Fisik

COMPETING FORCES
Carlos peñuela,Sandra lopez, Fernanda linares
Abstract
This practice consist in find the components of a force and check that the forces are summed in vector form. First took three forces for the ring was in the center, then took two forces with difference magnitudes and with a third force the system was in balance, finally took four forces with any magnitude for these were inbalance.
Keywords: Forces, magnitudes, balance, vectors.

Introduction
In this practice it is possible to observe the different masses in a table of force, hereby the balance of the bodies decides with regard to the angles in which he was thinking each of the masses, with the end that later one gives step to the vectors sum and somehow the theoretical calculations are corroborated by theexperimental ones in such a way that the equivalent sum of forces of zero.
In the competing forces all the physical magnitudes always are one to climb or a vector like it is the mass (m) the weight (w) and the force (F) .un to climb possesses magnitude direction and the point of application. The direction of in analytical form is indicated as the angle by the axis X.
Rectangular components of aforce in the plane, we use them in the previous sub subject of calculation of the resultant one of a system of vectors and we saw that the equations to using are:
Fx=Fcosθ (1)
Fy=F sinθ (2)
When on a particular action several forces, for example F1, F2, F3 these can be replaced by the alone certain resultant one F, which is equal to the vectorial sum of the forces:
F= F1 + F2 + F3 …..Fn =0(3)
What implicit that:
i=1N F X =0 (4)
i=1N F Y = 0 (5)

Particularly if on the body they operate three forces at the time:
F1 +F2 +F3 = where -F3 = F2 +F1 (6)
It is to say that the force F3 is equal to the resultant one of F1 and F2, of equal form-F1 is equal to the resultant one of F2 and F3 and-F2 is equal to the resultant one of F1 and F3.
To find the error percentage usingthe following equation:
E%= Vtheoretical-VpracticalVtheoretical*100

Materials and methods
First was applied a force of 350 g in the direction of 50° in the direction x and y, causing the ring is centered.
After were applied two forces, the first of 300 g in direction 50° and the second of 500 g and 130°, then with third forces.

Photography N° 1: Table of forces.
Source: Authors.
Source:Authors.

Photography N° 2: Masses.
Source: Authors.
Source: Authors.

Tables
TABLE N° 1: Experiment 1.
FORCES | DEGREES ( °) | WEIGHT (g) |
1 | 50 | 350 |
2 | 180 | 230 |
3 | 270 | 280 |
Source: Authors.
Source: Authors.

TABLE 2: Experiment N° 2
FORCES | DEGREES ( °) | WEIGHT (g) |
1 | 50 | 300 |
2 | 130 | 500 |
3 | 275 | 630 |
Source: Authors.
Source: Authors.TABLE 3: Experiment N° 3
FORCES | DEGREES ( °) | WEIGHT (g) |
1 | 40 | 295 |
2 | 120 | 310 |
3 | 230 | 382 |
4 | 320 | 250 |
Source: Authors.
Source: Authors.

Results
1. Applied to a force of 350g ring in a direction 50°, by means of forces in the X and Y, then the ring is in equilibrium and perfectly centered in the center of the table. What do the values ​​of these forces?Calculate the rectangular components of the given force. Do they match the calculated values ​​obtained experimentally with that?

To find the strength, you must first convert grams to kilograms; this is done by dividing the data in 1000 and then multiplied by the gravity (9.8 m/s2); so this way obtained the following results:

* F1=0.35 Kg*9.8ms2=3.43 N

* Fx=0.23 Kg*9.8ms2=2.25 N* Fy=0.28 Kg*9.8ms2=2.74 N

To find to theoretical components of the force, multiply the force vector by the cosine and sine of the angle, where we find the components in X and Y respectively.

* F1=3.43N (cos50 i+ sin50 j )
F1=2.21 i+2.62 j
Represents the vector component in the X axis and the vector j represents the vector component en the Y axis.
To find the experimental value of...