Question #98018

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Answer 1 · 2017-06-30T19:56:27

$F_x = 15.3$ $"N"$

$F_y = 12.9$ $"N"$

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

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 $color(green)(6.84$ $sfcolor(green)("newtons"$ .