Distance between the force and origin

In summary, the conversation discusses the use of different vectors for calculating the moment in a given scenario. It is clarified that the vector r must begin at point O and end somewhere on the line of action of the force, which can be OD, OC, or any point in between. The confusion was caused by mixing up the vectors r and F, and it is pointed out that the line of action of F does not necessarily end at point D. The conversation also mentions different equations for calculating the moment, with different results depending on which vector is used for r.
  • #1
goldfish9776
310
1

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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  • #2
Note the presence of the cross product when calculating the moment. Would any of those choices give a different result?
 
  • #3
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?
 
  • #4
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.
 
  • #5
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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  • #6
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 ?
 
  • #7
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.
 
  • #8
goldfish9776 said:
Since Force at C is moved towards D , why not Vector r_CD is used ?
Vector r begins at pt O.
 
  • #9
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 .
 
  • #10
goldfish9776 said:
sorry, I'm still confused now . IMO, vector r _OD is not along the vector F .
Can you explain further?
 
  • #11
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 ?
 
  • #12
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)
 
  • #13
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.
 
  • #14
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 ...
 
  • #15
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.
 
  • #16
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?
 
  • #17
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!
 
  • #18
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.
 
  • #19
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 ?
 
  • #20
sorry , i didnt mean to spam , can someone help me with this ?
 
  • #21
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.
 
  • #22
Doc Al said:
That cannot be correct. The point is that rOC X F = rOD X F.
Which part of my working is wrong?
 
  • #23
goldfish9776 said:
Which part of my working is wrong?
Show your work. What are rOC, rOD, and F?
 
  • #24
OD=0.6i +0.8j
 
  • #25
uploadfromtaptalk1445003234597.jpg
 
  • #26
goldfish9776 said:
OD=0.6i +0.8j
I believe that's the unit vector uOD, not rOD.
 
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  • #27
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 )
 
  • #28
goldfish9776 said:
sorry , i mean rOC= 0.4j + 0.5k ( as shown in the working above )
What are the others?
 
  • #29
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 ?
 
  • #30
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.
 
  • #31
Doc Al said:
That doesn't look right. What are the coordinates of C?The diagram's a bit hard to read, but that looks right.
0.4j +0.3k
 
  • #32
goldfish9776 said:
0.4j +0.3k
Looks to me like you're missing the i component of that vector.
 

FAQ: Distance between the force and origin

1. What is the distance between the force and origin?

The distance between the force and origin is the length of the straight line connecting the two points. It is measured in units such as meters or feet.

2. How is the distance between the force and origin calculated?

The distance between the force and origin can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse (the longest side) of a right triangle is equal to the sum of the squares of the other two sides. In this case, the hypotenuse is the distance between the force and origin, and the other two sides represent the x and y coordinates of the points.

3. What is the significance of the distance between the force and origin in physics?

The distance between the force and origin is an important concept in physics as it helps determine the magnitude and direction of a force. It also plays a role in calculating work, energy, and torque.

4. Can the distance between the force and origin be negative?

No, the distance between the force and origin cannot be negative. It is always a positive value as it represents the length of the line connecting the two points.

5. How does the distance between the force and origin affect the strength of the force?

The distance between the force and origin is inversely proportional to the strength of the force. This means that as the distance increases, the force decreases, and vice versa. This relationship is described by the inverse square law, which is a fundamental principle in physics.

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