# Thread: A basic Engineering Physics (vectors/forces) problem

1. ## A basic Engineering Physics (vectors/forces) problem

Woo, alright. So, I'll throw this out there right off the bat: I'm taking a college level Physics course, but I've had no experience with anything Physics prior. So forgive me if I seem like I don't know what I'm doing (because I probably don't). I have a lab that I'm working on involving forces, and using vectors to graphically depict/analytically calculate said forces. Here is my dilemma:

Given two forces, F1 and F2:

F1 =
Force: 2.452 N
Mass: .25 kg
Direction: 30 degrees

F2 =
Force: 3.433 N
Mass: .35 kg
Direction: 130 degrees

[As cos 130 yields a negative value, I added 90 to it to get my angle measure back into Q1 for a positive value and used that for my calculations]

, I am to find the resultant |R| (its magnitude and direction).

Using the head-to-tail method of moving vectors, I moved F2 on the graph without rotating it to the tail of F1, and constructed |R| by drawing a straight line from the tail of F2 to the head of F1. This forms a triangle (not a right triangle, just a triangle). The first part of this exercise asks me to graphically depict and measure the resultant force. Measuring things out with a ruler and protractor, I come to the result:

|R|:
Force(N) = 3.813 N
Direction: 93.5 degrees.

The next step is to calculate the resultant analytically. The formula below is given to accomplish this:

Rx = ([F1]x+[F2]x) = [F1]cos(angle[F1]) + [F2]cos(angle[F2])
Ry = ([F1]y+[F2]y) = [F1]sin(angle[F1]) + [F2]sin(angle[F2])
|R| = sqrt([Rx]^2+[Ry]^2)

Plugging in my values I get:

Rx = ([F1]x+[F2]x) = [2.452]cos(30) + [3.433]cos(40)
Ry = ([F1]y+[F2]y) = [2.452]sin(30) + [3.433]sin(40)

Rx = 4.7533
Ry = 3.4327
Rx^2 = 22.594
Ry^2 = 11.783

|R| = sqrt([Rx]^2+[Ry]^2)

|R| = sqrt(22.594+11.783)
|R| = sqrt(34.377)
|R| = 5.8632

So, analytically:
|R|:
Force(N) = 5.863 N
Direction: 72.2 degrees.
Mass: 0.60 kg

My analytical and graphical results are extremely different. I can't figure out where I'm going wrong.

2. ## Re: Some basic Engineering Physics (vectors/forces)

The bolded part is likely your problem. I'm not sure what the "head to tail" method is, but if one method of rearranging triangles is a right triangle (you use Pythagorean theorem) and one isn't, they aren't the same triangles and you will get different magnitudes and directions.

3. ## Re: Some basic Engineering Physics (vectors/forces)

On my phone and I briefly skimmed the problem -

with vectors, split them into horizontal and vertical components and add those individually to get the horizontal/vertical components of your resultant. Use trig and Pythag theorem to find the magnitude and angle of your resultant.

I am reading something about adding 90 to an angle - BAD. DONT DO THIS. If you get a negative that's a-ok. Vectors are a magnitude and direction (think of despicable me); sign typically indicates direction.

4. ## Re: Some basic Engineering Physics (vectors/forces)

i'm assuming it looks similar to:

f1=2.452N
m1 = .25kg
theta1 = 30 deg(from the 0)

f2= 3.433N
m2= .25kg
theta 2 = 50 deg(reversed from the 180)

X component:(minus as the 2 forces are working against each other in the x direction I think)
f1cos(30)-f2cos(50)= resultant force in x direction
2.123-2.206= 0.083

Y component:(+ since they are working in the same y direction)
f1sin(30)+f2sin(50) = resultant force in y direction
1.226+2.630=3.856

so the final vector has an x component of 0.083, and a y component of 3.856

use Pythagorean theorem to find F of the hypotenuse(the actual resultant force amount)

making a triangle, that would be quadrant 1 where tan(thetaresultant) = 3.856/.083
or theta(r) ~= 88.77 degrees

finding |r| should be easy point on

didn't do any numberchecking(in bed atm in the dark), but that's how I would have approached the problem

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