a force of 10 newton makes an angle of 30 degree with y axis. the magnitude of x-component will be
Answers
Answer:
Everyone has given the “sin = opp/hyp” answer (aside from Victor Avasi who suggests scale drawings - a great idea!) Anyway, I thought I’d offer a different way of thinking about it. When decomposing a vector
∣
∣
V
⃗
∣
∣
=V
|V→|=V
, we always have that one of the legs is
V
s
=Vsin(θ)
Vs=Vsin(θ)
and one is
V
c
=Vcos(θ)
Vc=Vcos(θ)
. Note that I’m avoiding using subscripts x & y here because the legs needn’t be parallel to the x & y axes - you are welcome to decompose a vector into any coordinate system.
For complex cases the angle that you aren’t given might not even be “in the triangle” of the vector that you are decomposing and its legs, which can make the problem hard to solve geometrically. But, if you know that
sin(θ)→0
sin(θ)→0
as
θ→0
θ→0
then you can reason without doing geometry. For example, in your case the angle is with the y-axis. If that angle were to shrink (head towards zero) then the length that you are interested in, the x-component, would also shrink. That means that the x-component (in this case) is
Vsin(θ)
Vsin(θ)
. Note that if the angle were instead with the x-axis, that if the angle were to go to zero then the x-component would actually grow. That is the behavior of cosine, so in that case we'd have
V
x
=Vcos(θ).
Vx=Vcos(θ).
Here’s the kind of problem where this technique begins to really shine. Imagine a hill, inclined at
θ
θ
to the horizontal. A rocket is launched from the hill with speed v, perpendicular to the hill. What is the y-component of the launch velocity?
You could do geometry, but you’d have to move the angle around to figure out the appropriate component. But if you realize that were the angle to go to zero (the hill to “go flat”) then the launch would be purely vertical (i.e., in this case as the angle goes to zero the component we are interested in gets bigger) then we must be dealing with cosine:
v
y
=vcos(θ)
Explanation: