Confusion with force components

[rudransh verma]

Homework Statement

Three forces causes a net acceleration of $3 \frac {m}{sec^2}$ . Two of the forces $F_1$ and $F_2$ are 10N and 20N. Find the $F_3$ in unit vector notation?
Mass of the body is 2 kg.

Relevant Equations

$\vec F= m\vec a$
$\vec{F_1} +\vec{F_2}+\vec{F_3} = m \vec a$

x component of $F_3$
$F_{3x}= m a_x- F_{1x}-F_{2x}$
= $ma\cos 50-F_1\cos(-150)-F_2\cos90$

y component of $F_3$
$F_{3y}= m a_y-F_{1y}-F_{2y}$
=$ma\sin50-F_1\sin(-150)-F_2\sin90$
And so on…
My question how we can represent it in diagram $F_1\sin(-150)$. I suppose what is $\sin(-150)$ : $\sin(-30)$ and -30 is measured from position X axis. So what is the component now . Is it now in Forth Quadrant from 3rd.
I can understand the component $F_1\sin30$ or $F_1\cos30$ in diagram but the above component, I don’t.
I could have done it via reference angle taking a right triangle in each quadrant and taking acute angles and put a sign accordingly to each component after calculating magnitude of each component. But I choose to do it more systematically by textbook way.
I was told you can do it by unit circle convention. That’s what I am aiming for, for doing all the vector questions. Taking all the angles from + direction of x-axis and straight away putting in $\cos and \sin$. In this way I don’t have to worry about the signs. It is dealt automatically.

 

[Lnewqban]

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html#veccon

http://hyperphysics.phy-astr.gsu.edu/hbase/ttrig.html#c2

 

[Steve4Physics]

I don't understand your difficulty. It might help if you answer these questions:

In which quadrant is each force:
Q1 Force $F_A$ has a positive x-component and a negative y-component
Q2 Force $F_B$ has a negatvei x-component and a negative y-component
Q3 Force $F_C$ has a negative x-component and a positive y-component
Q4 Force $F_D$ has a positive x-component and a positive y-component

What can you say about the signs of the x and y components of these vectors:
Q5 Force $F_E$ points into the 2nd quadrant,
Q6 Force $F_F$ points into the 3rd quadrant?
Q7 Force $F_G$ points into the 4th quadrant

Q8 On a scale of 1-10 (10=best) how well would you say you understand the meaning of a component?
Q9 On a scale of 1-10 (10=best) how well do you understand why sine and cosine are used (and which way round they used) to find components?

 

[rudransh verma]

@Steve4Physics
4,3,2,1
-x,+y
-x,-y
+x,-y
10
10
My question is simple: plot $F_1\sin(-150) , F_1\cos(-150)$ on xy plane.

 

[Steve4Physics]

@Steve4Physics
4,3,2,1
-x,+y
-x,-y
+x,-y
10
10

Good.

My question is simple: plot F1sin(-150) , F1cos(-150) on xy plane.

That is not a question. Maybe you mean ‘How do you plot… on the xy plane?’

Taking F1 = 10N:
F1 has x-component = 10N x cos(-150º) = -8.66 N
F1 has y-component = 10N x sin(-150º) = -5N

With the xy axes marked in units of Newtons, one way to plot these is:
For the x-component, draw an arrow starting at (0,0) and ending at (-8.66, 0)
For the y-component, draw an arrow starting at (0,0) and ending at (0, -5).

Note that if you add these two components (e.g. using the parallelogram method), you get F1 back.

EDIT: Typo's corrected. I got x and y interchanged.

 

[rudransh verma]

Taking F1 = 10N:
F1 has x-component = 10N x cos(-150º) = -8.66 N
F1 has y-component = 10N x sin(-150º) = -5N

Which right triangle did you take who’s sides are -5 and -8.66 ?
Surely there is no right triangle whos one angle is $150^\circ$

 

[Steve4Physics]

Which right triangle did you take who’s sides are -5 and -8.66 ?
Surely there is no right triangle whos one angle is $150^\circ$

When trig’ functions are taught in school at an introductory level, right-triangles are used. For example, in a right triangle, for one of the acute angles $θ$:

$sin(θ) = \frac {length \space of \space opposite \space side}{length \space of \space hypotenuse}$.

$cos(θ) = \frac {length \space of \space adjacent \space side}{length \space of \space hypotenuse}$.

Lengths of the sides of the triangle are always positive values. You can’t have negative lengths.

So using right-triangles to find trig’ functions of larger angles isn’t the best approach (in my opinion). It leads to the problems you are struggling with. But it can be done by allowing $θ$ to be an exterior angle sometimes, and having additional rules to attach signs (+ or -) to the lengths of the 'opposite' and 'adjacent' sides.
_____

Consider a point P moving round the unit circle, centre O. Let Q be the projection of P onto the x-axis. ΔPOQ is right-angled; θ is the angle between OP and the x-axis; θ can be interior or exterior to the triangle. Draw/label it for yourself for various angles.

One way to express the additional rules are:
θ in 1st quadrant: QP is taken as positive; OQ is taken as positive
θ in 2nd quadrant: QP is taken as positive; OQ is taken as negative
You should work out the rules for the 3rd and 4th quadrants for yourself.

After allocating the correct signs, you can use, for example $sin(θ) = \frac {QP}{OP}$.

 

[rudransh verma]

When trig’ functions are taught in school at an introductory level, right-triangles are used. For example, in a right triangle, for one of the acute angles $θ$:

$sin(θ) = \frac {length \space of \space opposite \space side}{length \space of \space hypotenuse}$.

$cos(θ) = \frac {length \space of \space adjacent \space side}{length \space of \space hypotenuse}$.

Lengths of the sides of the triangle are always positive values. You can’t have negative lengths.

So using right-triangles to find trig’ functions of larger angles isn’t the best approach (in my opinion). It leads to the problems you are struggling with. But it can be done by allowing $θ$ to be an exterior angle sometimes, and having additional rules to attach signs (+ or -) to the lengths of the 'opposite' and 'adjacent' sides.
_____

Consider a point P moving round the unit circle, centre O. Let Q be the projection of P onto the x-axis. ΔPOQ is right-angled; θ is the angle between OP and the x-axis; θ can be interior or exterior to the triangle. Draw/label it for yourself for various angles.

One way to express the additional rules are:
θ in 1st quadrant: QP is taken as positive; OQ is taken as positive
θ in 2nd quadrant: QP is taken as positive; OQ is taken as negative
You should work out the rules for the 3rd and 4th quadrants for yourself.

After allocating the correct signs, you can use, for example $sin(θ) = \frac {QP}{OP}$.

I think you are talking about reference angles. Taking right triangles in quadrants and after finding the components attaching +- signs according to which quadrant they are in.

What I want is to be able to solve any problem by just putting the angle given relative to +x axis , not the reference angle and without worrying about the signs, get the right component in any quadrant.

You can see in the question they just put -150 in the eqn without converting it to acute angle and using the right triangle approach.

 

[Steve4Physics]

What I want is to be able to solve any problem by just putting the angle given relative to +x axis , not the reference angle

EDITED

The angle measured anticlockwise relative from the +x axis is the standard position angle. [EDIT - I incorrectly called this the reference angle pre-edit.]

If you are given a different angle, it is trivial to calculate the standard position angle. Then using your calculator will give the correct sign for the trig' function.

 

[rudransh verma]

If you are given a non-reference angle, it is trivial to calculate the reference angle. Then using your calculator will give the correct sign for the trig' function.

I just need one angle that is relative to + x-axis to calculate component in any quadrant. I don’t want to solve by reference angle (which is angle relative to horizontal )method.
So by directly putting -150 like in this example I’ll get the component. But the thing is how do you represent this component in figure?

 

[Steve4Physics]

My apologies. I misused the term ‘reference angle’. I’ll add an edit-correction to my post.

The angle, measured anticlockwise from the +x axis, is called the ‘standard position angle’. I have also seen it called the terminal angle.

It is simple to find the standard position angle from whatever angle you are given. Then use these to find the components.

For example, for F1, the standard position angle is 180º+30º=210º. So using cos(210º) and sin(210º) will give the correct answer for the x and y components. No triangle required.

 

[rudransh verma]

My apologies. I misused the term ‘reference angle’. I’ll add an edit-correction to my post.

You do it often and I start to think where I went wrong.

For example, for F1, the standard position angle is 180º+30º=210º. So using cos(210º) and sin(210º) will give the correct answer for the x and y components. No triangle required.

But how do you know which components they are in figure if no triangle is there?

 

[Steve4Physics]

But the thing is how do you represent this component in figure?

I already explained how to represent the components in Post #5.
Their representations are the same whatever method you use to calculate them,

Another way to think about it is that components are simply projections onto the x and y axes.

 

[rudransh verma]

@Steve4Physics By the way how do you calculate trigonometric ratios of any angle. Do you use calculator ? Is it right?

 

[Steve4Physics]

@Steve4Physics By the way how do you calculate trigonometric ratios of any angle. Do you use calculator ? Is it right?

What method do you use?

 

[Chestermiller]

They asked you to solve this in unit vector notation. Why didn't you do that? In terms of the unit vectors in the x and y directions, what are your equations for the. two unit vectors in the directions of the two known forces?

 

[rudransh verma]

What method do you use?

By identities.

 

[rudransh verma]

They asked you to solve this in unit vector notation. Why didn't you do that? In terms of the unit vectors in the x and y directions, what are your equations for the. two unit vectors in the directions of the two known forces?

I don’t need to.

 

[Steve4Physics]

By identities.

How would you evaluate, for example, sin(47º) using identities?

 

[Chestermiller]

I don’t need to.

Too bad you feel that way. Very arrogant attitude. Based on your struggles with this problem, apparently working with unit vectors would have help. Certainly, when you get to more complicated problems, working with unit vectors helps.

 

[rudransh verma]

How would you evaluate, for example, sin(47º) using identities?

No I mean something like sin150, sin235,cos330.

 

[Steve4Physics]

No I mean something like sin150, sin235,cos330.

EDIT: In Post #14, you specifically stated *any* angle. So you are now asking a different question.

I was made to learn (at school):
1) 'standard' values (which you should be able to work out anyway) for the sine, cosine and tangent of angles 0º, 30º, 45º, 60º and 90º;
2) the graphs of sine, cosine and tangent over -360º to +360º;
3) how to use the above toegether, if needed.

In post #8 you wrote:
"What I want is to be able to solve any problem by just putting the angle given relative to +x axis , not the reference angle and without worrying about the signs, get the right component in any quadrant."

But that's exactly what happens if you use a calculator, inputting the angle (measured anticlockwise) relative to the +x axis. You will get the + or - sign automatically. The sign gives the component's direction: e.g. a negative value for the x-component means the component acts in the -x direction (to the left).

However, without a calculator, you need to do some additional work (e.g. using the 3 points noted above) to get the signs.