1. Normal Force Force on an object perpendicular to the surface (Fn)
It may equal the weight (Fg), as it does here.
It does not always equal the weight (Fg), as in the second example.
Fn = mg cos 
Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

2. Normal Force
An object placed on a tilted surface will often slide down the surface.
The rate at which the object slides down the surface is dependent upon how tilted the surface is; the greater the tilt of the surface, the faster the rate at which the object will slide down it.
In physics, a tilted surface is called an inclined plane.
Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

3. Normal Force
As shown in the diagram, there are always at least two forces acting upon any object that is positioned on an inclined plane - the force of gravity and the normal force.
The force of gravity (also known as weight) acts in a downward direction
The normal force acts in a direction perpendicular to the surface (in fact, normal means "perpendicular").
Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

4. Normal Force
Up to this point in the course, we have always seen normal forces acting in an upward direction, opposite the direction of the force of gravity.
But this is only because the objects were always on horizontal surfaces and never upon inclined planes.
The normal forces is not always upwards, but rather that it is directed perpendicular to the surface that the object is on.
Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

5. Normal Force
The force of gravity will be resolved into two components of force - one directed parallel to the inclined surface and the other directed perpendicular to the inclined surface.
Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

6. Normal Force
The perpendicular component of the force of gravity is directed opposite the normal force and as such balances the normal force
Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

7. Normal Force
The parallel component of the force of gravity is not balanced by any other force. This object will subsequently accelerate down the inclined plane due to the presence of an unbalanced force
Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

8. Normal Force
It is the parallel component of the force of gravity that causes this acceleration. The parallel component of the force of gravity is the net force.
Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

9. Normal Force
The task of determining the magnitude of the two components of the force of gravity is a mere manner of using the equations. The equations for the parallel and perpendicular components are: .
Point out that the equation for normal force applies to the first example also. Because cos(0)=1, the equation reduces to Fn = mg when the forces are directly opposite one another.

10. Practice problems
A 25 kg chair initially at rest on a horizontal floor require a 365 N horizontal force to set it in motion. Once the chair is in motion , a 327 N horizontal force keeps it moving at a constant velocity.
Find the coefficient of static and kinetic friction between the chair and the floor.

11. Practice problems
A box of book weighting 325 N moves with a constant velocity across the floor when it is pushed with a force of 425 N exerted downward at an angle of 35.2° below the horizontal. Find the μk between the box and the floor.

12. Practice problems
A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 185 N at an angle of 25° above the horizontal. The box has a mass of 35 kg, and μk between the box and the floor is Find the acceleration of the box.

13. Practice problems
The student in the above item moves the box up a ramp inclined at 12° with the horizontal. If the box starts from rest at the bottom of the ramp and is pulled at an angle of 25° with respect to the incline and with the same 185 N force, what is the acceleration up the ramp? μk = .27

14. Practice problems
A 75 kg box slides down a 25° ramp with an acceleration of 3.6 m/s2.
Find μk
What acceleration would a 175 kg box have on the same ramp?