physics question please!?

Question

Three forces are applied to an object. Force 1 has a magnitude of 27.0 newtons (27.0 N) and is directed 30.0° to the left of the +y axis. Force 2 has a magnitude of 18.0 N and points along the +x axis. What must be the magnitude and direction of the third force 3 such that the vector sum of the three forces is 0 N? (force 3 points south of west)

Answer 1

Convert each of the degrees to standard degrees with
0 degrees at due east.

force 1:
30deg left of the y axis = 120 deg-> 60deg reference angle
force 2:
0 deg


Resolve the vectors to vertical and horizontal components.

force1:
f1vertical=27*sin(60deg)
=23.383 n
f1horizontal=-(27*cos(60deg))
=-13.5 n

force2:
f2vertical=0 n
f2horizontal=18*cos(0)
=18 n

Combine the known forces and then determine the counter
force to establish equilibrium.

ftvertical=23.383 n (north)
fthorizontal=18-13.5
=4.5 n (east)

Find the magnitude:
h=sqrt( x^2 + y^2 )
h=sqrt( 4.5^2 + 23.383^2 )
=23.812 n

Find the direction:
tan(theta)=23.383/4.5
theta=arctan(23.383/4.5)
=79deg ; east 79deg north

Find the counter direction by adding 180deg and finding
the coterminal angle if necessary.

thetacounter=79+180
=259deg

Find the directional in terms of north, south, east, west
259deg is greater than 180deg and less than 270deg
it must be in the 3rd quadrant. The reference angle will
be 259deg minus 180deg.

thetacounter: west 79deg south

The counterforce must have magnitude of 23.812 newtons
and direction west 79deg south.

Answer 2

The approach to take is to break the forces down into their x and y components first.

Force 1 has y of 27*cos(30) and x of -27*sin(30)
Force 2 has 0 y and x=18

The third force must have x and y such that the sum of each equals zero:
for y
0=27*cos(30)+F3y

F3y=-27*cos(30)

x:

0=18-27*sin(30)+F3x
F3x= 27*sin(30)-18

the magnitude of F3 is
sqrt(F3x^2+F3y^2)
and the direction is
ATAN(F3y/F3x) south of west

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Answer 3

Break the two forces into components along teh x and y axes.

F1 --> F1x = -27*sin(30) = =-13.5 N
F1y = 27*cos(30) = 13.5*sqrt(3) N =23.4N

F2 --> F2x = 18 N

Now you want a force F3 that has an x component:

F3x = -(F1x + F2X) = 4.5 N

F3y = -F1y = -23.4N

F3 = sqrt(F3x^2+f3y^2) = 23.8 N

angle = tan^(-1)(F3y/F3x) = 79.2 deg --> convert to 3rd quadrant angle

angle +180 = 259.3 deg