1. Inclined Planes

3. Purpose of an inclined plane is to lessen the force needed
Less force but larger distance

4. Inclined Plane
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.
Objects are known to accelerate down inclined planes because of an unbalanced force.

5. 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").
The normal force pushes on an object to prevent it from going through

6. When the surface is horizontal the Normal Force is equal to the Force of Gravity (Weight)
This is not true on an inclined plane

7. Determining the net force on an inclined plane is a difficult since the two (or more) forces are not directed in opposite directions.
One (or more) of the forces will have to be resolved into perpendicular components so that they can be easily added to the other forces acting upon the object.
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.

8. The perpendicular component of the force of gravity is directed opposite the normal force and as such balances the normal force.
The parallel component of the force of gravity is not balanced by any other force (unless friction is present).
This object will accelerate down the inclined plane due to the presence of an unbalanced force.
The parallel component of the force of gravity is the net force ( when there is no friction).

9. The angle of the inclined ramp is equal to the angle between Fgrav and F⊥

11. What happens to the components of gravitational force as the angle of the incline is increased?

12. As the angle is increased, the acceleration of the object is increased.
As the angle increases, the component of force parallel to the incline increases and the component of force perpendicular to the incline decreases.
It is the parallel component of the weight vector that causes the acceleration. Thus, accelerations are greater at greater angles of incline.

14. a 1000-kg roller coaster on the first drop of two different roller coaster rides. Use the principles of vector resolution to determine the net force and acceleration of the roller coaster cars for each drop. Assume a negligible effect of friction and air resistance.

15. Car A The Fgrav can be calculated from the mass of the object.
Fgrav = m • g = (1000 kg) • (9.8 m/s/s) = 9800 N The parallel and perpendicular components of the gravity force can be determined from their respective equations:
Fparallel = m • g • sin (45 degrees) = 6930 N Fperpendicular = m • g • cos (45 degrees) = 6930 N
The forces directed perpendicular to the incline balance each other. Thus Fnorm is equal to Fperpendicular.
Fnorm = 6930 N There is no other force parallel to the incline to counteract the parallel component of gravity. Thus, the net force is equal to the Fparallel value.
Fnet = 6930 N, down the incline The acceleration is can be found from a = Fnet / m :
 a = Fnet / m = (6930 N) / (1000 kg) a = 6.93 m/s/s, down the incline

16. Car B The Fgrav can be calculated from the mass of the object.
The Fgrav can be calculated from the mass of the object.
Fgrav = m • g = (1000 kg) • (9.8 m/s/s) = 9800 N The parallel and perpendicular components of the gravity force can be determined from their respective equations:
Fparallel = m • g • sin (60 degrees) = 8487 N Fperpendicular = m • g • cos (60 degrees) = 4900 N
The forces directed perpendicular to the incline balance each other. Thus Fnorm is equal to Fperpendicular.
Fnorm = 4900 N There is no other force parallel to the incline to counteract the parallel component of gravity. Thus, the net force is equal to the Fparallel value.
Fnet = 8487 N, down the incline The acceleration is can be found from a = Fnet / m :
 a = Fnet / m = (8487 N) / (1000 kg) a = 8.49 m/s/s, down the incline

17. Problem
The free-body diagram shows the forces acting upon a 100-kg crate that is sliding down an inclined plane. The plane is inclined at an angle of 30 degrees. The coefficient of friction between the crate and the incline is 0.3. Determine the net force and acceleration of the crate.

18. Answer
Begin the above problem by finding the force of gravity acting upon the crate and the components of this force parallel and perpendicular to the incline. The force of gravity is 980 N and the components of this force are Fparallel = 490 N (980 N • sin 30 degrees) and Fperpendicular = 849 N (980 N • cos30 degrees). Now the normal force can be determined to be 849 N (it must balance the perpendicular component of the weight vector). The force of friction can be determined from the value of the normal force and the coefficient of friction; Ffrict is 255 N (Ffrict = "mu"*Fnorm= 0.3 • 849 N). The net force is the vector sum of all the forces. The forces directed perpendicular to the incline balance; the forces directed parallel to the incline do not balance. The net force is 235 N (490 N N). The acceleration is 2.35 m/s/s (Fnet/m = 235 N/100 kg).

19. Find the Normal Force and what would the Ffrict have to be in order to keep the ball rolling at a constant speed?

20. Answer
433 N – normal force
The parallel component is 250 N (m*g*sine 30 degrees) so an equal but opposite frictional force of 250N is needed to move at a constant speed.

21. a 100-kg box is sliding down a frictional surface at a constant speed of 0.2 m/s.

22. The Fgrav can be calculated from the mass of the object.
Fgrav = m • g = (100 kg) • (9.8 m/s/s) = 980 N
The parallel and perpendicular components of the gravity force can be determined from their respective equations:
Fparallel = m • g • sin (45 degrees) = 693 N
Fperpendicular = m • g • cos (45 degrees) = 693 N
For a constant speed motion, the forces parallel to the incline must balance each other. Thus, the Ffrict is equal to the Fparallel.
Ffrict = 693 N
The forces directed perpendicular to the incline balance each other. Thus Fnorm is equal to Fperpendicular.
Fnorm = 693 N