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.

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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)