1. Gravity and Friction

2. Kepler’s Laws
In the early 1600’ s Johannes Kepler mathematically analyzed known astronomical data in order to develop three laws to describe the motion of planets about the sun.
The paths of the planets about the sun are elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses)
An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas)
The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)

3. The cause for how the planets moved as they did was never stated
The cause for how the planets moved as they did was never stated. Kepler could only suggest that there was some sort of interaction between the sun and the planets that provided the driving force for the planet's motion.
Newton was troubled by the lack of explanation for the planet's orbits. Newton knew that there must be some sort of force that governed the heavens
Circular and elliptical motion were clearly departures from the inertial paths (straight-line) of objects. And as such, these celestial motions required a cause in the form of an unbalanced force.

4. universal gravitation
the notion of gravity as the cause of all heavenly motion was instigated when he was struck in the head by an apple while lying under a tree in an orchard in England.
Whether it is a myth or a reality, the fact is certain that it was Newton's ability to relate the cause for heavenly motion (the orbit of the moon about the earth) to the cause for Earthly motion (the falling of an apple to the Earth) that led him to his notion of universal gravitation.  

5. It was known at the time, that the force of gravity causes earthbound objects (such as falling apples) to accelerate towards the earth at a rate of 9.8 m/s2. And it was also known that the moon accelerated towards the earth at a rate of m/s2.
If the same force that causes the acceleration of the apple to the earth also causes the acceleration of the moon towards the earth, then there must be a plausible explanation for why the acceleration of the moon is so much smaller
Newton had to be able to show how the affect of gravity is diluted with distance.

6. What is it about the force of gravity that causes the more distant moon to accelerate at a rate of acceleration that is approximately 1/3600-th the acceleration of the apple?
The moon’s orbit is approximately 60 times further from the earth's center than the apple
The force of gravity between the earth and any object is inversely proportional to the square of the distance that separates that object from the earth's center
The moon, being 60 times further away than the apple, experiences a force of gravity that is 1/(60)2 times that of the apple.

7. The force of gravity follows an inverse square law.
The force of gravity is inversely related to the square of the distance

8. Newton suggests that the force of gravity acting between any two objects is inversely proportional to the square of the separation distance between the object's centers.
an increase in one quantity results in a decrease in the value of the other quantity
if the separation distance is doubled (increased by a factor of 2), then the force of gravity is decreased by a factor of four (2 raised to the second power).

9. Questions
Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is doubled, what is the new force of attraction between the two objects?
Answer: F = 4 units
If the distance is increased by a factor of 2, then force will be decreased by a factor of 4 (22). The new force is then 1/4 of the original 16 units.
F = (16 N) / 4 = 4 units

10. Questions
Suppose that two objects attract each other with a gravitational force of 16 units. If the distance between the two objects is reduced in half, then what is the new force of attraction between the two objects?
Answer: F = 64 units
If the distance is decreased by a factor of 2, then force will be increased by a factor of 4 (22). The new force is then 4 times the original 16 units.
F = (16 N) • 4 = 64 units

11. distance is not the only variable affecting the magnitude of a gravitational force. Consider Newton's famous equation F = m*a
the force of gravity acting between the earth and any other object is directly proportional to the mass of the earth, directly proportional to the mass of the object, and inversely proportional to the square of the distance that separates the centers of the earth and the object.

12. The UNIVERSAL Gravitation Equation
Since the gravitational force is directly proportional to the mass of both interacting objects, more massive objects will attract each other with a greater gravitational force.
If the mass of one of the objects is doubled, then the force of gravity between them is doubled.
If the mass of one of the objects is tripled, then the force of gravity between them is tripled.

15. universal gravitation constant
Knowing the value of G allows us to calculate the force of gravitational attraction between any two objects of known mass and known separation distance.

16. G - the universal gravitation constant.
The value of G was not experimentally determined until nearly a century after Newton(1798) by Lord Henry Cavendish
The value of G is an extremely small numerical value. Its smallness accounts for the fact that the force of gravitational attraction is only appreciable for objects with large mass.

17. Question
Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance of 6.38 x 106 m from earth's center.

18. Answer

19. Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is in an airplane at feet above earth's surface. This would place the student a distance of 6.39 x 106 m from earth's center.

21. Question
Suppose that you have a mass of 70 kg (equivalent to a 154-pound person). How much mass must another object have in order for your body and the other object to attract each other with a force of 1-Newton when separated by 10 meters?

22. Answer
m = 2.14 x 1010 kg
Use the equation Fgrav = G • m1 • m2 / d where m1 = 70 kg, d = 10 m and G = x N•m2/kg2
Note that the object is equivalent to an approximately 23 million ton object!! It takes a large mass to have a significant gravitational force.

24. Friction Ff=μFN Friction is always opposite the direction of motion
Friction is a reaction to motion (about to start motion or already in motion)
Friction converts mechanical energy into heat
Friction can be from direct contact or in a fluid (air)
Friction is always opposite the direction of motion
Ff=μFN

25. Friction depends on: 1) Directly related to the normal force
A- Is friction less on a flat surface or an inclined surface?
Inclined: less normal force
B- What does increasing the mass of an object due to friction?
Increases friction – increased mass causes an increase in the normal force
2)Type of material (coefficient of friction)

26. Type of surface
ters+phone+book+friction

27. Sliding versus Static Friction
Sliding friction (kinetic friction) results when an object slides across a surface.
Static friction results when the surfaces of two objects are at rest relative to one another and a force exists on one of the objects to set it into motion relative to the other object.
Suppose you were to push with 5-Newtons of force on a large box to move it across the floor. The box might remain in place. A static friction force exists between the surfaces of the floor and the box to prevent the box from being set into motion.

28. Coefficient of Friction (μ)
Values of μ have been experimentally determined for a variety of surface combinations and are often listed in books
The values of μ provide a measure of the relative amount of adhesion or attraction of the two surfaces for each other
The more that surface molecules tend to adhere to each other, the greater the coefficient values and the greater the friction force
static friction coefficients are greater than the values of sliding (kinetic) friction coefficients for the same two surfaces. It typically takes more force to budge an object into motion than it does to maintain the motion once it has been started.

29. An applied force of 20 N is used to accelerate an object to the right across a frictional surface. The object encounters 10 N of friction. Use the diagram to determine the normal force, the net force, the coefficient of friction (μ) between the object and the surface, the mass, and the acceleration of the object. (Neglect air resistance.)

30. Fnorm = 100 N; m = 10. 2 kg; Fnet = 10 N, right; "mu" = 0. 1; a =0
Fnorm = 100 N; m = 10.2 kg; Fnet = 10 N, right; "mu" = 0.1; a = m/s/s, right
Since there is no vertical acceleration, the normal force is equal to the gravity force. The mass can be found using the equation Fgrav = m * g.
Using "mu" = Ffrict / Fnorm, "mu" = (10 N) / (100 N) = 0.1.
The Fnet is the vector sum of all the forces: 100 N, up plus 100 N, down equals 0 N. And 20 N, right plus 10 N, left = 10 N, right.
Finally, a = Fnet / m = (10 N) / (10.2 kg) = m/s/s.
close

31. A 5-kg object is sliding to the right and encountering a friction force that slows it down. The coefficient of friction (μ) between the object and the surface is Determine the force of gravity, the normal force, the force of friction, the net force, and the acceleration. (Neglect air resistance.)

32. Fgrav = 49 N; Fnorm = 49 N; Ffrict = 4. 9 N; Fnet = 5 N, left; a = 0
Fgrav = 49 N; Fnorm = 49 N; Ffrict = 4.9 N; Fnet = 5 N, left; a = m/s/s, left
Fgrav = m • g = (5 kg) • (9.8 m/s/s) = 49 N. Since there is no vertical acceleration, the normal force equals the gravity force.
Ffrict can be found using the equation Ffrict ="mu"• Fnorm.
The Fnet is the vector sum of all the forces: 49 N, up plus 49 N, down equals 0 N. And 4.9 N, left remains unbalanced; it is the net force.
Finally, a = Fnet / m = (4.9 N) / (5 kg) = 0.98 m/s/s.

33. Edwardo applies a 4. 25-N rightward force to a 0
Edwardo applies a 4.25-N rightward force to a kg book to accelerate it across a tabletop. The coefficient of friction between the book and the tabletop is Determine the acceleration of the book.
Since there is no vertical acceleration, normal force = gravity force. Each of these forces can be determined using the equation Fgrav = m • g = (0.765 kg) • (9.8 m/s/s) = N
The force of friction can be determined using the equation Ffrict = mu • Fnorm. So Ffrict = (0.410) • (7.497 N) = ( N)
The Fnet is the vector sum of all the forces: 4.25 N, right plus N, left = N, right.
Finally, a = Fnet / m = ( N) / (0.765 kg) = 1.54 m/s/s. 

34. A rightward force is applied to a 10-kg object to move it across a rough surface at constant velocity. The coefficient of friction between the object and the surface is 0.2. Use the diagram to determine the gravitational force, normal force, applied force, frictional force, and net force. (Neglect air resistance.)

35. Fnet = 0 N; Fgrav = 98 N; Fnorm = 98 N; Ffrict = 19.6 N; Fapp = 19.6 N
When the velocity is constant, a = 0 m/s/s and Fnet = 0 N
Since the mass is known, Fgrav can be found (Fgrav = m • g = 10 kg • 9.8 m/s/s). Since there is no vertical acceleration, normal force = gravity force. Once Fnorm is known, Ffrict can be found using Ffrict = "mu" •Fnorm. Since there is no horizontal acceleration, Ffrict = Fapp = N

36. A rightward force of 25 N is applied to a 4-kg object to move it across a rough surface with a rightward acceleration of 2.5 m/s/s. Use the diagram to determine the gravitational force, normal force, frictional force, net force, and the coefficient of friction between the object and the surface. (Neglect air resistance.)

37. Fnet = 10 N, right; Fgrav = 39. 2 N; Fnorm = 39
Fnet = 10 N, right; Fgrav = 39.2 N; Fnorm = 39.2 N; Ffrict = 15 N; "mu"=
Fnet can be found using Fnet = m • a = (4 kg) • (2.5 m/s/s) =10 N, right.
Since the mass is known, Fgrav can be found: Fgrav = m • g = 4 kg • 9.8 m/s/s = 39.2 N.
Since there is no vertical acceleration, the normal force equals the gravity force.
Since the Fnet=10 N, right, the rightward force (Fapp) must be 10 N more than the leftward force (Ffrict); thus, Ffrict must be 15 N.
Finally, "mu"= Ffrict / Fnorm = (15 N) / (39.2 N) =