Question

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

Answer 1

`F_x = 100N`

`F_y = 100\sqrt 3 \ N`

Step-by-step 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 * (\sqrt 3)/(2)`

`F_y = 100\sqrt 3 \ N`