# Thread: Problem I have been struggling with

1. ## Problem I have been struggling with

I am new to this forum and I really could use some help if any one has the answer please let me know how you got it

The magnitudes FA=40 N and FB=50 N FC=40 N
the angles alpha=50 deg and beta = 80 degrees
Determine the magnitude of FA+FB+FC

The book has the answer as 83 N and I have not clue on how they got this answer.

Can anyone help

2. Originally Posted by jaycas21
I am new to this forum and I really could use some help if any one has the answer please let me know how you got it

The magnitudes FA=40 N and FB=50 N FC=40 N
the angles alpha=50 deg and beta = 80 degrees
Determine the magnitude of FA+FB+FC

The book has the answer as 83 N and I have not clue on how they got this answer.

Can anyone help

1. What does the N denote (Newtons)?

2. The angles alpha and beta, what are they the angles between?

RonL

3. I got your question in my PM.
But you need to draw a triangle and explain better.

4. ## OK here is the clearification for the question posted

OK attached to the link is the question I have 2.3 and 2.6 . I am having a hard time setting up the problem. Please click on the question link
Question

5. Remember that if vectors $\displaystyle \vec{V_1},\vec{V_2}$ have magnitudes $\displaystyle a,b$ respectively, then,
$\displaystyle ||\vec{V_1}+\vec{V_2}||^2=a^2+b^2-2ab\cos \alpha$ where $\displaystyle \alpha$ is one of the angles. (this is the law of cosines).

6. Originally Posted by jaycas21
I am new to this forum and I really could use some help if any one has the answer please let me know how you got it

The magnitudes FA=40 N and FB=50 N FC=40 N
the angles alpha=50 deg and beta = 80 degrees
Determine the magnitude of FA+FB+FC

The book has the answer as 83 N and I have not clue on how they got this answer.

Can anyone help
That is suppopsed to be your problem/question, which is 2.4 per your attached figure below, but in that posting below, you said your problems are 2.3 and 2.6.
So, which or what is really the problem you want solved?
2.4, or 2.3 and 2.6 ?

--------------------------------
Let us say it is 2.4.

It is mentioned there that you solve it graphically, meaning, using drawing instruments and scales, plot those data on paper, then:
---get the resultant of FA+FB using the parallelogram method.
---then get the resultant of [(FA+FB) resultant] +FC, using the parallelogram method again.
This second resultant is (FA+FB+FC). Using the scale, it should be about 83N according to the book.

Analytically, or using Trigonometry, we can get the exact (FA+FB+FC).
Let us that resultant R. Or, R = (FA+FB+FC), in vectors.

In the figure, FA looks horizontal, so I assume it to be horizontal.
We put the origin (0,0) of the x,y axes setup at the ring of the eye-screw or at the common point of the 3 vectors. So if FA is on the x-axis, then FB is 50 degrees counterclockwise from the x-axis, and FC is (50+80=) 130 degrees counterclockwise from the x-axis.
Or, FA is on the x-axis, FB is in the 1st quadrant, and FC is in the 2nd quadrant.

---Horizontal component of R = Sum of horizontal components of FA,FB,FC.
Rh = (FA)h +(FB)h +(FC)h
where h is read "sub h" and it signifies the horizontal component of the vector.
Rh = FA +(FB)cos(50deg) -(FC)cos(50deg)
FC is 50 degrees clockwise from the negative side of the x-axis.
Rh = 40N +(50N)cos(50deg) -(40N)cos(50deg)
The author of this problem hates 50.
Using a calculator,
Rh = 40N +32.1394N -25.7115N
Rh = 46.4279N ---------------------***

---Vertical component of R = Sum of vertical components of FA,FB,FC.
Rv = (FA)v +(FB)v +(FC)v
Rv = (FB)sin(50deg) +(FC)sin(50deg)
Since FA is horizontal then it has no vertical component.
Rv = (50N)sin(50deg) +(40N)sin(50deg)
Rv = (90N)sin(50deg)
Rv = 68.944N --------------***

Then, by Pythagorean Theorem,
R^2 = (Rh)^2 +(Rv)^2
R = sqrt[(46.4279N)^2 +(68.944N)^2]