Trigonometric Solution for Resultant Vector Magnitude and Direction

[pennywise1234]

Homework Statement

.
Using a trigonometric solution, find the magnitude and direction of the resultant vector from the following
forces 6.5m/s at 45°, 9m/s at 180°, 14km/hr at 205°, and 8.7m/s at 305° The vector directions are referenced to a co-ordinate system where 0° represents the right horizontal and positive angles are in a counterclockwise direction.

Homework Equations

how to solve properly?
when drawing out the horizontal and vertical lines if it points left or down do you let your final answer be negative?

The Attempt at a Solution

i converted 14 km/hr into m/s which is 3.88

i got all my vertical and horizontal numbers and added them up. did the Pythagorean theorem but my answer is off. so I was wondering if it how my triangles were drawn. i.e my 305 degree triangle had horizontal moving left and vertical moving up, leaving my final vertical number to be positive and horizontal to be negative? correct.

 

[TSny]

You referred to your vectors as force vectors, but I guess you meant velocity vectors.

my 305 degree triangle had horizontal moving left and vertical moving up

No, those directions are not correct. What quadrant are you in when you have an angle of 305^o?

If you post the details of your work, then we can pinpoint specific errors.

 

[pennywise1234]

i have my 45 degree and 6m/s triangle in the upper right quadrant with the vertical moving down and horizontal to the left
giving me a vertical answer of 4.6m/s and horizontal at -4.6m/s

i have the 205 degree in the left lower quadrant and 3.89 m/s triangle (the one that need conversion to km/hr) with the vertical down and horizontal to the left giving me a vertical of -1.64 and horizontal of -3.53

my 305 degree triangle is in the bottom right quadrant with a vertical of - 7.12 and horizontal of -4.99

and there is no vertical for the 180 degree at 9m/s so the horizontal is at 9m/s

i know the answer is wrong looking at the answer key, but where did i go wrong?

 

[SammyS]

i have my 45 degree and 6m/s triangle in the upper right quadrant with the vertical moving down and horizontal to the left
giving me a vertical answer of 4.6m/s and horizontal at -4.6m/s

i have the 205 degree in the left lower quadrant and 3.89 m/s triangle (the one that need conversion to km/hr) with the vertical down and horizontal to the left giving me a vertical of -1.64 and horizontal of -3.53

my 305 degree triangle is in the bottom right quadrant with a vertical of - 7.12 and horizontal of -4.99

and there is no vertical for the 180 degree at 9m/s so the horizontal is at 9m/s

i know the answer is wrong looking at the answer key, but where did i go wrong?

Isn't right horizontal a positive number ?

 

[TSny]

You have the correct quadrants, but it appears that you are drawing the directions for each vector in the opposite direction that it should be. If you draw each vector at the origin, then each vector should point away from the origin. Thus, a vector in the first quadrant should point up and toward the right. The horizontal and vertical components for a vector in the first quadrant will be positive.

 

[pennywise1234]

so, you have to draw each vector away from the origin? i.e the vector with the 8.7 m/s at 305 degrees would have a left horizontal and down vertical when drawing it? so you do not have to draw with the "head to tail" rule?

 

[TSny]

When you add the vectors, you draw them head to tail. However, you first need to get the direction of each vector correct. A vector that points at 305^o will point down and to the right. If you were to place the tail of the vector at the origin, the vector would point away from the origin in a direction down and to the right.

 

[pennywise1234]

ok makes senses, thank you. Just to make sure I comprehend fully when trying to get the vertical and horizontal components of the 45 m/s at 305 degrees did you get a vertical of -7.12 and horizontal of +4.99 ?

 

[TSny]

Yes. (I actually get -7.13 m/s for the vertical component. But you are only working with 2 significant figures. So, it won't make any difference when you get your final answer to 2 significant figures.)

 

[pennywise1234]

thank you, so i have my new vertical -4.16 and horizontal is -2.94 i used Pythagorean theorem to get the reluctant which as 5.10 m/s (the right answer). I need to get the new angle, i used inverse trig to get the answer tan=-4.16/-2.94 and my answer is way off the from the answer key (234.86 degrees), any reason for this ?

 

[TSny]

Your answers look good. When you use your calculator to do an arctan operation, your answer will not necessarily be in the correct quadrant. Your calculator is programmed to give you only the "principal values" of the arctan function. These principal values lie between -90^o and +90^o. That is, it only gives you angles in the first and fourth quadrants. So, if your vector actually lies in the second or third quadrant, the calculator is not giving you the angle you want. Can you see how to modify the answer that your calculator gives so that it represents the correct angle?

The result of adding vectors is usually referred to as the "resultant" rather than the "reluctant".

 

[pennywise1234]

no i cannot see it, so there is no other way to get the answer (234.86 degrees) ? the way i was taught to do (in lecture) was inverse trig. Do you know of another way ?

 

[CWatters]

Make a sketch showing x and y axis, the horizontal and vertical components and resultant vector.

Hint: You might find it easier to calculate the required angle as the sum of two angles.

 

[pennywise1234]

i drew out the diagram i was just not able to get the angle that was similar to the answer key

 

[TSny]

Consider $θ_1 = \tan^{-1} \left( \frac{-4}{-3} \right)$ and $θ_2 = \tan^{-1}\left( \frac{4}{3} \right)$. Can you see why your calculator gives you the same answer for both $θ_1$ and $θ_2$?

$θ_1$ is the angle for a vector $\vec{A}$ with x component = -3 and y component = -4.
$θ_2$ is the angle for a vector $\vec{B}$ with x component = +3 and y component = +4.

Graph these two vectors to see how their directions are related.

Likewise, you can try the example $θ_1 = \tan^{-1} \left( \frac{-4}{+3} \right)$ and $θ_2 = \tan^{-1}\left( \frac{+4}{-3} \right)$.

 

[CWatters]

i drew out the diagram i was just not able to get the angle that was similar to the answer key

Post your diagram.