Vectors Directions: Where is this Resultant Vector Pointing?

Ok. My problem is what angle to choose when adding vector. Statement does not tell me which one is the "first" force vector. So, when using the law of sine formula I get two results.

First, using cosine to get the magnitude:
$$\vec c = \sqrt{a^2 + b^2 +2ab\cos\theta},$$
$$\vec c = \sqrt{15^2 + 35^2 +2(15(35) \cdot \cos(40)},$$
$$\vec c = 47~\text{lbs}$$

NOW; how and what angle do I choose for the resultant? (I know the problem is not asking for it but I want to practice). Picture of it attached.

An angle depends on a reference direction. Like 15 degrees above the horizontal. You need to choose a suitable reference in your case.

PeroK said:

An angle depends on a reference direction. Like 15 degrees above the horizontal. You need to choose a suitable reference in your case.

So, for this problem is difficult to say? Does it need to specify which force is the first one? I, most of the time, choose the ##x\text{-axis}## as a reference but that too gave me 12 degrees and 28 degrees.

Remle said:

So, for this problem is difficult to say? Does it need to specify which force is the first one? I, most of the time, choose the ##x\text{-axis}## as a reference but that too gave me 12 degrees and 28 degrees.

It can't be both. The order you add vectors doesn't matter.

The question only asks for the magnitude of the resultant vector. What angle are you calculating?

PeroK said:

The question only asks for the magnitude of the resultant vector. What angle are you calculating?

I know is asking only for the magnitude of the vector. I believe the question is, which vector connects head-to-tail to whom to get the angle from the ##x\text{-axis}##?

The angle between two vectors is always the smaller angle that you get when you draw the arrows representing them with their tails together (see diagram drawn to scale.) It is less than or equal to 180°.

kuruman said:

The angle between two vectors is always the smaller angle that you get when you draw the arrows representing them with their tails together (see diagram drawn to scale.) It is less than or equal to 180°.

View attachment 341067

Using your diagram and ##x\text{-axis}## for the reference angle I get 28 degrees. Am I right?

Remle said:

Using your diagram and ##x\text{-axis}## for the reference angle I get 28 degrees. Am I right?

If you have two vectors, then you can calculate the resultant angle relative to either. I suspect that's what you've done in this case.

For example, if you put the longer vector along the x-axis, then the resulting angle satisfies:
$$\tan \theta =\frac{15\sin(40)}{35+15\cos(40)}$$And, if you put the shorter vector along the x-axis, then you can swap the 15 and 35 in that calculation.

Remle said:

Using your diagram and ##x\text{-axis}## for the reference angle I get 28 degrees. Am I right?

You don't show your work, so I cannot tell if you are right. I prefer to add the vectors by the component method.
##\mathbf{A}=(15,0)##
##\mathbf{B}=[35\cos(40^{\circ}),35\sin(40^{\circ})]##
##\mathbf{A}+\mathbf{B}=[15+35\cos(40^{\circ}),0+35\sin(40^{\circ})].##
Then the magnitude of the resultant is
##\vert \mathbf{A}+\mathbf{B}\vert =\sqrt{\left[15+35\cos(40^{\circ})\right]^2+\left[35\sin(40^{\circ})\right]^2}##
and the tangent of the angle between the resultant and the x-axis is the ratio of the resultant's y-component to the x-component
##\tan\theta=\dfrac{35\sin(40^{\circ})}{15+35\cos(40^{\circ})}## which is, of course, what @PeroK said you would get "if you put the shorter vector along the x-axis."

(Edited to insert missing parentheses.)

@kuruman @PeroK
Yeah, sorry for not showing work. Here it is:
If ##\text{vector-15}## is along the ##x\text{-axis}##, then:
$$\theta = \sin^{-1} {(\frac {15 \sin{140}} {47})}$$
$$\theta = \text{28°}$$

If ##\text{vector-35}## is along the ##x\text{-axis}##, then:
$$\theta = \sin^{-1} {(\frac {35 \sin{140}} {47})}$$
$$\theta = \text{12°}$$

And 28 + 12 = 40, which is the angle between the original vectors. So, the resultant vector is 12 degrees from the longer vector and 28 degrees from the shorter vector.

Note that ##\frac{28}{12} = \frac{35}{15}##.

PeroK said:

Note that ##\frac{28}{12} = \frac{35}{15}##.

Just in case the OP mistakenly thinks the above is a general rule, it’s worth adding this.

If we limit rounding errors in intermediate steps (e.g. use the more precise value that resultant = 47.48 lbf, not 47 lbf) then the 2 angles turn out to be about 28.3º and 11.7º.

##\frac{28.3}{11.7} \ne \frac{35}{15}##.

(Rounding the angles to 28º and 12º when giving ‘final answers’ is OK of course.)

However, the inequality is an approximate equality here. That's a result of the small angle approximation '##\sin \theta \approx \theta##' being valid within 10% even for angles up 0.75rad (43º).

A general message to the OP is to work to sufficient precision in intermediate steps. E.g. stating that the resultant is 47 lbf is OK; but in actual calculations 47.48 lbf should be used.

Edit - minor rewording.

Yes, it was something of a coincidence! In general, it's the ratio of sines that equals the ratio of the sides. In this case:
$$\frac{\alpha}{\beta} \approx \frac{\sin \alpha}{\sin \beta} = \frac A B $$