Vector-Force image and resolve force

In summary, to resolve non-perpendicular forces, draw perpendicular lines from the tip of the force vector to the two axes and use trigonometry to find the components. The term "image" most likely refers to finding the projections of the force vector on the axes.
  • #1
52
7
Homework Statement
Force image and resolve force
Relevant Equations
vector
The components of the force of 800 lb F along the non-perpendicular axes Get a, b. Also determine the image F on the axes a, b

1677864645339.png


could some one help me how to resolve forces non-perpendicular.And also what is the meaning of image F one the a,b axes
 
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  • #2
sHatDowN said:
Homework Statement:: Force image and resolve force
Relevant Equations:: vector

The components of the force of 800 lb F along the non-perpendicular axes Get a, b. Also determine the image F on the axes a, b

View attachment 323140

could some one help me how to resolve forces non-perpendicular.And also what is the meaning of image F one the a,b axes
You find the components along ##a## and ##b## the usual way. Draw perpendicular lines from the tip of ##F## to the two axes. This will give you two right triangles. Then proceed as usual.

I suspect the "image of ##F##" means find the components of ##-\vec F## along axes ##a## and ##b##. That is consider the reflection of ##\vec F## about a mirror plane that is perpendicular to it. Others may have a different opinion.
 
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  • #3
kuruman said:
Others may have a different opinion
Wouldn't one of the two be such that $$\vec F = F_a\;\hat a + F_b\; \hat b \qquad ? $$

but I would leave it to OP's textbook (or teacher) which is which ... 🤔

##\ ##
 
  • #4
BvU said:
Wouldn't one of the two be such that $$\vec F = F_a\;\hat a + F_b\; \hat b \qquad ? $$

but I would leave it to OP's textbook (or teacher) which is which ... 🤔

##\ ##
Without ever having encountered such a thing before I would be thinking in terms of solving a set of simultaneous equations to find the unique scalars for ##F_a## and ##F_b## which would satisfy that vector equation.

That feels like it ought to be a linear mapping between the components of the vector expressed in some original coordinate system and the components of the same vector expressed in the new coordinate system.

Google says that "image" is the result of applying a linear map to a vector, so this interpretation kinda sorta feels right.
 
  • #5
kuruman said:
You find the components along and the usual way. Draw perpendicular lines from the tip of to the two axes. This will give you two right triangles. Then proceed as usual.
Thanks sir,
I Find F to a =692.82
and F to b =772.74

is that correct?
kuruman said:
I suspect the "image of " means find the components of along axes and . That is consider the reflection of about a mirror plane that is perpendicular to it. Others may have a different opinion.
I have same opinion but still i dont know what should i do.
 
  • #6
sHatDowN said:
Is that correct?
It doesn't look correct. Please show how you got these numbers and include your drawings.
 
  • #7
kuruman said:
It doesn't look correct. Please show how you got these numbers and include your drawings.
1677869686220.png
 
  • #8
The vector should be the hypotenuse. Try to draw this more to scale. Your 75° angle looks like a right angle. Your trig ratios are incorrect. You need to review them.
 
  • #9
If you want the component of F in the direction of a, you want the portion of F that is parallel to a, not the portion of F that is perpendicular to a.
 
  • #10
jbriggs444 said:
If you want the component of F in the direction of a, you want the portion of F that is parallel to a, not the portion of F that is perpendicular to a.
Then its 400
is that correct?
 
  • #11
A few thoughts…

The required components of ##\vec F## are ##\vec a## and ##\vec b##. This means ##\vec a + \vec b = \vec F##.

(The question requires only the magnitudes of the components, i.e. ##a## and ##b##.)

To find the components, a simple approach is to draw the vector-addition triangle (or parallelogram). This gives a triangle with all three angles known and one side of known length. (The triangle is not a right-triangle.) The rest is simple trig’.

For the question about the ‘image’ I’d guess:
“… determine the image F on the axes a, b”
is very badly/incorrectly written. The intended meaning is possibly:
“… determine the projections of ##\vec F## on the axes of ##\vec a## and ##\vec b##”

The reason for the above interpretation is that the question could be about making the student aware of the differences between perpendicular and non-perpendicular components of a vector.
 
  • #12
Steve4Physics said:
A few thoughts…

The required components of ##\vec F## are ##\vec a## and ##\vec b##. This means ##\vec a + \vec b = \vec F##.
No, that's only true for perpendicular components. E.g. consider ##\vec a, \vec b## collinear.
 
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  • #13
kuruman said:
You find the components along ##a## and ##b## the usual way. Draw perpendicular lines from the tip of ##F## to the two axes. This will give you two right triangles. Then proceed as usual.

I suspect the "image of ##F##" means find the components of ##-\vec F## along axes ##a## and ##b##. That is consider the reflection of ##\vec F## about a mirror plane that is perpendicular to it. Others may have a different opinion.
No, the image of ##F## on the axis ##a## just means the component of ##F## in the direction ##a##.
"Get a, b" sounds like it is asking for a vector representation of those, and that is reinforced in my mind by the overall skewed depiction. But there is no specified coordinate system.
@sHatDowN , are you sure you have quoted the whole question exactly?
 
  • #14
@sHatDowN What is the source of this question? From where did you get the image you posted in the OP?
 
  • #15
nasu said:
@sHatDowN What is the source of this question? From where did you get the image you posted in the OP?
i find out in internet about 2 week ago
 
  • #16
haruspex said:
@sHatDowN , are you sure you have quoted the whole question exactly?
yes its what question said
 
  • #17
haruspex said:
No, the image of ##F## on the axis ##a## just means the component of ##F## in the direction ##a##.
"Get a, b" sounds like it is asking for a vector representation of those, and that is reinforced in my mind by the overall skewed depiction. But there is no specified coordinate system.
@sHatDowN , are you sure you have quoted the whole question exactly?
1677875453253.png

and its B part of question
and AO is cable which is 700N
Calculate the components of this force at A located on the beam along the t, n axes.
still i have confusion:(
 
  • #18
sHatDowN said:
its B part of question
Ok, but I struggle to relate the two parts. Are you sure you quoted part B in post #1 word for word? "The components of the force of 800 lb F along the non-perpendicular axes Get a, b." doesn’t read like that to me.
Any chance of providing an image of the whole question?
sHatDowN said:
AO is cable which is 700N
You mean AB.
 
  • #19
haruspex said:
You mean AB.
no AO

haruspex said:
Ok, but I struggle to relate the two parts. Are you sure you quoted part B in post #1 word for word? "The components of the force of 800 lb F along the non-perpendicular axes Get a, b." doesn’t read like that to me.
Any chance of providing an image of the whole question?
i past exactly what is the question
 
  • #20
haruspex said:
You mean AB.
the AB cable prevents the rotation of the beam OA clockwise around O. If the tension created in the cable is equal to 700 newtons, calculate the components of this force at A located on the beam along the t, n axes.

and this is what is the B part
 
  • #21
haruspex said:
No, that's only true for perpendicular components. E.g. consider ##\vec a, \vec b## collinear.
If we resolve a vector ##\vec F## into two components ##\vec a## and ##\vec b##, then by definition ##\vec a + \vec b = \vec F##. The directions of the components do not have to be orthogonal (though for conveience they often are).

In general, components can not be collinear as they can not then span the required space.
 
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  • #22
sHatDowN said:
the AB cable prevents the rotation of the beam OA clockwise around O. If the tension created in the cable is equal to 700 newtons, calculate the components of this force at A located on the beam along the t, n axes.

and this is what is the B part
So how did you arrive at the diagram in post #1? Is that provided with the question or did you draw it? Where does the angle of 45° come from?
In post #1 you wrote "how to resolve forces non-perpendicular", but ##t, n## are perpendicular.

It really would help enormously if you would post the entire question in a single image.
sHatDowN said:
AO is cable
The cable is AB. AO is a beam, not a cable.
 
  • #23
haruspex said:
So how did you arrive at the diagram in post #1? Is that provided with the question or did you draw it? Where does the angle of 45° come from?
In post #1 you wrote "how to resolve forces non-perpendicular", but ##t, n## are perpendicular.

It really would help enormously if you would post the entire question in a single image.

The cable is AB. AO is a beam, not a cable.
1677878150397.png
 
  • #24
Steve4Physics said:
If we resolve a vector ##\vec F## into two components ##\vec a## and ##\vec b##, then by definition ##\vec a + \vec b = \vec F##. The directions of the components do not have to be orthogonal (though for conveience they often are).

In general, components can not be collinear as they can not then span the required space.
Sorry, you are right.
 
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  • #25
sHatDowN said:
Ok, thanks, that explains it. I read your posts as saying the parts were in the other order, but I see now it was ambiguous.

Also, "get a, b" is not in the actual question. That clears up the puzzle that there is no provided coordinate system.

Neither does it say anything about resolving into components. It only asks you to find components in two given directions.
Various responses in this thread have been correct in themselves but addressing different tasks. I'll try to clarify.

You can find the component in the direction of ##\vec a## as @kuruman described in post #2. Independently, you can find the component in the direction of ##\vec b## by the same method. But those two will not add up to equal the original vector, so it does not constitute resolving into components. You added that word in post #1, which has created all the confusion.

You can resolve the force into components in the directions of non perpendicular ##\vec a, \vec b##, but it requires a different method. You cannot do it independently for the two directions. I don’t think this is what the question is asking for.

As an aside, resolving into non-perpendicular components drops out nicely in three dimensions because you get to use the cross product. If ##\vec v=\alpha_1\vec x_1+\alpha_2\vec x_2+\alpha_3\vec x_3## then ##\alpha_1=\frac{[\vec x_2, \vec x_3, \vec v]}{[\vec x_1, \vec x_2, \vec x_3]}##, etc., where ##[\vec a, \vec b, \vec c]=\vec a\cdot(\vec b\times\vec c)##, the triple scalar product.
 
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  • #26
haruspex said:
Ok, thanks, that explains it. I read your posts as saying the parts were in the other order, but I see now it was ambiguous.

Also, "get a, b" is not in the actual question. That clears up the puzzle that there is no provided coordinate system.

Neither does it say anything about resolving into components. It only asks you to find components in two given directions.
Various responses in this thread have been correct in themselves but addressing different tasks. I'll try to clarify.

You can find the component in the direction of ##\vec a## as @kuruman described in post #2. Independently, you can find the component in the direction of ##\vec b## by the same method. But those two will not add up to equal the original vector, so it does not constitute resolving into components. You added that word in post #1, which has created all the confusion.

You can resolve the force into components in the directions of non perpendicular ##\vec a, \vec b##, but it requires a different method. You cannot do it independently for the two directions. I don’t think this is what the question is asking for.
Thanks sir, then what your opinion about B part
 
  • #27
sHatDowN said:
what your opinion about B part
Since the only connection with part A is subject matter, it belongs in a separate thread.
But the first step is to draw a free body diagram for the beam. Note all the forces acting on it and consider what equations are implied by equilibrium.

Edit: can you do part A now?
 
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  • #28
Steve4Physics said:
The required components of ##\vec F## are ##\vec a## and ##\vec b##. This means ##\vec a + \vec b = \vec F##.

(The question requires only the magnitudes of the components, i.e. ##a## and ##b##.)

To find the components, a simple approach is to draw the vector-addition triangle (or parallelogram). This gives a triangle with all three angles known and one side of known length. (The triangle is not a right-triangle.) The rest is simple trig’.

For the question about the ‘image’ I’d guess:
“… determine the image F on the axes a, b”
is very badly/incorrectly written. The intended meaning is possibly:
“… determine the projections of ##\vec F## on the axes of ##\vec a## and ##\vec b##”

The reason for the above interpretation is that the question could be about making the student aware of the differences between perpendicular and non-perpendicular components of a vector.
Actually, both pairs are technically called "components". The ones obtained by projection on the two axes are the contarvariant components and the other ones are the covariant components of the vector. Only for orthogonal axes they are the same. But we are so used to the orthogonal case that is even hard to find a definition of "components" without reference to orthogonal axes, at least in the context of basic physics problems. This is a problem where we are forced to think about the difference about the two types of "projections".
 

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