1. Welcome to Physics 101! Lecture 01: Introduction to Forces
Newton’s Laws
Forces
Free Body Diagrams

2. Newton’s 1st Law of Motion
If the sum of all external forces on an object is zero, then its speed and direction will not change. (Know as the Law of Inertia.)
Example: air hockey table
When the puck travels with no net force on it, it moves in a straight line with constant speed.
The only time the speed or direction changes is when a net force (from the walls or mallets) is exerted on the puck!

3. Newton’s 2nd Law of Motion
If a nonzero net force is applied to an object its motion will change. (The change is described by the equation F= ma.)
Example: dropping a book
When you drop a Calculus book from your 10th story dorm window, there is a net force acting on it (gravity) thus it begins to accelerate toward the street below!

4. Newton’s 3rd Law of Motion
In an interaction between two objects, the forces that each exerts on the other are equal in magnitude and opposite in direction.
Example: rowing a boat
When you row, you are pushing the water backwards…
Thus the water pushes your paddle (and therefore you) forward!

5. Forces Quantifies “interactions” between objects. 4 Fundamental Forces
Gravity
Electromagnetic
Strong Nuclear
Weak Nuclear
Forces are Vectors
Have magnitude and direction
SI units are Newtons (N)
Blow up hydrogen balloon here?

6. Gravity Any two objects with mass are attracted.
On the surface of the earth, the force of attraction is referred to the force of gravity or weight.
rearth = 6.4x106 m, mearth = 6x1024 kg, G = 6.67x10-11 N-m2/kg2
Near the earth, the radius of the earth can be used for r.
Example: a person’s weight
Consider a person with a typical mass of 80 kg.
Near the earth, his weight (or force of gravity pulling him down toward the center of the earth) would be Fg = 784 N.

7. Contact Forces
An object in contact with a surface may have a normal force and force of friction acting on it.
The normal force is always perpendicular to the surface. There is no equation for the normal force – its magnitude depends on the situation.
The frictional force is always parallel to the surface (opposing motion). The frictional force depends on the “coefficient of friction” between the object and surface.
Kinetic friction (object sliding on a surface): Ff = k*FN
Static friction (no sliding): there is no definite equation for static friction, but its maximum value is given by: Ff  s*FN
Example: a sled sliding across a snowy field
The ground is pushing up on the sled (to keep it from falling into the earth), and the ground is pushing against the sled’s motion (slowing it down a little).

8. Free Body Diagrams The key to success in Physics 101! FN Ff Fp Fg
Simple picture of just the object of interest.
Choose coordinate system (x,y).
Identify all forces acting on object and draw arrows showing the directions.
Example: a dresser being pushed down the hall of your apartment as you move in.
There are four forces: gravity, the normal force, friction, and you pushing on the dresser. FN Ff Fp y x Fg

9. Summary Newton’s Laws of Motion Forces: Free Body Diagrams Inertia
F=ma
Equal and Opposite Force Pairs
Forces:
Non-Contact: Gravity
Contact: Friction and Normal
Free Body Diagrams
Isolate One Object
Each Direction is Independent

10. Example: The Dresser FN Ff Fp Fg
Let’s return to the dresser FBD and do some calculations…(I will solve the horizontal direction and you will solve the vertical direction)… FN
Given: the dresser has a mass of 120 kg and you are pushing on it with a force of 345 N to make it slide down the hall at a constant velocity. Ff Fp y x Fg

11. Example: The Dresser FN Ff Fp Fg
m = 120 kg and Fp = 345 N
Once the FBD is drawn and we have picked a coordinate system we must apply Newton’s Second Law: FN
F = ma
We know the horizontal acceleration is zero (the velocity is constant).
F = 0 Ff Fp y x Fg

12. Example: The Dresser Now start to substitute in what is known: FN Ff
m = 120 kg and Fp = 345 N
Now start to substitute in what is known: FN
F = 0
We know there are two forces in the x-direction (the push and friction).
Fp – Ff = 0 Ff Fp y x Fg

13. The force of the push is given as 345 N.
Example: The Dresser
m = 120 kg and Fp = 345 N
Solve for the unknown:
Fp – Ff = 0
The force of the push is given as 345 N.
(345 N) – Ff = 0
Ff = 345 N FN Ff Fp y x Fg

14. Example: The Dresser
Follow-up: now we can find the coefficient of kinetic friction between the dresser and the floor… FN
Ff = k*FN
(345 N) = k*(1176 N)
k = 0.29 Ff Fp y x Fg