1. Warm-up m = 54 kg g = 9.81 m/s² θ= 15° Sin θ = Fx/mg Fx = mg sin θ
Rose is sliding down an ice covered hill inclined at an angle of 15° with the horizontal. If Rose and the sled have a combined mass of 54.0 kg, what is the force pulling them down the hill? (neglect friction) mg Fn Fx
15°
m = 54 kg
g = 9.81 m/s²
θ= 15° Fn
Sin θ = Fx/mg
Fx = mg sin θ Fx mg
Fx = (54 kg)(9.81 m/s²) sin 15°
Fx = 137 N
θ = 15°

2. Friction
There are two types of frictional forces in the physical world that affects us.
- Static/Kinetic Friction
- Air Resistance

3. Air Resistance
Is a resistance force that acts in the opposite direction of gravitational forces.
Help!!
Force of Gravity
Force of Gravity
Air Resistance
SPLAT!!

4. Air Resistance Can also be a horizontal force Fx Fr (Air Resistance)mg

5. Static and Kinetic Friction
Are forces that oppose motion between two surfaces that are touching each other. Fr Fa
Fr (non-moving) = Static Friction (Fs)
Fr (moving) = Kinetic Friction (Fk)
Fk = ∑F

6. The force of friction is proportional to the size and mass of an object that you are pushing across a surface. Fk
Fapplied Fk
Fapplied
The amount of friction that occurs between an object and the surface can depend on the type of surface that the object is in contact with.

7. Fk
Fapplied
Tile Floor Fk
Carpet

8. The quantity that expresses the dependence of frictional forces on the particular surface the object is in contact with is called the coefficient of friction (µ).
It is the ratio between the normal force and the force of friction between two surfaces.

9. Coefficient of Kinetic Friction
μk = Fk/Fn coefficient of kinetic friction
μs = Fs,max/Fn coefficient of static friction
Ff = μFn frictional force

10. m = 82 kg g = 9.81 m/s² μk = Fk/Fn w = mg μk = 320 N/804 N
Practice problem
While redecorating her apartment, Suzy slowly pushes an 82 kg cabinet across a wooden dining room floor, which resists the motion with a force of friction of 320 N. What is the coefficient of kinetic friction between the cabinet and the floor? Fn
m = 82 kg
g = 9.81 m/s²
Fk = 320 N
Fapplied mg
μk = Fk/Fn
w = mg
μk = 320 N/804 N
w = (82 kg) (9.81 m/s²)
w = 804 N
w = Fn = 804 N
μk = 0.40

11. Practice Problem 2
A student moves a box of books by attaching a rope to the box by pulling with a force of 90 N at an angle 30° to the horizontal. The box of books has a mass of 20.0 kg and the coefficient of kinetic friction between the bottom of the box and the sidewalk is What is the acceleration of the box? (Pg. 146)

12. Step 1: Solve for Weight and X & Y components of applied force
m = 20.0 kg
g = 9.81 m/s² Fn
Fapplied = 90 N Fk
μk = 0.50
30° mg
w = mg = 196 N
Fapp,y
30°
90 N
Fapp, x
Fapp, y = (90N) (sin 30°) = 45.0 N
Fapp, x = (90N) (cos 30°) = 77.9 N

13. Step 2: Find the Kinetic Friction
Fapp, y = 45.0 N
μk = 0.50
Fapp, x = 77.9 N
w = 196 N
Fapp,y
30°
90 N
Fapp, x
∑Fy = Fn + w + Fapp, y = 0
0 = Fn – 196 N N
Fn = 196 N – 45 N = 151 N
μk = Fk/Fn
Fk = (μk)(Fn) = (0.50)(151 N) = 75.5 N

14. Fk = 75.5 N ∑F = ma a = ∑F/m a = (Fapp, x – Fk) / m
Step 3: Solve for acceleration
m = 20.0 kg
Fapp,y
30°
90 N
Fapp, x
Fk = 75.5 N
∑F = ma
Fapp, x = 77.9 N
a = ∑F/m
a = (Fapp, x – Fk) / m
a = (77.9 N – 75.5 N) / 20.0 kg
a = 0.12 m/s² to the right