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

- anonymous

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

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- anonymous

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- anonymous

I know how to find it about point A, but I am having difficulty seeing how to find it for point B.

- anonymous

Seperately Solve the hinge forces...:)

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## More answers

- anonymous

I don't understand what you mean.

- 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

You guys have missed the question. What is being asked is what the moment due to each force.
Can you help the questioner?

- JamesJ

I'll do it then.
What is the definition of the moment of a force?

- anonymous

The moment of a force is its ability to create rotation.

- JamesJ

Yes, but mathematically and precisely, with a formula?

- anonymous

M = F*d, where d is the distance between the point of application and the pivot

- anonymous

or M = r x F

- anonymous

(btw thanks for your help, I have been lost for hours)

- JamesJ

Yes. Now, let's start with F3. What's the moment due to that force?

- 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

ah!! that's what I thought! but I thought it would be wrong bcs it was 0

- JamesJ

Now, what about F2?

- anonymous

bcs the position vector at B is <0,0,0> and crossing it with the F3 vector would give zero.

- JamesJ

yes

- anonymous

But one question, would not the perpendicular component of F3 try to produce some moiton that vertical bar?

- 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

ahh i see

- 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

yes

- 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

yes

- anonymous

awesome.

- anonymous

for f2. I think it goes like this:

- anonymous

the position vector would be : <-3, 4 , 0>

- JamesJ

No, you want the positive vector RELATIVE TO B

- JamesJ

Which will be just r = <0,4,0>

- anonymous

hmmm would not that be moving to the right of B and up from B?

- anonymous

oh yes! sorry I was doing F1 mistakenly !

- JamesJ

oh right. yes, F2. Doing them backwards because F1 is the most complicated.

- anonymous

r = <0, 4 0 > ; F2 = < -300*sin(60) , -300 *cos(60) , 0 >

- JamesJ

check your sin and cos there

- anonymous

ah, i see that I did. I was putting it on the wrong side

- anonymous

< -300*cos(60) , -300 *sin(60) , 0 >

- JamesJ

Yes.
Ok, you've got it. I'll leave you with F1 now by yourself.

- anonymous

I have dyslexia and its hard to see things sometimes! Thanks for your patience

- anonymous

for F1, r = < -3 , 4 , 0 >

- JamesJ

yes

- anonymous

F1 = < 250*sin(30) , -250*cos(3), 0 >

- anonymous

r x F1 = 149.519

- JamesJ

Looks right (30 obviously). Now calculate the cross products

- anonymous

r2 x f2 = 600

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