Question 98018

Jun 30, 2017

${F}_{x} = 15.3$ $\text{N}$

${F}_{y} = 12.9$ $\text{N}$

${F}_{n} = 6.84$ $\text{N}$ (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, $\text{N}$.

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

${20}^{\text{o" + 20^"o" = 40^"o}}$

relative to the horizontal. It has a magnitude of $20$ $\text{N}$, and its components are thus

${F}_{x} = F \cos \theta = \left(20 \textcolor{w h i t e}{l} \text{N")(cos40^"o}\right) = \textcolor{red}{15.3}$ color(red)("N"

${F}_{y} = F \sin \theta = \left(20 \textcolor{w h i t e}{l} \text{N")(sin40^"o}\right) = \textcolor{b l u e}{12.9}$ 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 ${20}^{\text{o}}$ with the plane, so it's vertical component ${F}_{n}$ is

${F}_{n} = F \sin \theta = \left(20 \textcolor{w h i t e}{l} \text{N")(sin20^"o}\right) = \textcolor{g r e e n}{6.84}$ color(green)("N"

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