# Thread: Vectors F and G

1. ## Vectors F and G

The vectors F and G represent two forces acting on an object as indicated by the attached picture. Compute, to two decimal places, the magnitude and direction of the resultant. Give the direction of the resultant by specifying the angle theta between vector F and the resultant.

Note: Alpha = 60°.

2. ## Re: Vectors F and G

Originally Posted by nycmath
The vectors F and G represent two forces acting on an object as indicated by the attached picture. Compute, to two decimal places, the magnitude and direction of the resultant. Give the direction of the resultant by specifying the angle theta between vector F and the resultant.

Note: Alpha = 60°.
this one is pretty basic.

Let $\vec{G}=(10,0)$ then $F=(8\cos(60\deg),8\sin(60\deg))=(4,4\sqrt{3})$

$F+G=(14,4\sqrt{3})$

the length is given by $|F+G|=\sqrt{14^2+(4\sqrt{3})^2}=\sqrt{244}=2\sqrt {61}$

$\theta=\arctan\left(\dfrac{4\sqrt{3}}{14}\right)= 26.3\deg$

3. ## Re: Vectors F and G

Originally Posted by nycmath
The vectors F and G represent two forces acting on an object as indicated by the attached picture. Compute, to two decimal places, the magnitude and direction of the resultant. Give the direction of the resultant by specifying the angle theta between vector F and the resultant.

Note: Alpha = 60°.
Good morning!

1. Use the Cosine rule to determine the magnitude of the resultant force R. Use the angle and the lengthes of the indicated triangle:

$\vec R^2 = \vec F^2 + \vec G^2 - 2 \cdot |\vec F| \cdot | \vec G | \cdot \cos(180^\circ - \alpha)$

2. To determine the value of $\theta$ use the Cosine rule again:

$\cos(\theta) = \frac{\vec G^2 - \vec R^2 - \vec F^2}{-2 \cdot |\vec R| \cdot |\vec F|}$

4. ## Re: Vectors F and G

Thank you so much.

5. ## Re: Vectors F and G

Originally Posted by romsek
this one is pretty basic.

Let $\vec{G}=(10,0)$ then $F=(8\cos(60\deg),8\sin(60\deg))=(4,4\sqrt{3})$

$F+G=(14,4\sqrt{3})$

the length is given by $|F+G|=\sqrt{14^2+(4\sqrt{3})^2}=\sqrt{244}=2\sqrt {61}$

$\theta=\arctan\left(\dfrac{4\sqrt{3}}{14}\right)= 26.3\deg$

I applied your steps to the follow question and was able to determine the magnitude but my answer for theta was not correct.

|F| = 5N, |G| = 4N and alpha = 80°.

I got |F+G| = 6.92 N.

For theta, I got 45.29° but the correct answer in the textbook is 34.67°.
Can you tell me how to get the correct answer for theta?

6. ## Re: Vectors F and G

I should warn you that finding the angle $\theta$ isn't always quite so straightforward. You have to examine the signs of the x and y components of the vector and determine what quadrant it lies in. Then you can adjust the arctan result as necessary to place it in the correct quadrant.

For example suppose $\vec{v}=(-1,-1)$

$\arctan\left(\dfrac{-1}{-1}\right)=\arctan(\dfrac{1}{1})=\dfrac{\pi}{4}$

but the angle of $\vec{v}$ is actually $\dfrac{5\pi}{4}\text{ or }-\dfrac{3\pi}{4}$

7. ## Re: Vectors F and G

Originally Posted by earboth
Good morning!

1. Use the Cosine rule to determine the magnitude of the resultant force R. Use the angle and the lengthes of the indicated triangle:

$\vec R^2 = \vec F^2 + \vec G^2 - 2 \cdot |\vec F| \cdot | \vec G | \cdot \cos(180^\circ - \alpha)$

2. To determine the value of $\theta$ use the Cosine rule again:

$\cos(\theta) = \frac{\vec G^2 - \vec R^2 - \vec F^2}{-2 \cdot |\vec R| \cdot |\vec F|}$

I love your picture replies. I have seen your work and love the geometry as part of each reply.

8. ## Re: Vectors F and G

Originally Posted by nycmath
I applied your steps to the follow question and was able to determine the magnitude but my answer for theta was not correct.

|F| = 5N, |G| = 4N and alpha = 80°.

I got |F+G| = 6.92 N.

For theta, I got 45.29° but the correct answer in the textbook is 34.67°.
Can you tell me how to get the correct answer for theta?
I get the same answer you do.

9. ## Re: Vectors F and G

Thank you, romsek. By the way, what is the meaning of your username? Look for one or two more questions later in the precalculus and calculus forums. Having so much fun learning math with you and the other tutors.

10. ## Re: Vectors F and G

Originally Posted by nycmath
Thank you, romsek. By the way, what is the meaning of your username? Look for one or two more questions later in the precalculus and calculus forums. Having so much fun learning math with you and the other tutors.
It means about as close to nothing as is possible. It originated from a randomly generated selection of pre-approved names for a game I used to play.

11. ## Re: Vectors F and G

Hi,
Here's another slightly different way to solve the problem: