2.  Force on objects whose surfaces are in contact  Acts in the opposite direction of motion  Acts parallel to the surface

3. F f = μF N F f = force of friction (N) μ = coefficient of friction  Unitless (yes, unitless…)  Always less than 1  Describes roughness of surfaces in contact  Rougher surfaces have larger μ

4. F a = applied force F N = normal force = surface reaction force F W = F g = gravitational force = weight F f = frictional force = contact force that opposes motion F T = tension = force in rope, string, cord, etc.

5. Step 1: Draw a diagram with all forces labeled Step 2: Write a net force equation Step 3: Replace F net with ma or 0 Step 4: Substitute and solve

6. Determine the force necessary to pull a 10 kg box across a horizontal surface with an acceleration of 3 m/s 2 if a 25 N frictional force is also acting upon it. Step 1: Draw a diagram with all forces labeled FWFW FaFa FNFN F f = 25 N **Everything will have a weight **If the object is on a surface there must be a normal force **There is only an applied force if they tell you there is one **Friction always opposes motion

7. Determine the force necessary to pull a 10 kg box across a horizontal surface with an acceleration of 3 m/s 2 if a 25 N frictional force is also acting upon it. Step 2: Write a net force equation F net = F a - F f FWFW FaFa FNFN F f = 25 N Choose horizontal or vertical

8. Determine the force necessary to pull a 10 kg box across a horizontal surface with an acceleration of 3 m/s 2 if a 25 N frictional force is also acting upon it. Step 3: Replace F net with ma or 0 F net = F a - F f ma = F a - F f FWFW FaFa FNFN F f = 25 N

9. Determine the force necessary to pull a 10 kg box across a horizontal surface with an acceleration of 3 m/s 2 if a 25 N frictional force is also acting upon it. Step 4: Substitute and solve F net = F a - F f ma = F a - F f (10 kg)(3 m/s 2 ) = F a -25 N FWFW FaFa FNFN F f = 25 N 30 N = F a -25 N +25 N 55 N= F a