# physics lab help needed

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#### quantumnumber

##### Guest
ok, the lab I am working on deals with vectors. I am stuck on a part of the lab dealing with the component method. Here is a link to the lab, this might help a little more in explaining what I am trying to do:

http://faculty.genesee.edu/macrittenden/b/lab/3.htm

I have part 1 done, finding the resultant graphically.

What I am stuck on is part 2 - the component method.

Force 1 is 6.0 newtons at an angle of 0 degrees - right on the x-axis. Force 2 is 6.1 newtons at 85 degrees. Force 3 is 9.0 newtons at 223 degrees.

Q

#### quantumnumber

##### Guest
oh, and try not to laugh too hard, this is my first physics class ever

O

#### origin

##### Guest
What part are you stuck on?

Edited to add - your data looks good. Did you look at the example calculation? Basically all you have to do is substitute your numbers into the equation and 'plug and chug'. Just remeber the apparatus isn't moving so the forces should add up to zero and don't get confused between radians and degrees.

S

#### Shpaget

##### Guest
Well basicly to calculate a x component of a force vector you use:
Fx = Fxcosß
Note that:
x in Fx is subscript
x in Fxcosß stands for multiplication
ß is angle between positive part of the x axis and the direction of the vector (measuring counterclockwise).

For y component you use:
Fy = Fxsinß

When you have multiple force vectors you use:
Fx = F1xcosß1 + F2xcosß2+ F3xcosß3 + ... + Fnxcosßn
Fy = F1xsinß1 + F2xsinß2 + F3xsinß3 + ... + Fnxsinßn

These two formulas are great because you don't need to change + and -, sin and cos are going to tell you whether it's + or -.
For example:
You have two forces:
F1 = 10 N ß1=30°
F2 = 15 N ß2 = 200°

Fx = 10xcos30° + 15xcos200°
Fy = 10xsin30° + 15xsin200°

Fx = 8,66 + (-14,1)
Fy = 5 + (-5,13)

Fx = -5,44 N
Fy = -0,13 N

Since you got both components negative your vector will be in the third quadrant.

To get the intensity (module, scalar whatever you want to call it) of that vector use:
|FR| = square root (Fx^2 + Fy^2)
Note that |FR| has an arrow above F to indicate that it's a vector, R is subscript, | | means module.

|FR| = 5,4416 N

To calculate the direction (angle ßR, resulting angle) you can use:
tg ßR = Fy/Fx
tg ßR = 0,239
ßR = 1,369°
IMPORTANT:
You must add another 180° to get correct angle, since you know your vector is in III. quadrant.

FR = 5,4416 N
ßR = 181,369°

O

#### origin

##### Guest
Shpaget":2zti4ttq said:
Well basicly to calculate a x component of a force vector you use:
Fx = Fxcosß
Note that:
x in Fx is subscript
x in Fxcosß stands for multiplication
ß is angle between positive part of the x axis and the direction of the vector (measuring counterclockwise).

For y component you use:
Fy = Fxsinß

When you have multiple force vectors you use:
Fx = F1xcosß1 + F2xcosß2+ F3xcosß3 + ... + Fnxcosßn
Fy = F1xsinß1 + F2xsinß2 + F3xsinß3 + ... + Fnxsinßn

These two formulas are great because you don't need to change + and -, sin and cos are going to tell you whether it's + or -.
For example:
You have two forces:
F1 = 10 N ß1=30°
F2 = 15 N ß2 = 200°

Fx = 10xcos30° + 15xcos200°
Fy = 10xsin30° + 15xsin200°

Fx = 8,66 + (-14,1)
Fy = 5 + (-5,13)

Fx = -5,44 N
Fy = -0,13 N

Since you got both components negative your vector will be in the third quadrant.

To get the intensity (module, scalar whatever you want to call it) of that vector use:
|FR| = square root (Fx^2 + Fy^2)
Note that |FR| has an arrow above F to indicate that it's a vector, R is subscript, | | means module.

|FR| = 5,4416 N

To calculate the direction (angle ßR, resulting angle) you can use:
tg ßR = Fy/Fx
tg ßR = 0,239
ßR = 1,369°
IMPORTANT:
You must add another 180° to get correct angle, since you know your vector is in III. quadrant.

FR = 5,4416 N
ßR = 181,369°

Tsk, tsk, now why would you do the lab for him? He ain't goin to leant nuttin dat way....

Q

#### quantumnumber

##### Guest
Yes but this is an example! different numbers were used
I already took a brief look at this but I am going to take a further look at this after my precalc class. I was just studying for a physics quiz that I have today. oh, and I am a she, not a he

O

#### origin

##### Guest
quantumnumber":1v7k5q9g said:
Yes but this is an example! different numbers were used
I already took a brief look at this but I am going to take a further look at this after my precalc class. I was just studying for a physics quiz that I have today. oh, and I am a she, not a he

Sorry, young lady. I hope you excel in your Physics/Astronomy degree work!

M

#### MeteorWayne

##### Guest
origin":3jy2914k said:
Sorry, young lady. I hope you excel in your Physics/Astronomy degree work!

But having us do your work for you is not the way to accomplish that
Wayne

S

#### Shpaget

##### Guest
But I didn't do the work for her (is a girl!!! on a science forum :shock: )
I just provided formulas (which by the way can be found anywhere), and made up some numbers to show how they work.

Q

#### quantumnumber

##### Guest
Shpaget":1y2qocmb said:
But I didn't do the work for her (is a girl!!! on a science forum :shock: )
I just provided formulas (which by the way can be found anywhere), and made up some numbers to show how they work.

Thank you very much for your help. I got it! I am like an exponential function. It takes me a while to get it, but once I understand it.....

S

#### Shpaget

##### Guest
You're welcome. Glad I could help.

Q

#### quantumnumber

##### Guest
I got a 92 on my first physics test

S

#### SpeedFreek

##### Guest
Well done, keep up the good work!

Q

#### quantumnumber

##### Guest
Thank you

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