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caculate the component of a force of 200 ns
at a direction of 60° to the force​

Sagot :

MrRoyalidn

Answer:

[tex]F_x = 100N[/tex]

[tex]F_y = 100\sqrt 3 \ N[/tex]

Explanation:

Given

[tex]F = 200N[/tex]

[tex]\theta = 60^o[/tex]

Required

The component of the force in F direction

To do this, we simply calculate the force in the vertical and horizontal direction.

This is calculated as:

[tex]F_x = F * \cos(\theta)[/tex] --- Horizontal

[tex]F_y = F * \cos(\theta)[/tex] ---- Vertical

So, we have:

[tex]F_x = F * \cos(\theta)[/tex] --- Horizontal

[tex]F_x = 200N * \cos(60^o)[/tex]

[tex]F_x = 200N * 0.5[/tex]

[tex]F_x = 100N[/tex]

[tex]F_y = F * \cos(\theta)[/tex] ---- Vertical

[tex]F_y = 200N * \sin(60^o)[/tex]

[tex]F_y = 200N * \frac{\sqrt 3}{2}[/tex]

[tex]F_y = 100\sqrt 3 \ N[/tex]

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