1. Aim: How do we explain motion along an inclined plane?

2. Review of the normal force

3. An object at rest on an Inclined Plane
Label the following forces on the diagram above: (
Ignore the blue lines)
Friction
Gravity (Weight)
Normal Force

4. Vector Resolution of Weight Force
The weight force can be resolved into two components: One component is parallel to the surface of the incline (mgsinθ) One component is perpendicular to the surface of the incline (mgcosθ)

5. Motion along an inclined plane (Problem 1)
A crate has a mass of 56 kg and is at rest on a plane inclined at 30 degrees above the horizontal.
a) Find the weight of crate. Fg=mg=56kg(9.8m/s2)=548.8N
b) Find the components of the weight force.
Fgcosθ=(548.8N)cos30=475.27N
Fgsinθ=(548.8N)sin30=274.7N
c) Find the normal force.
FN =475.27N
d)Find the force of friction.
Ff=274.7N
e) Would this freebody diagram
look any different if the mass
was sliding down the incline? No, the directions of the forces would remain the same

6. Effect of angle on motion down an incline
Why do objects fall down an incline when the angle is increased?
mgsinθ increases

7. Incline Example (Problem 2)
Stacie who has a mass of 45 kg , moves down a slide at constant velocity that is inclined at a 45 degree angle with the horizontal.
Find Stacie’s weight.
Fg=mg=45kg(9.8m/s2)=441 N
b) Find the components of Stacie’s weight
Fgcosθ=441Ncos45=311.8N
Fg sinθ =441Nsin45=311.8N
c) Find the normal force acting on Stacie.
FN=311.8N

8. Draw a freebody diagram for an object being pulled up an incline
If this load is being pushed up an incline, draw all forces acting on crate. Are there any differences from before?
Redraw freebody Diagram for this situation

9. Problem 1 Regents Physics
Just find the parallel component of the weight using Fgsinθθ

10. Problem 2 Regents Physics
Mgsinθ=2kg(9.8m/s2)sin30=9.8N

11. Problem 3 Regents Physics
F=Ff +mgsinθ
F=3N+10Nsin30
F=3N+5N=8N

12. Problem 4 Regents Physics
Ff =mgsinθ because the net force is zero
Ff=20Nsin30=10 N up the ramp