Can anyone help me with this asap thanks?

Question

Answer 1

The sum of forces=

8[cos(30*)i+sin(30*)j]+

6[cos(-60*)i+sin(-60*)j]+

4[cos(90*+15*)i+sin(90*+15*)j]

=

8.893i+2.668j

=>

[magnitude=

sqr(8.893^2+2.668^2)~9.285,

angle=tan^-1(2.668/8.893)~16.7*]

=

9.285[cos(16.7*)i+sin(16.7*)j]

=

9.3 n in the direction of 16*1'59"

from the x-axis.

Answer 2

you want the vector sum of the three vectors shown?

F1x = 6 cos 60

F1y = –6 sin 60

F2x = 8 cos 30

F2y = 8 sin 30

F3x = –4 sin 15

F3y = 4 cos 15

total is

Fx = F1x + F2x + F3x

Fy = F1y + F2y + F3y

F = √(Fx² + Fy²)

θ = arctan (Fy/Fx)

surely you can do the arithmetic...

edit:

I'll fill in some of the numbers

F1x = 6 cos 60 = 3

F1y = –6 sin 60 = –6(√3/2) = –3√3 = –5.20

F2x = 8 cos 30 = 8√3/2 = 4√3 = 6.93

F2y = 8 sin 30 = 4

F3x = –4 sin 15 = –4((√6–√2)/4) = –√6 + √2 = –1.035

F3y = 4 cos 15 = 4 ((√6+√2)/4) = √6 + √2 = 3.86

Fx = 3 + 6.93 – 1.04 = 8.89

Fy = –5.20 + 4 + 3.86 = 2.66

F = √(8.89² + 2.66²) = √(79.0321 + 7.0756) = 9.28

θ = arctan (Fy/Fx) = arctan (2.66/8.89) = 16.7º

but check the math.

Answer 3

I believe it's about 8.4N at 28°