Question

Question #98018

Answer 1

`F_x = 15.3` `"N"`

`F_y = 12.9` `"N"`

`F_n = 6.84` `"N"` (normal force)

Explanation:

I'll assume the vector shown with magnitude `20` is the force, and I can't say what the unit `m` is, although I'll treat it as Newton's, `"N"`.

We're asked to find the horizontal and vertical components of a force, and the force normal to the incline plane.

Components:

The force is at angle of

`20^"o" + 20^"o" = 40^"o"`

relative to the horizontal. It has a magnitude of `20` `"N"`, and its components are thus

`F_x = Fcostheta = (20color(white)(l)"N")(cos40^"o") = color(red)(15.3)` `color(red)("N"`

`F_y = Fsintheta = (20color(white)(l)"N")(sin40^"o") = color(blue)(12.9)` `color(blue)("N"`

The force normal to the plane is the vertical component of the force relative to the plane, which is perpendicular to it.

It makes an angle of `20^"o"` with the plane, so it's vertical component `F_n` is

`F_n = Fsintheta = (20color(white)(l)"N")(sin20^"o") = color(green)(6.84)` `color(green)("N"`

Thus, the force exerted normal to the incline plane has a magnitude of `color(green)(6.84` `sfcolor(green)("newtons"`.