Question

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

Answer 1

Answer:

$F_x = 100N$

$F_y = 100\sqrt 3 \ N$

Explanation:

Given

$F = 200N$

$\theta = 60^o$

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:

$F_x = F * \cos(\theta)$ --- Horizontal

$F_y = F * \cos(\theta)$ ---- Vertical

So, we have:

$F_x = F * \cos(\theta)$ --- Horizontal

$F_x = 200N * \cos(60^o)$

$F_x = 200N * 0.5$

$F_x = 100N$

$F_y = F * \cos(\theta)$ ---- Vertical

$F_y = 200N * \sin(60^o)$

$F_y = 200N * \frac{\sqrt 3}{2}$

$F_y = 100\sqrt 3 \ N$