Distance between the force and origin

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Homework Statement


In this note , I was told that either vector OD , OC , AD and AC can be used as the distance between the Force and OA axis ... why ? To produce moment , the distance r must be originate from the origin of force ( point C ) , am i right ? So , IMO, only r AC can be used...

Homework Equations

The Attempt at a Solution

 

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Note the presence of the cross product when calculating the moment. Would any of those choices give a different result?
 
Doc Al said:
Note the presence of the cross product when calculating the moment. Would any of those choices give a different result?
Sorry, I still don't understand. Can you explain further?
 
Realize that when you take the cross product ##\vec{r} \times \vec{F}##, where ##\vec{r}## is a vector starting at pt O and extending anywhere along the line of ##\vec{F}##, that you get the same answer. Essentially you are finding the perpendicular distance between pt O and the line.

In simpler terms: r X F = r sinθ F. Realize that when r changes, θ also changes so that r sinθ remains the same. Play around with this until you get it.
 
Doc Al said:
Realize that when you take the cross product ##\vec{r} \times \vec{F}##, where ##\vec{r}## is a vector starting at pt O and extending anywhere along the line of ##\vec{F}##, that you get the same answer. Essentially you are finding the perpendicular distance between pt O and the line.

In simpler terms: r X F = r sinθ F. Realize that when r changes, θ also changes so that r sinθ remains the same. Play around with this until you get it.
from the figure , it show that the vector r _OD is not along the vector F ...
By saying ##\vec{r}## is a vector starting at pt O and extending anywhere along the line of ##\vec{F}## , do you mean like this ? the r can be any vector ( in different colours ) along the vector F ?
 

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Doc Al said:
Realize that when you take the cross product ##\vec{r} \times \vec{F}##, where ##\vec{r}## is a vector starting at pt O and extending anywhere along the line of ##\vec{F}##, that you get the same answer. Essentially you are finding the perpendicular distance between pt O and the line.

In simpler terms: r X F = r sinθ F. Realize that when r changes, θ also changes so that r sinθ remains the same. Play around with this until you get it.
Since Force at C is moved towards D , why not Vector r_CD is used ?
 
goldfish9776 said:
from the figure , it show that the vector r _OD is not along the vector F ...
By saying ##\vec{r}## is a vector starting at pt O and extending anywhere along the line of ##\vec{F}## , do you mean like this ? the r can be any vector ( in different colours ) along the vector F ?
Yes.
 
goldfish9776 said:
Since Force at C is moved towards D , why not Vector r_CD is used ?
Vector r begins at pt O.
 
Doc Al said:
Vector r begins at pt O.
sorry, I'm still confused now . IMO, vector r _OD is not along the vector F .
 
goldfish9776 said:
sorry, I'm still confused now . IMO, vector r _OD is not along the vector F .
Can you explain further?
 
goldfish9776 said:
sorry, I'm still confused now . IMO, vector r _OD is not along the vector F .
That's true, it is not. Why would you think it would be?

Are you confusing it with vector CD?
goldfish9776 said:
Since Force at C is moved towards D , why not Vector r_CD is used ?
 
Doc Al said:
That's true, it is not. Why would you think it would be?

Are you confusing it with vector CD?
Becoz I was told that the vector r must originate from anywhere along line of action ( force)
 
goldfish9776 said:
Becoz I was told that the vector r must originate from anywhere along line of action ( force)
The vector r originates at point O and terminates anywhere along the line of action of the force.
 
Doc Al said:
The vector r originates at point O and terminates anywhere along the line of action of the force.
the line of action refers to the line along F , right ? Vector CD isn't on this line , so i think vector CD is not suitable ...
 
goldfish9776 said:
the line of action refers to the line along F , right ?
Right.

goldfish9776 said:
Vector CD isn't on this line , so i think vector CD is not suitable ...
Looks to me like F points along the line CD.
 
Doc Al said:
Right.Looks to me like F points along the line CD.
Ya, it points along cd? Why vector cd is not used? But vector od is used?
 
goldfish9776 said:
Ya, it points along cd? Why vector cd is not used? But vector od is used?
You seem to be mixing up ##\vec{r}## with ##\vec{F}##. The vector r must begin at O and end somewhere on the line of action of F. So r could be OD or OC or anything in between, but not CD!
 
Doc Al said:
You seem to be mixing up ##\vec{r}## with ##\vec{F}##. The vector r must begin at O and end somewhere on the line of action of F. So r could be OD or OC or anything in between, but not CD!
oh , yeah . I forgot that F can be longer . It doesn't necessarily ends at point D. that's why I'm confused.
 
Doc Al said:
You seem to be mixing up ##\vec{r}## with ##\vec{F}##. The vector r must begin at O and end somewhere on the line of action of F. So r could be OD or OC or anything in between, but not CD!
if i use MOA = uOA . (rOC X F) , then my ans = 108 , if i use MOA = uOA . (rOD X F) , then the ans = 100 , is it correct ?
 
sorry , i didnt mean to spam , can someone help me with this ?
 
goldfish9776 said:
if i use MOA = uOA . (rOC X F) , then my ans = 108 , if i use MOA = uOA . (rOD X F) , then the ans = 100 , is it correct ?
That cannot be correct. The point is that rOC X F = rOD X F.
 
Doc Al said:
That cannot be correct. The point is that rOC X F = rOD X F.
Which part of my working is wrong?
 
goldfish9776 said:
Which part of my working is wrong?
Show your work. What are rOC, rOD, and F?
 
goldfish9776 said:
OD=0.6i +0.8j
I believe that's the unit vector uOD, not rOD.
 
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Doc Al said:
I believe that's the unit vector uOD, not rOD.
sorry , i mean rOC= 0.4j + 0.5k ( as shown in the working above )
 
goldfish9776 said:
sorry , i mean rOC= 0.4j + 0.5k ( as shown in the working above )
What are the others?
 
Doc Al said:
What are the others?
rOD = 05i +0.5 k , according to transmisbility of forces , the r can be anywhere on the line of action of F ( force ) , am i right ? so , no matter what is the value of r , as long as it has contact with the line of action of F , then the value of MOA should be the same , right ? why the ans for MOA is different if i am using uOA . (rOC x F) ?
to find MOA , i gt my ans = 108 if i am using uOA . (rOC x F)
to find MOA , i gt my ans = 100 if i am using uOA . (rOD x F)
why are they different ?
 
goldfish9776 said:
sorry , i mean rOC= 0.4j + 0.5k
That doesn't look right. What are the coordinates of C?

goldfish9776 said:
rOD = 05i +0.5 k ,
The diagram's a bit hard to read, but that looks right.