An object with a mass of 5 kg is on a plane with an incline of pi/8 . If the object is being pushed up the plane with 3 N of force, what is the net force on the object?

1 Answer
Aug 14, 2017

sumF_x = 15.8 "N" directed down the ramp

Explanation:

We're asked to find the net force acting on an object on an incline plane.

![upload.wikimedia.org](upload.wikimedia.org)

The vertical forces (perpendicular to the incline) cancel out, because the normal force equals the perpendicular component of the weight force; thus, we are only looking at forces parallel to the incline.

There are two forces acting on the object (assuming the surface is frictionless):

  • the gravitational force (acting down the ramp), equal to mgsintheta

  • the applied force directed up the ramp

The net force equation is thus

sumF_x = overbrace(F_"applied")^"upward force" - overbrace(mgsintheta)^"downward force"

(taking positive direction to be up the ramp)

We know:

  • m = 5 "kg"

  • g = 9.81 "m/s"^2

  • theta = pi/8

  • F_"applied" = 3 "N"

Plugging these in:

sumF_x = 3color(white)(l)"N" - (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)sin(pi/8) = color(red)(ulbar(|stackrel(" ")(" "-15.8color(white)(l)"N"" ")|)

(negative because it is directed down the ramp)