1. What is a force? An interaction between TWO objects. For example, pushes and pulls are forces. We must be careful to think about a force as acting on one object from (or due to ) another object.

2. Adding Forces  Forces are vectors (They have both magnitude and direction) and so add as follows:  In one dimension, note direction using a + or – sign then add like scalar quantities (regular numbers with no direction associated with them)  Examples: + = + = +3 N +6 N +3 N -3 N 0 N

3. “Consider a body on which no net force acts…”  An important word here is NET. It means “total” or “sum of all” (forces).  It is not that no force at all can act on the body. It is just that all the forces must add to zero (cancel each other out).

4. Under this condition (no net force acting on the body): If the body is at rest, it will remain at rest. If the body is moving with constant velocity, it will continue to do so. What if the body is moving with a velocity which is not constant? Why isn’t this discussed?

5. Commonly shortened to “F=ma”. Correctly, it is :  Only forces which act on that object affect the acceleration of the object.  Forces exert by the object on another object do not. Newton’s Second Law in One Dimension

6. Using Newton’s 2 nd Law to Solve Problems 1.Identify all forces acting on the object -Pushes or Pulls -Frictional forces -Tension in a string -Gravitational Force (or weight = mg where g is 9.8 m/s 2 ) - “Normal forces” (one object touching another). 2.Draw a “Freebody Diagram” -draw the object, show all forces acting on that object as vectors pointing in the correct direction. Show the direction of the acceleration. 3.Chose a coordinate system. 4.Translate the freebody diagram into an algebraic expression based on Newton’s second law.

7. Free fall and weightlessness  An elevator is accelerating downward at 9.8 m/sec 2.  The scale feels no force because it is falling away from your feet at the same rate you are falling.  As a result, you are weightless.

8. Consider an elevator moving downward and speeding up with an acceleration of 2 m/s 2. The mass of the elevator is 100 kg. Ignore air resistance. What is the tension in the cable? v 1.Identify Forces: Tension in cable, weight of the elevator 2.Draw freebody diagram 3.Chose coordinate system: Let up be the +y direction and down –y. Then : 4.Translate the FBD into an algebraic expression. T- W = m(-a) so T-(100 kg)(9.8 m/s 2 ) = (100 kg)(-2 m/s 2 ) T W=F g earth  elevator. a Note: No negative sign