Question #98018

1 Answer
Jun 30, 2017

Answer:

#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"#.