Question

The magnitude of F is 384 newtons and it points at 344o measured counterclockwise from the positive x-axis. What is the x component (in newtons) of F?

Answer 1

Draw a picture and the answer should be fairly simple:\[|A|=16^{\circ}\]\[Ax=|A|cos(\theta)\]\[Ay=|A|sin(\theta)\]

Answer 2

but i don't kno what to do.. Do i subtract 360-344?

Answer 3

Yes, to find the angle above (as I did) you do 360-344. That gives you theta

Answer 4

So to find the amount of force in just the x direction you would have:\[F_x=(384N)cos(16)=369.12N\]

Answer 5

oh

Answer 6

thanks! so much.. Could i ask u another question if u dont mind?

Answer 7

ok

Answer 8

Question 2 of 6
-F points in the opposite direction as F and has the same magnitude as F
True/False

Question 3 of 6
F1 = (8.4,3.4) and F2 = (-7.1,-5.8) where all components are in newtons. What is the exact x-component of F1 + F2 (in newtons)?

Answer 9

The first one is true.

For the second one you have to do vector addition:
F1=(8.4,3.4) which states that the x value is 8.4 and the y value is 3.4.
F2=(-7.1,-5.8) which states that the x value is -7.1 and the y value is -5.8.

The resultant vector (in component notation) will simply be:
Fresult=[(8.4+(-7.1)), (3.4+(-5.8))]Just perform the math and you'll know what the answer to the second question is.

Answer 10

dont i just need the value of x? so it will be 8.4+-7.1 = 1.3

Answer 11

If it's not clear, I'm just adding the x's and y's. You should end up with Fresult=(1.3, -2.4)

Answer 12

Yes

Answer 13

ok thanks!. how would u do this question. F = (30.8,44.4), where all components are in newtons. If a vector's direction is measured counterclockwise from the positive x-axis, what angle (in degrees, from 0-360) does the negative vector -F make?

Answer 14

Can I do opp/adj and then find the inverse tangent?

Answer 15

to get the angle

Answer 16

Yes

Answer 17

hey, @Shane_B when you're done, wanna help me? :)