Magnitude and angle of vector sum

[ineedhelpnow]

help!
find the magnitude of the resultant force and the angle it makes with the positive x-axis.

View attachment 2971

i don't have any examples in my book like this one

 

[I like Serena]

help!
find the magnitude of the resultant force and the angle it makes with the positive x-axis.

i don't have any examples in my book like this one

Can you find the x-component and the y-component of each force?
And add them separately?
And find the magnitude of the result?
And the angle?

 

[ineedhelpnow]

the part I am having trouble with is the first part. finding the components. if you can help with that then i might be able to figure out the rest on my own. I've literally been staring at this with no clue for like 3 days.

 

[I like Serena]

the part I am having trouble with is the first part. finding the components. if you can help with that then i might be able to figure out the rest on my own. I've literally been staring at this with no clue for like 3 days.

Well, the 200 N force has an x-component of $200\text{ N}$ times $\cos 60^\circ$...

 

[Dethrone]

If you notice, the second vector can be broken down into its x and y components. Drawing a right angle triangle and applying "SOH CAH TOA", we can determine the forces acting in those directions.

 

[ineedhelpnow]

so I am guessing it has a y component of 200*sin(60)

and is it the same for 300 N only negative? x=-300*cos(60)) and same for y only sin?

 

[I like Serena]

so I am guessing it has a y component of 200*sin(60)

Yep.

and is it the same for 300 N only negative? x=-300*cos(60)) and same for y only sin?

The 300 N is aligned with the x-axis, so it has an $x=-300\text{ N}$...

 

[ineedhelpnow]

i don't really understand the second part for the 300 N. so what's y in that case

 

[Dethrone]

http://www.physics.fsu.edu/users/ng/Courses/phy2053c/Hw/Ch03/ch03-sum1.gif

I think this picture might be useful. We are splitting the second vector into its constituent components, namely $x=200\cos\left({60}\right)$ and $y=200\sin\left({60}\right)$.

 

[ineedhelpnow]

yes but what about for b=300 N? what are the x and y components then?

 

[Dethrone]

300N is for the first vector, which is purely in the negative x direction. After this, we sum up all the forces acting in the x direction, and in the y direction.

 

[ineedhelpnow]

ooooh i see. no y component. ok thanks guys!

- - - Updated - - -

wait so do i multiply the two components? 200cos(60)*200sin(60)

 

[Dethrone]

You add up all the forces in the x-direction, which we have two, and we add up all the forces in the y-direction, which we have one. But then, that's just the magnitude of the resultant; we still need to find its direction.

 

[ineedhelpnow]

200cos(60)+(-300) and y is just -300. to find the magnitude of the resultant force now don't you square both of them and add them then take the square root?

 

[Dethrone]

That is correct, and that will give you its magnitude.

 

[Dethrone]

You're solving this problem by component method, but you could also solve it by using the law of cosines, combined with the parallelogram law. Completing the parallelogram, the diagonal will be $F_1 + F_2$, and its magnitude is computed by, in the most correct notation:

$$\left|\vec{F_{net}}\right|^2=\left|\vec{F_1} \right|^2+\left| \vec{F_2}\right|^2-2\left| \vec{F_1} \right|\left| \vec{F_2} \right|\cos\left({180-\theta}\right)$$

$\theta$ is the angle between the force vectors. After that, you can use the law of sines to compute its direction.

 

[ineedhelpnow]

here's how i did it:

$\left| \vec{F_1} + \vec{F_2}\right| = \sqrt{(200cos(60)+(-300cos(0)))^2+(200sin(60))^2}=100 \sqrt{7}$

is that right?

 

[ineedhelpnow]

and for the angle:

$\arcsin\left({\theta}\right)=\frac{200}{264.6}$ so $\theta=49.1$

49.1+90=139.1

 

[Dethrone]

The magnitude seems correct, but can you explain why you used arcsine for the angle? Breaking down the vectors into components, we can construct a triangle purely in the x and y-directions, which you have found, since you used Pythagorean's Theorem. We have x, which is the adjacent side, and we have y, which is the opposite side. The hypotenuse is the resultant vector.

If you're using the other method I suggested above, then you'll have to use the sine law since you're not working with a right-angle triangle.

 

[ineedhelpnow]

i used sin because i tried it other ways but the only way i could get the same answer as the book was when i did it this way

 

[Prove It]

I have to make a comment: ALWAYS simplify everything you can BEFORE trying to make any calculations. I don't know why the OP keeps carrying expressions like $\displaystyle \begin{align*} \sin{ \left( 60^{\circ} \right) } \end{align*}$ and $\displaystyle \begin{align*} \cos{ \left( 60^{\circ} \right) } \end{align*}$ when they so easily simplify to $\displaystyle \begin{align*} \frac{\sqrt{3}}{2} \end{align*}$ and $\displaystyle \begin{align*} \frac{1}{2} \end{align*}$ respectively. These would be sure to make the calculations easier!

 

[ineedhelpnow]

i used sin(60) cos(60) because i didnt feel like doing the whole latex thing and when you attempt at writing square root or making fractions without that latex format it gets confusing.

 

[Dethrone]

I'm not sure why you're using sine, but I will work out the problem now which should agree with the answer in your book.

I will drop the vector arrows and absolute values to make $\LaTeX$ easier.

$$F_x=200\cos\left({60}\right)-300=-200N$$ or $200N [W]$$$
F_y=200\sin\left({60}\right)=100\sqrt{3} N [North]$$

Applying pythagorean theorem, the magnitude is $ 100\sqrt{7}N$ and the direction:
$$\tan\left({x}\right)=\frac{o}{a}=\frac{100\sqrt{3}}{200}$$
$$\tan^{-1}\left({\frac{100\sqrt{3}}{200}}\right)\approx 40.9^o $$or $$W 40.9^o N$$

The angle can also be N 49.1 degrees W.

 

[MarkFL]

...i didnt feel like doing the whole latex thing...

Using $\LaTeX$ should not be looked upon as a chore, but rather as a way to improve your presentation. Like any language, the more you use it, the more fluent you become, and the easier it flows. You begin to think in $\LaTeX$...formatting expressions in your dreams.

Perhaps I exaggerated a bit there (Giggle) (or did I?), but in seriousness, presenting your work so that it is easy to read and understand is an effort others immediately recognize and appreciate. (Yes)

 

[ineedhelpnow]

Perhaps I exaggerated a bit there (Giggle) (or did I?)

:) yeah a bit. "formatting expressions in your dreams" that's very catching though (Giggle)

 

[ineedhelpnow]

ok so I am reviewing for my test and i just looked back at this question. i think i understand how to get the angle now. i get what rido did (except with all the north west stuff. no clue about that). so you do tangent (which is y/x) inverse of the y component over the x component and then add 180 and you get 139.1 as the angle

 

[Dethrone]

ok so I am reviewing for my test and i just looked back at this question. i think i understand how to get the angle now. i get what rido did (except with all the north west stuff. no clue about that). so you do tangent (which is y/x) inverse of the y component over the x component and then add 180 and you get 139.1 as the angle

You don't need the northwest stuff, just have a way to indicate direction, maybe with +/-. Not sure why you are adding 180, the angle you get from the inverse tangent should be between -pi/2 to pi/2, and if you get a negative answer, adjust it depending on the quadrant you are in. There are two angles you can use as your answer, either 40.9 or 49.1

 

[ineedhelpnow]

$arctan(100\sqrt{3}/-200)$ is -40.8934. so add 180 to get +139.107

 

[Dethrone]

Calculating net forces, we have in the x direction -200, and y direction 173.1. These are vectors, so a negative simply flips the direction. Drawing a right angle triangle, we have 200N going in the left direction, and 173.1N going upwards. Calculating tan theta with arctan(173.21/200), which is 40.9 from the horizontal, or 49.1 degrees from the vertical. If you really don't want to drop the negative, so arctan(173.21/-200), then we have to pay attention to which quadrant we are in. If you draw the appropriate net diagram (triangle), the forces lie in the 2nd quadrant. Tan theta returns the value of -40.9, which is in the 4th quadrant. So we subtract that angle from 180 to get 139.1N. So yes, 139.1 is right, but the answer probably wants the acute angle 40.9 or 49.1.

 

[ineedhelpnow]

Rido the back of the book says 139.1 (Dull) lol