyp and adj/hyp. What do you see. You will see that just dividing by the legs "is in fact" the same thing as finding the angle and then using sin and cosines. Except, you do 1 or 2 steps more. You see dividing by the legs(if you have all 3 defined) is exactly the same thing as if you had the angle. In some cases, it's faster and easier to check back your written arithmetic.
Don't know if you're still active but, why is the (500 x 3/4) force taken as 4 meteres. I thought a moment was M=Fd(perpedicular) where d is distance from pivot point A?
alternatively you can use -2 x 250sin120 if that helps you visualize it better. a lot of physics people use trig angle identities from the unit circle and it confuses the hell out of people (including myself)
So you use the X plane distance in the moment multiplication for the y force and the Y plane distance from the point for the multiplication of the force for the X direction. It's confusing.
I am really confused. Who were telling you the truth? On the other video I watched, he said that the counter clockwise was the negative and the clockwise was the positive.
Force × its perpendicular distance . For Fx, since it is a horizontal force its perpendicular distance from point A should be the vertical length which is 4m in the given example.
Did you understand Victor's answer above? He's obviously thinking two steps ahead of us. Basically, let's consider "both components". F1(x-dir) and F1(y-dir). The moment at A of force F1 is the following: F1(y-dir) x 2 meters= F1(cos30) x 2meters F1(x-dir) x 0 meters = F1(sin30) x 0 meters = 0 I know for some students it's a no brainer, but I can understand people getting confused by the rules/conventions. You must be absolutely 100% certain/confident of the rules to apply vector statics. If you try to mix some logic you will quickly get lost
neeeegaat iam sorry positive
Something that I have come to think of when looking for the direction of moment in each component is thinking of each force as it moves along its own line of action, and then considering how I would have to 'steer' the arrow head in order to reach the point that I are trying to find the moment around. This has worked well for me in 2D problems, I am not sure how it would hold up in 3D problems.
Legend.
Thank you! This makes the components make more sense when talking about moments.
A number of questions, if I may.
1) Why didn't you break the first two into the vertial and horizontal?
2) Why is it -5(500 4/5)?
+Rozalind Aria Since the horizontal component of the first two line up perfectly with point A... as in they pass through it, there is no moment created by that component of the force. So you could break them up into components but you'd find the horizontal components come out to 0. The second question: The triangle is used to resolve that force into components. If you're trying to find the Y component, the leg of the triangle parallel to the Y axis is used. Sin = y/r cos = x/r Same thing with the X component of that force using the triangle. The part that is parallel is 3, so the X component of that force would be 3/5 x 500.
Did you understand the answer by Rusty. Another smart guy who is thinking 2 steps ahead of us.
1.) Your question is valid. You can break the first forces(F1 and F2) into it's component. When you do that you will get F1(horiz) x 0 meters = 0 and F2(horiz) x 0 meters = 0. So basically, do the extra work to write it out completely and you can't. Later it becomes instinctive.
2.) Work out the trigonometry of F3 directions. In fact, find the angle between the 4 and the 5 legs. What do you get? Then review your trigonometry again. You know opp/hyp and adj/hyp. What do you see. You will see that just dividing by the legs "is in fact" the same thing as finding the angle and then using sin and cosines. Except, you do 1 or 2 steps more. You see dividing by the legs(if you have all 3 defined) is exactly the same thing as if you had the angle. In some cases, it's faster and easier to check back your written arithmetic.
lol i was looking for 2D from the Gorillaz I clicked because I thought 2D was teaching Statics XD
same lmao
Better than my teacher... 😂Thank you
how did you get cosine and sine for the first 2 forces?
Nice video, Mark Holdhusen.
Don't know if you're still active but, why is the (500 x 3/4) force taken as 4 meteres. I thought a moment was M=Fd(perpedicular) where d is distance from pivot point A?
Why are you using cosine for the first one?
The angle is formed at the vertical instead of the horizontal. So, the force in the y direction is a measurement of the side adjacent to the angle.
alternatively you can use -2 x 250sin120 if that helps you visualize it better. a lot of physics people use trig angle identities from the unit circle and it confuses the hell out of people (including myself)
A wonderful video, sir if you please upload some videos of dynamics , i will be very thankful
Where r u from?
So you use the X plane distance in the moment multiplication for the y force and the Y plane distance from the point for the multiplication of the force for the X direction. It's confusing.
thanks for your effort fully understand the Moment
Understand it clearly! Thanks man!
how does a force going to the right produce a CCW rotation about a?
CW = negative
CCW = positive
COMponent
sir - you are the best on this topic
why is the horizontal 4/5 shouldn't it be 3/5? and the vertical 4/5
Because 4/5 is perpendicular to the horizontal distance. 😁
i thought Fy should be sin not cos?
I am really confused. Who were telling you the truth? On the other video I watched, he said that the counter clockwise was the negative and the clockwise was the positive.
in moment_A1 it should be a 250 sin 30
because its respect to y axis
clockwise should be positive?
Jhen Dela Rosa no
yes, the vid was wrong
Clockwise is negative, counter clockwise is positive. Seems backwards but that’s the way it is
I thought counter clockwise is the same as anticlockwise
Why are the 2 and 3 negative?
thank u sir for excelent presentation
THANK YOU!
well explained thank you man :)
Should the y component of F1 be 250sin(30) ?
nope because in the first one, Fy should be 250×cos30 according to the Pythagoras
why is Fx component 4 meters? i don´t quite get that
Force × its perpendicular distance . For Fx, since it is a horizontal force its perpendicular distance from point A should be the vertical length which is 4m in the given example.
thanks post more videos please
Why is the first one not sin 30?????
If you take sin30, you will be finding the Fx which is not perpendicular to the distance from point A.
Did you understand Victor's answer above? He's obviously thinking two steps ahead of us.
Basically, let's consider "both components". F1(x-dir) and F1(y-dir). The moment at A of force F1 is the following:
F1(y-dir) x 2 meters= F1(cos30) x 2meters
F1(x-dir) x 0 meters = F1(sin30) x 0 meters = 0
I know for some students it's a no brainer, but I can understand people getting confused by the rules/conventions.
You must be absolutely 100% certain/confident of the rules to apply vector statics. If you try to mix some logic you will quickly get lost
trexinvert yea I agree I approached in a logical manner I remember this problem haha
you can use sin but you will use 60 instead of 30.
haha
cos 30 is cos 30 radians - silly not writing the degree symbol. Feeble students make this mistake.
but you get the same answer with or without it when you pput it in calculator
I got a different answer 2349.05Nm
Why are you using cosine for the first one?
Because that creates the vertical force on the body. The horisontal force (sin 30) goes right theough A and is therefore 0.