x
=0.150gN
F
y
=0.260gN
F
x
=0.150gN
F
y
=0.260gN
Vector addition IV
F
1
=(0.100)gN,
1
=30
o
F
2
=(0.200)gN,
2
=90
o
F
3
=(0.300)gN,
3
=225
o
F
R
=0.131gN
r
=163
o
F
R
=0.131gN
r
=163
o
F
R
=0.08gN
r
=141.5
o
Vector addition V
F
1
=(0.225)gN,
1
=56
o
F
2
=( 0.240)gN,
2
=-
62
o
F
R
=0.238gN
r
=-5.4
o
F
R
=0.239gN
r
=-6.0
o
F
R
=0.265gN
r
=-5.5
o
References
See Figure
1. Graph
section
Please see
Formulas
and
Calculation
s section
Data
Gathered
directly from
Force table
Formulas and Calculations
The weight of an object is the force of gravity on the object and may be
calculated as the mass times the acceleration of gravity. Since the weight
is a force or magnitude, its SI units is the Newton (N). For an object in free
fall, so that gravity is the only force acting on it, then the expression for
weight follows from Newton's second law is:
F = m. g
(1)
F
→
Force in Newton's (N)
m
→
Mass in Kilograms (g)
g
→
Gravity acceleration (9.8 m/s²) – this factor is a
constant and is left as a constant in calculations as “g”.
When adding force vectors, the resultant force may be determined by the
analytical method of components. For two concurrent forces (F1 and F2)
acting on the center ring on the force table, the resultant force, FR =
F1+F2. To add these force vectors by analitycal method, the magnitude
and direction of the resultant force is determined using the law of cosines
and the law of sines for every force vector in order to find their
components (X and Y).

F1x = F1cos
F2x = F2cos
(2)
F1y = F1sin
F2y = F2sin
By the addition of the like components in this case FX and FY, the
components of the resultant force FR can be determined.
Frx = F1x + F2x
Fry = F1y + F2y
(3)
Pythagorean theorem can be used to find the magnitude of the resultant
force of the vector FR, assuming that the X component lies on the X- axis
with an angle of 0°, and the Y component lies on the Y- axis forming an
angle of 90°, therefore by adding those vectors tip to tail graphically the
hypotenuse can be determined by drawing a diagonal line from the origin
point (0,0) to the tip of the component Y vector creating a triangle
rectangle, hence the magnitude of the resultant force vector (Fr) can be
calculated.
Fr² = Frx² + Fry²,
F
r
=
√
F
x
2
+
F
y
2
(4)
Knowing the components X and Y of the resultant force vector, the
direction of the vector also can be computed using the law of arc tangent
in order to find the angle of the triangle rectangle, but in reality the
resultant direction of the arc tangent law is not going to be the
equilibrated vector for all the other forces. The equilibrated force will be
the opposite of the direction of the vector force resultant.
¿
tan
−
1
F
Ry
F
Rx
(5)
Calculations
Vector Addition I.
forces
X component (2)
Y component (2)
F
1
=.200 gN,
1
=30
o
F1x=0.200cos
30
=0.173
gN
F1y=0.200sin30
=0.100
gN
F
2
= .200 gN,
2
=120
o
F2x=0.200cos120
=-
0.100gN
F2y=0.200 sin12
0
=
0.173gN
Sum of
components (3)
Fx=0.173+(-
0.100)=0.073gN
Fy=0.100+0.173=0.273
gN

Resultant vector, Fr
at angle,
F
r
=
√
0.073
2
+
0.273
2
= 0.283gN
¿
tan
−
1
0.273
0.073
= 75
Vector Addition II.
forces
X component (2)
Y component (2)
F
1
=.200 gN,
1
=20
o
F1x=0.200cos
20
=0.188
gN
F1y=0.200sin20
=0.068
gN
F
2
= .150 gN,
2
=80
o
F2x=0.150cos80
=0.026
gN
F2y=0.150sin80
=
0.148gN
Sum of
components (3)
Fx=0.188+0.026=0.214
gN
Fy=0.068+0.148=0.216
gN
Resultant vector, Fr
at angle,
F
r
=
√
0.214
2
+
0.216
2
= 0.304gN
¿
tan
−
1
0.216
0.214
= 45.3
Vector addition III.
forces
X component (2)
Y component (2)
F
1
=.200 gN,
1
=0
o
F1x=0.200cos
0
=0.2gN
F1y=0.200sin0