1. Physics: Principles with Applications, 6th edition
Lecture PowerPoints
Chapters 2, 3.1, 4, 5.10
Physics: Principles with Applications, 6th edition
Giancoli
© 2005 Pearson Prentice Hall
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2. Constant Force Apply Newton’s Second Law Force and Acceleration
Types of Forces
Equilibrium
Analyze Friction

3. Constant Force Newton’s Second Law
In an inertial reference frame, the sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration a of the object 
F = ma.
Where:
F is force in newtons
m is mass in kilograms
a is acceleration in m/s2

4. Constant Force Newton’s Second Law
Another way to state this is that if you apply an unbalanced (nonzero net) force to an object with mass, it will accelerate in the direction of the applied force

5. Constant Force Newton’s Second Law
Relationships between acceleration, force and mass
Rewrite F = ma to solve for a
F/m = a
So force is directly proportional to acceleration (for the same mass, more force = more acceleration)
And mass is inversely proportional to acceleration (for the same force, more mass means less acceleration)

6. Constant Force Newton’s Second Law
Graphical relationships between acceleration, force and mass

7. Constant Force Newton’s Second Law Types of forces (review)
Field force
Includes gravitational, electromagnetic
Contact force
Includes normal, tension, friction

8. Constant Force Newton’s Second Law Equilibrium Nonequilibrium Σ =ma
When net force on an object is zero
Σ =0
Nonequilibrium
Σ =ma

9. Constant Force Examples
Force to cause a car to accelerate
500kg car goes from rest to 25m/s in 5s. What is the net force required?
a = v – v0 / t = 25m/s – 0m/s / 5s = 5m/s2
F = ma = 500kg x 5m/s2 = 2500N

10. Constant Force Examples
Force to cause a car to deaccelerate
500kg car slows from 25m/s to rest in 5s. What is the net force required for it to stop?
a = v – v0 / t = m/s / 5s = -5m/s2
F = ma = 500kg x -5m/s2 = -2500N

11. Constant Force Examples
Elevator and counterweight
Apply Newton’s 3rd Law to both objects
FT – mEg = mEaE = -mEa
FT – mCg = mCaC = +mCa
Solve for a
a = g (mE – mC) / (mE + mC)
Elevator and counterweight accelerate at same rate but in different directions

12. Constant Force Examples
Advantage of using a pulley
With one pulley, 2 x FT exerted on piano with pull of 1 FT by man on roof

13. Constant Force Examples
Getting a car out of the mud
Student pushes on small section of rope. Force diagram is for that point.
FP=300N, ϴ=5°
FT=FP / 2 sin(ϴ)
FT=300N / 2 sin(5°)
=1700N

14. Constant Force Examples
Pulling means less normal force therefore less friction than pushing

15. Constant Force Examples

16. Constant Force Examples
Two boxes and a pulley For box A: FN = mag = (5.0 kg)(9.8 m/s2) = 49N Ffr = μK FN = 0.20(49N) = 9.8N ∑FAx = FT – Ffr = mAa For box B: ∑FBy = mBg - FT = mBa Tension is the same in all parts of the cord. FT = Ffr + mAa

17. Constant Force Examples
Inclines

18. Constant Force Examples

19. Constant Force Examples