Private Sub Physics_Normal_Force(m As Single, g As Single, Angle As Single, N As Point_2D)
N.X = m * g * Cos(Angle * PI / 180)
N.Y = m * g * Sin(Angle * PI / 180)
End Sub
Calculating Normal Force
Author
I'm trying to create a function that calculates the normal force of an object, and was wondering if my code is right or not?
From what you posted it seems correct. But without any other context there's no way for us to tell for sure [smile]
Author
Well I wasn't too sure if it was correct cause if the angle is 0, then N.Y = 0 and N.X was not 0, which didn't make any sense. If I reversed the cos and sin's around, then it sorta would make sense.
Fx = mg * sin ( angle )
Fy = mg * cos ( angle )
Rn= Fy
Ft = Cf * Rn = Cf * Fy = Cf * mg * cos ( angle )
Fx -Ft = m * a
a= ( Fx-Ft ) / m
s= 0.5 *a *t*t +v0 *t + s0
ds/dt = v = a* t+ v0
ds2/dt2 = dv/dt = a
Fy = mg * cos ( angle )
Rn= Fy
Ft = Cf * Rn = Cf * Fy = Cf * mg * cos ( angle )
Fx -Ft = m * a
a= ( Fx-Ft ) / m
s= 0.5 *a *t*t +v0 *t + s0
ds/dt = v = a* t+ v0
ds2/dt2 = dv/dt = a
Quote:Original post by Jacob Roman
Well I wasn't too sure if it was correct cause if the angle is 0, then N.Y = 0 and N.X was not 0, which didn't make any sense. If I reversed the cos and sin's around, then it sorta would make sense.
In that case it depends what you consider 0°. If 0° is when the ground is facing up and N.X is to the right then your results would seem weird. However let's say you consider a flat ground 90°, or in other words use the direction of the normal as the orientation of the ground. Then the results would make more sense. Since sine and cosine are 90° out-of-phase, it explains why you would think of switching them.
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