anonymous
  • anonymous
Determine the moment of each of the three forces about point B. Please see attached figure.
Physics
katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
I know how to find it about point A, but I am having difficulty seeing how to find it for point B.
anonymous
  • anonymous
Seperately Solve the hinge forces...:)

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anonymous
  • anonymous
I don't understand what you mean.
anonymous
  • anonymous
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
JamesJ
  • JamesJ
You guys have missed the question. What is being asked is what the moment due to each force. Can you help the questioner?
JamesJ
  • JamesJ
I'll do it then. What is the definition of the moment of a force?
anonymous
  • anonymous
The moment of a force is its ability to create rotation.
JamesJ
  • JamesJ
Yes, but mathematically and precisely, with a formula?
anonymous
  • anonymous
M = F*d, where d is the distance between the point of application and the pivot
anonymous
  • anonymous
or M = r x F
anonymous
  • anonymous
(btw thanks for your help, I have been lost for hours)
JamesJ
  • JamesJ
Yes. Now, let's start with F3. What's the moment due to that force?
JamesJ
  • JamesJ
..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.
anonymous
  • anonymous
ah!! that's what I thought! but I thought it would be wrong bcs it was 0
JamesJ
  • JamesJ
Now, what about F2?
anonymous
  • anonymous
bcs the position vector at B is <0,0,0> and crossing it with the F3 vector would give zero.
JamesJ
  • JamesJ
yes
anonymous
  • anonymous
But one question, would not the perpendicular component of F3 try to produce some moiton that vertical bar?
JamesJ
  • JamesJ
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.
anonymous
  • anonymous
ahh i see
anonymous
  • anonymous
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!
JamesJ
  • JamesJ
yes
anonymous
  • anonymous
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
JamesJ
  • JamesJ
yes
anonymous
  • anonymous
awesome.
anonymous
  • anonymous
for f2. I think it goes like this:
anonymous
  • anonymous
the position vector would be : <-3, 4 , 0>
JamesJ
  • JamesJ
No, you want the positive vector RELATIVE TO B
JamesJ
  • JamesJ
Which will be just r = <0,4,0>
anonymous
  • anonymous
hmmm would not that be moving to the right of B and up from B?
anonymous
  • anonymous
oh yes! sorry I was doing F1 mistakenly !
JamesJ
  • JamesJ
oh right. yes, F2. Doing them backwards because F1 is the most complicated.
anonymous
  • anonymous
r = <0, 4 0 > ; F2 = < -300*sin(60) , -300 *cos(60) , 0 >
JamesJ
  • JamesJ
check your sin and cos there
anonymous
  • anonymous
ah, i see that I did. I was putting it on the wrong side
anonymous
  • anonymous
< -300*cos(60) , -300 *sin(60) , 0 >
JamesJ
  • JamesJ
Yes. Ok, you've got it. I'll leave you with F1 now by yourself.
anonymous
  • anonymous
I have dyslexia and its hard to see things sometimes! Thanks for your patience
anonymous
  • anonymous
for F1, r = < -3 , 4 , 0 >
JamesJ
  • JamesJ
yes
anonymous
  • anonymous
F1 = < 250*sin(30) , -250*cos(3), 0 >
anonymous
  • anonymous
r x F1 = 149.519
JamesJ
  • JamesJ
Looks right (30 obviously). Now calculate the cross products
anonymous
  • anonymous
r2 x f2 = 600

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