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dainel40
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?
Draw a picture and the answer should be fairly simple:|dw:1359421352236:dw|\[|A|=16^{\circ}\]\[Ax=|A|cos(\theta)\]\[Ay=|A|sin(\theta)\]
but i don't kno what to do.. Do i subtract 360-344?
Yes, to find the angle above (as I did) you do 360-344. That gives you theta
So to find the amount of force in just the x direction you would have:\[F_x=(384N)cos(16)=369.12N\]
thanks! so much.. Could i ask u another question if u dont mind?
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)?
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.
dont i just need the value of x? so it will be 8.4+-7.1 = 1.3
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)
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?
Can I do opp/adj and then find the inverse tangent?
hey, @Shane_B when you're done, wanna help me? :)