Net force using parallelograms

[Sifuejos]

Homework Statement

There are only 3 forces acting on an object. The forces are 20.0 dynes E15°, 30.0 dynes W15°, and 40.0 dynes S15°W


2. Relevant equation
I've tried using the parallelogram method

The Attempt at a Solution

When i did the parallelogram method i got 30^2+20^2-2*30*20*cos(115) because 15° for the drawn parallelogram 15+15=30 for 2 of the 4 total angles in every corner, then 360-30=330/2=165 for the two remaining angles that will let the Resultant be 50 dynes. After that i tried doing the same thing for the parallelogram figure but i just couldn't figure it out. I've tried looking for different equations but it wouldn't budge or maybe I'm just not seeing it simple?

 

[SteamKing]

Homework Statement

There are only 3 forces acting on an object. The forces are 20.0 dynes E15°, 30.0 dynes W15°, and 40.0 dynes S15°W


2. Relevant equation
I've tried using the parallelogram method

The Attempt at a Solution

When i did the parallelogram method i got 30^2+20^2-2*30*20*cos(115) because 15° for the drawn parallelogram 15+15=30 for 2 of the 4 total angles in every corner, then 360-30=330/2=165 for the two remaining angles that will let the Resultant be 50 dynes. After that i tried doing the same thing for the parallelogram figure but i just couldn't figure it out. I've tried looking for different equations but it wouldn't budge or maybe I'm just not seeing it simple?

Your manner of representing vectors is not clear.

Take the first vector: 20.0 dynes E15° Is this supposed to mean 15° E of north? Or what?

Also, the parallelogram method is best used when trying to find the resultant of two vectors graphically. Using the P-method to find a resultant by calculation is very complex and prone to error, and you probably have to make a sketch to keep things straight.

If you are trying to find the resultant of three or more vectors by calculation, it's easier to find the components of each vector, add them together, and then convert the components of the resultant back into the vector form of your choice.

 

[haruspex]

How did you get 115?

 

[Adjoint]

Maybe you can upload a diagram or restate your the problem more clearly. Because like SteamKing said, its not possible to understand how your vectors are oriented from the way you described them.

 

[Sifuejos]

Yes I'm sorry, it's 30 dynes North west, 20 dynes North east, and 40 dynes South west, in the problem in my book, they used the parallelogram method in order to solve it, connecting the 30 and 20 dynes together as a parallelogram and then connected it as a parallelogram (the 40 dynes) back to the point of 0 which it's where 30 dynes north west and 20 dynes north east. I've only been doing physics for a week so my understanding of equations is very dim, I'm sorry.

 

[Sifuejos]

Net force problem

Homework Statement

The following were given, 30 Dynes 15° North west, 20 Dynes 15° North East, and 40 Dynes 15° South west. Fine the Net force

Homework Equations

Parallelogram Method, Law of cosines

The Attempt at a Solution

You have to find the force between 30 dynes and 20 dynes. So i used parallelogram method to add 15+15=30 degrees. I inserted that in The law of cosines formula c^2=a^2+b^2=2*a*b*cos(30°) and i got 16.4, you do the same for the other side and get 32.8 in total. I followed to draw vectors connecting them and i used, again, The law of cosines. Plugging in 40^2+32.8^2-2*40*32.8*cos(48°) so i added the product 30.33+40 dynes South west and got 70.33 dynes and subtracted 32.8 dynes that is the force that's applying north. This is the part where i got lost. I guesstimated it was 50 Degrees because the way the angle applied and was drawn as it could be divided by 3 (using parallelogram method 15+15=30*2=60, 360-60=300/2=150 divided by 3 because of how the vectors are aligned) End point is that i just played with the degrees so that it could make me obtain 37 dynes South west. This problem was off a book and it didn't offer any equations to solve it, Just a picture of it and the result. Which was 37 dynes South west. I took 2 hours to solve this problem and i really wished for an answer or help. Please

 

[Adjoint]

I don't think your description of the problem is same as the book's.
From your problem statement

The following were given, 30 Dynes 15° North west, 20 Dynes 15° North East, and 40 Dynes 15° South west. Fine the Net force

the diagram will be as below:

Is this the diagram given in your question?

If so then the correct answer should be about 48 dyne. And the direction is 87 degree West of North.
But as these never agree with your book's answer, I guess your question doesn't describe the books diagram properly.

You should upload a diagram describing your problem.

 

[NascentOxygen]

Homework Statement

The following were given, 30 Dynes 15° North west, 20 Dynes 15° North East, and 40 Dynes 15° South west. Fine the Net force

Homework Equations

Parallelogram Method, Law of cosines

The Attempt at a Solution

You have to find the force between 30 dynes and 20 dynes. So i used parallelogram method to add 15+15=30 degrees. I inserted that in The law of cosines formula c^2=a^2+b^2=2*a*b*cos(30°) and i got 16.4, you do the same for the other side and get 32.8 in total. I followed to draw vectors connecting them and i used, again, The law of cosines. Plugging in 40^2+32.8^2-2*40*32.8*cos(48°) so i added the product 30.33+40 dynes South west and got 70.33 dynes and subtracted 32.8 dynes that is the force that's applying north. This is the part where i got lost. I guesstimated it was 50 Degrees because the way the angle applied and was drawn as it could be divided by 3 (using parallelogram method 15+15=30*2=60, 360-60=300/2=150 divided by 3 because of how the vectors are aligned) End point is that i just played with the degrees so that it could make me obtain 37 dynes South west. This problem was off a book and it didn't offer any equations to solve it, Just a picture of it and the result. Which was 37 dynes South west. I took 2 hours to solve this problem and i really wished for an answer or help. Please

Hi Sifuejos. Mathematics as a narrative is not a gripping read. I think you should scan your neat maths solution and attach it here.

I get your 16.148 magnitude when 2 are summed, but you haven't indicated the resultant angle. Then adding the 40 I seem to get a different result.

 

[Sifuejos]

Problem in the book

It's in the attachments

 

[Sifuejos]

F1+F2 is equal to the resultant. I use the resultant, 32.8 and 40 into the law of cosines equation to get F1+F2+F3 which then i follow to add with 40 and then subtract the original North force 32.8 from the south force that would be 70.33 and is equal to 37.53 (The answer that is given in the book)

 

[Sifuejos]

Is it the book that stated that problem incorrectly? I still don't know how or why they got that answer

 

[Adjoint]

Am I missing something here? How did the angle between $\overline{F3}$ and $\overline{F1}$+$\overline{F2}$+$\overline{F3}$ became 15 degree?

 

[Sifuejos]

I'm guessing that's the angle between F1+F2+F3 and F3, That's exactly how the problem was stated and drawn as. I really don't know what's going on

 

[Adjoint]

You know what... If I were you I would leave the Parallelogram Method and the Law of cosines.
Just find what angles F1, F2 and F3 make with X and Y axises and find their x components and y components. Then finding net force and direction is easy.

And if this problem is making you too mad, why not just leave it for now. Ha?

 

[Sifuejos]

I guess..I'll leave it for my physics teacher to solve next year! Thank you for your help though ^-^

 

[Adjoint]

Hey I said so because I think the problem statement, diagram, textbook answer all were creating a confusing situation.
Never mind, did you? I was just trying to be funny. :tongue2:

 

[Sifuejos]

If there was a formula for this situation then it'll be great. I tried searching up problems like these but they're college problems and really confusing stuff. I'm not even in a physics class yet

 

[Adjoint]

Just find what angles F1, F2 and F3 make with X and Y axises and find their x components and y components. Then finding net force and direction is easy.

I guess you know how to do this ? Or not yet?

 

[Sifuejos]

Actually no...

 

[Adjoint]

Okey, you are given three forces and their directions (by angles) right?
Now look at the way I tried to draw your problem in a diagram. Are my angles correct?

For F1 and F2 it seems that your diagram agrees with mine.

But for F3, can you tell me, what's the degree of this red angle OR what's the degree of this blue angle?

 

[Sifuejos]

It doesn't say anything in the book. The only degree that it gives me for the South side is 15 Degrees that it makes in comparison to F1+F2+F3

 

[Adjoint]

 

[Sifuejos]

Pic

Straight out of the textbook

 

[NascentOxygen]

It doesn't say anything in the book. The only degree that it gives me for the South side is 15 Degrees that it makes in comparison to F1+F2+F3

It isn't given to you, you must work it out! When you add two vectors, the resultant has two parts: a magnitude, and a direction (or angle). Application of the cosine rule, and of the sine rule, allows you to determine both these parts. An answer which shows only one of those parts is incomplete or wrong.

At every intermediate step, you have to determine both the magnitude and the direction, when summing vectors.

BTW, pdfs are no use to me---either jpegs or nothing.

 

[Adjoint]

It doesn't say anything in the book. The only degree that it gives me for the South side is 15 Degrees that it makes in comparison to F1+F2+F3

After taking a look at your book from the pdfs you uploaded, I think you miss judged the situation. The force F3 makes an angle 15 degree NOT with the resultant force.
Force F3 is actually 15 degree West of South.

First, let's work with 20 and 30 dyne. For that, we create a parallelogram.

Using cosine rule to the ΔOBC, we find OC = 16.1 (you found that too, right?)

Now, we can take the third force of 40 dyne and add with 16.1

Here, we have to find OD from triangle ODE. But for applying cosine rule to ODE we should find the (green) angle ∠DEO, right?
So find ∠DOE [EDIT: should be ∠DEO]

Then we can apply the law of cosines in ΔDEO to get OD
I got OD = 32.5

To find the direction, we can first use the law of sines in ΔDEO to find ∠DOE
Which gives direction of OD as (90 - ∠DOE - 15)
The answer I got is 52 degree South of West.

Finally, I find the resultant force is about 32.5 dyne and about 52 degree south of west.
This does not completely agree with the book though (the book seems to say 37 dyne and 54 degree south of west).
By the way, as far as I know, I did it right. But sorry in advance in case I did something wrong.

I omitted some calculations because PF prohibits posting complete solutions to homework problems.
But I hope you get the idea.

 

[NascentOxygen]

@Adjoint *Missing attachments* When you include a link to an attachment, we also need the attachment.

 

[Adjoint]

Oops... sorry.
Attachments included now.

 

[Sifuejos]

I agree with everything you did on the first problem. But on the second one, how can you apply law of cosines to find green angle <Doe If you need 3 sides to find an angle yet you are only given 2 sides?

 

[Adjoint]

How can you apply law of cosines to find green angle <Doe ?

I never said that I used the law of cosines to find ∠DOE. I said in order to apply cosine rule on the triangle ODE (to get OD) we need to find out ∠DOE first.

If you need 3 sides to find an angle yet you are only given 2 sides?

Do you know about the law of sines? http://www.mathsisfun.com/algebra/trig-sine-law.html

 

[Sifuejos]

Yes but you are needed to start with at least one angle and none are given except for the 15 degrees South West but by the diagram you used, it does not appear to have nothing to do with the triangle <Doe

 

[Adjoint]

Here, we have to find OD from triangle ODE. But for applying cosine rule to ODE we should find the (green) angle ∠DEO, right?
So find ∠DOE

I am really sorry. I made a typing mistake the second time. It should be ∠DEO both times.

Now this is getting a bit too long discussion. And everyone can see that you have tried for long and you seem to know all relevant equations but maybe just getting lost in the details. So I hope forum Moderators won't mind if I do a little rescue mission:

In ΔOAC the $\angle$CAO = 30 degree
Using the rule of sines in this tringle,
$\frac{OC}{sin30}$ = $\frac{AC}{sinAOC}$
→ $\frac{16.1}{sin30}$ = $\frac{20}{sinAOC}$
→ $\angle$AOC = 38.4 degree
Now ∠EOC = ∠AOC + 90
→ ∠EOC = 128.4 degree
So from parallelogram EOCD, ∠DEO = 180 - ∠EOC = 180 - 128.4 = 51.6 degree (that's the green angle)

Now we can apply the law of cosines in ΔDEO to get,
OD = (ED² + OE² - 2.OE.ED.cos(51.6))^1/2
Finally, OD = 32.5

Next, to calculate direction, take ΔDEO and apply law of sines.
OD/sin∠51.5 = DE/sin∠DOE
→ 32.5/sin51.6 = 16.1/sin∠DOE
→ ∠DOE = 23

Now it can be easily seen that OD is (90-23-15) = 52 degree South of West.