Determine the moment of each of the three forces about point B. Please see attached figure.

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Determine the moment of each of the three forces about point B. Please see attached figure.

Physics
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1 Attachment
I know how to find it about point A, but I am having difficulty seeing how to find it for point B.
Seperately Solve the hinge forces...:)

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I don't understand what you mean.
He means, solve for the force at each of the points where there is an angle and then sum them to see where the net force is
You guys have missed the question. What is being asked is what the moment due to each force. Can you help the questioner?
I'll do it then. What is the definition of the moment of a force?
The moment of a force is its ability to create rotation.
Yes, but mathematically and precisely, with a formula?
M = F*d, where d is the distance between the point of application and the pivot
or M = r x F
(btw thanks for your help, I have been lost for hours)
Yes. Now, let's start with F3. What's the moment due to that force?
..the moment of F3 about B? Well, what's the distance of the application of the force from B? Zero. Can F3 make something rotate *around* B? No. Therefore the moment due to F3 is zero.
ah!! that's what I thought! but I thought it would be wrong bcs it was 0
Now, what about F2?
bcs the position vector at B is <0,0,0> and crossing it with the F3 vector would give zero.
yes
But one question, would not the perpendicular component of F3 try to produce some moiton that vertical bar?
The question at a qualitative level is this: can a force applied at a point cause other things to rotate about that point? And the answer is no. It might make that point and the solid attached to it move, but it doesn't cause rotation around that point per se. Hence: A force applied to a point causes no rotation about that point.
ahh i see
its the same as if I go to the center of the bolt and try to make it rotate applying a force right on the center of rotations. no rotation would happen!
yes
I have to move at least an infinitesimally small amount away from that center of rotation for it to actually rotate (no matter how large the force). Motion may occur in the form of displacement, but no actual rotation about that point :D
yes
awesome.
for f2. I think it goes like this:
the position vector would be : <-3, 4 , 0>
No, you want the positive vector RELATIVE TO B
Which will be just r = <0,4,0>
hmmm would not that be moving to the right of B and up from B?
oh yes! sorry I was doing F1 mistakenly !
oh right. yes, F2. Doing them backwards because F1 is the most complicated.
r = <0, 4 0 > ; F2 = < -300*sin(60) , -300 *cos(60) , 0 >
check your sin and cos there
ah, i see that I did. I was putting it on the wrong side
< -300*cos(60) , -300 *sin(60) , 0 >
Yes. Ok, you've got it. I'll leave you with F1 now by yourself.
I have dyslexia and its hard to see things sometimes! Thanks for your patience
for F1, r = < -3 , 4 , 0 >
yes
F1 = < 250*sin(30) , -250*cos(3), 0 >
r x F1 = 149.519
Looks right (30 obviously). Now calculate the cross products
r2 x f2 = 600

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