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Solution :
Given :
Force, F = 500 N
Let [tex]$ \vec F = F_x\ \hat i + F_y\ \hat j$[/tex]
[tex]$|\vec F|=\sqrt{F_x^2+F_y^2}$[/tex]
∴ [tex]$F_x=F \cos 60^\circ = 500 \ \cos 60^\circ = 250 \ N$[/tex]
[tex]$F_y=-F \cos 30^\circ = -500 \ \cos 30^\circ = -433.01 \ N$[/tex] (since [tex]$F_y$[/tex] direction is in negative y-axis)
[tex]$F=250 \ \hat i - 433.01 \ \hat j$[/tex]
So scalar components are : 250 N and 433.01 N
vector components are : [tex]$250 \ \hat i$[/tex] and [tex]$-433.01\ \hat j$[/tex]
1. Scalar components along :
x' axis = 500 N, since the force is in this direction.
[tex]$F_{x'}= F \ \cos \theta = 500\ \cos \theta$[/tex]
Here, θ = 0° , since force and axis in the same direction.
So, cos θ = cos 0° = 1
∴ [tex]$F_{x'}=500 \times 1=500\ N$[/tex]
[tex]$F_{y'}= F \ \sin \theta = 500\ \sin 0^\circ=500 \times 0=0$[/tex]
[tex]$F_{y'}=F\ cos \theta$[/tex] but here θ is 90°. So the force ad axis are perpendicular to each other.
[tex]$F_{y'}=F\ \cos 90^\circ= 500 \ \cos 90^\circ = 500 \times 0=0$[/tex]
∴ [tex]$F_{x'}= 500\ N \text{ and}\ F_{y'}=0\ N$[/tex]
2. Scalar components of F along:
x-axis :
[tex]$F_x=F\ \cos \theta$[/tex], here θ is the angle between x-axis and F = 60°.
[tex]$F_x=500 \times \cos60^\circ=250\ N$[/tex]
y'-axis :
[tex]$F_{y'}=F\ \cos \theta$[/tex], here θ is the angle between y'-axis and F = 90°.
[tex]$F_{y'}=500 \times \cos90^\circ=500\times 0=0\ N$[/tex]
∴ [tex]$F_{x}= 250\ N \text{ and}\ F_{y'}=0\ N$[/tex]
