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CHAPTER 4 FORCES and the LAWS of MOTION

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Presentation on theme: "CHAPTER 4 FORCES and the LAWS of MOTION"— Presentation transcript:

1 CHAPTER 4 FORCES and the LAWS of MOTION

2 THE NORMAL FORCE Fn a force that acts on a surface in a direction perpendicular to the surface

3 THE NORMAL FORCE Fn the television is in equilibrium, so the normal force from the table must be equal in magnitude and opposite in direction to the gravitational force exerted on the television.

4 THE NORMAL FORCE Fn is ALWAYS perpendicular to the surface, but NOT ALWAYS opposite to the Fg Fn = mg ∙ cos 𝛉

5 The Net Force Is the vector sum of all forces acting on the body
4 forces acting on the car Add them as vectors The resultant is then the NET FORCE If the NET FORCE is 0, then you know that the car is travelling at constant velocity (zero acceleration)

6 Free Body Diagrams (FBDs)
are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.

7 Free Body Diagrams (FBDs)
4 forces acting on this object, but there MIGHT be more or less forces Represent the object as a SQUARE or a CIRCLE You MUST show all the forces acting on the object

8 Free Body Diagrams (FBDs)
A puck is sliding on a sheet of ice to the right FA Fapp Ffriction How many forces are acting on the puck? Fgravity Label all of them!

9 Free Body Diagrams (FBDs)
A puck is sliding on a sheet of ice to the right FA Fapp Ffriction If the object is moving, you MUST show the direction of acceleration Fgravity

10 Free Body Diagrams (FBDs)
A puck is sliding on a sheet of ice to the right a FA Fapp Ffriction If the object is moving, you MUST show the direction of acceleration Fgravity

11 Interpreting FBD (Quantitative = with numbers)
NET FORCE?? a FA Fapp Ffriction Fgravity

12 Interpreting FBD (Quantitative = with numbers)
NET FORCE?? a Fnet = 0 N in vertical direction 40 N 10 N 25 N Fnet = + 15 N in horizontal direction 40 N

13 Interpreting FBD (Quantitative = with numbers)
Fnet = 0 N in vertical direction 100 N 100 N Fnet = 0 N in horizontal direction 47 N If the Net Force is 0 (Fnet = 0 N), the object is in EQUILIBRIUM

14 4.1 The Newton’s FIRST Law:
Tells us what happens when all existing forces acting on a body are balanced If the Net Force is 0 (Fnet = 0 N), the object is in EQUILIBRIUM 47 N If the forces acting on a body are balanced, then the acceleration of the body will be m/s2 100 N 100N 47 N

15 4.2 The Newton’s Second Law:
Tells us what happens when an unbalanced force acts on a body If the Net Force is NOT 0 (Fnet ≠ 0 N), the object is NOT in EQUILIBRIUM a 47 N If an UNBALENCED force acts on a body, the body will ACCELARATE 100 N 200N 47 N

16 4.2 The Newton’s Second Law:
Tells us what happens when an unbalanced force acts on a body The rate at which the body will ACCELERATE depends DIRECTLY on the UNBALANCED FORCE (Fnet) and INVERSELY on the mass of the body. ∝ a ∝ Fnet a ∝ 1/m

17 4.2 The Newton’s Second Law:
Tells us what happens when an unbalanced force acts on a body The rate at which the body will ACCELERATE depends DIRECTLY on the UNBALANCED FORCE (Fnet) and INVERSELY on the mass of the body. ∝ a = Fnet / m Fnet = a · m

18 First Law vs. Second Law

19

20 Fnet = a · m Fg = m · g Fk = µk · FN
A 45.0 N block is being pushed along a floor, where the coefficient of kinetic friction is If a horizontal force of 25.0 N is applied, at what rate will the block accelerate? FBD Known? Unknown? Formula(s) Fg = 45.0 N µk = 0.333 Fapp = 25.0 N Fk = ??? FN = ??? Fk = ??? Fapp = 25.0 N Fnet = a · m m = Fg/g = 45.0/9.81 = 4.59 kg Fg = 45.0 N Fg = m · g Fk = µk · FN

21 A 45.0 N block is being pushed along a floor, where the coefficient of kinetic friction is If a horizontal force of 25.0 N is applied, at what rate will the block accelerate? FBD Known? Unknown? Formula(s)

22 Classwork Page: 112 Problems: all problems

23 QUIZ On Monday: 15 minutes Problems: from 4.1 and 4.2


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