Unit 5: Net Force Notes

Net Force, 2nd Law of Motion, Friction, Study Guide Problems

Review of Unit 4: Newton’s 1st & 3rd Law

Newton’s First Law

What is equilibrium?

• Objects that are either at rest or moving with constant

velocity are said to be in equilibrium.

• Equilibrium is the state in which there is no change in a

objects motion.

• An object is in equilibrium when the vector sum of the

forces acting on it is equal to zero.

Fnet = Sum of the forces (∑ F) X axis & y axis are treated separately 1st Law At constant velocity or speed, Fnet = 0

Review

1. There is no friction between the cart and the table. When the cart reaches point B the string breaks. Describe the motion of the cart across the frictionless table? 2. Describes the motion of the cart while it is pulled with a constant force across the table? The cart accelerates constantly (2nd law). 3. Now assume there is friction and the cart is moving. True or false. Explain. At constant velocity the pulling force is equals the friction forces. True. At constant velocity forces are balanced. (1st law). See the force diagram, next slide.

When the cart is speeding up or slowing down, it is acceleration and the forces are unbalanced. (2nd law). See the force diagram, next slide.

The cart moves at constant speed (1st law).

Review Force Diagrams A: Constant Velocity

FN FN,

Fg

FT Ff

B: Slowing down

Ff

Fg

FT

A: At constant velocity the forces are balanced & Fnet = 0.

B: the forces on the y-axis are balanced so the object is at constant velocity with respect to the y-axis. On the x-axis, the forces are unbalanced. Thus the cart is slowing down (accelerating to the left) and Fnet-x = mass • acceleration (F=ma)

Newton’s 2nd Law Explain Newton’s second law. • When an object is not at constant velocity, it is accelerating.

• Unbalanced forces cause acceleration.

• The amount by which the unbalanced forces acting on an object is called

the net force (Fnet).

Fnet = mass •acceleration or F = ma a) The acceleration of an object is directly proportional (∝) to

the net force acting on it, 𝒂 ∝ 𝑭𝒏𝒆𝒕 if mass is constant. If acceleration doubles, 𝑭𝒏𝒆𝒕 doubles

b) 𝒎 ∝ 𝑭𝒏𝒆𝒕 if acceleration is constant. If m is decreased by ¼, 𝑭𝒏𝒆𝒕 is decreased by ¼

c) Acceleration is inversely proportional to its mass if 𝑭𝒏𝒆𝒕 is

constant: 𝒂 ∝𝟏

𝒎𝒂𝒔𝒔

If m is decreased by ¼, ‘a’ increases by 4

If m is doubled, ‘a’ decreases by 1/2

∝ = proportional

Relationship Between Acceleration, Net Force and Mass

1. Which system has the greatest net force? Explain how you

know.

The tension in the string comes from the force of gravity which

equals the weight. W=mg. Thus Fnet = mg. C has the most mass

hanging down so it has the most weight so C has the most force.

2. Which system has the least inertia? Explain how you know. Inertia is resistance to change in motion, and is determined by the

total mass of the system, in other words, all the mass is added

together. System A has the least mass and the least inertia.

System B & system C have the same amount of system mass.

Relationship Between Acceleration, Net Force and Mass

3. Determine the acceleration for each system. Each block has M kilograms of mass. Net Force on the system = Fg = w =mg.

g = 10 m/s/s (or m/s2). System mass = total mass for each system.

System A,

System B,

System C,

Fnet = ma

Fnet = ma

Fnet = ma

Newton’s 1st Law, Fnet=0 Objective: When the forces acting on an object are balanced, the object move at constant velocity. 1. An elevator is moving up at a constant velocity of 2.5 m/s. The passenger

has a mass of 85 kg. Construct a force diagram for the passenger. Calculate the force the floor exerts on the passenger (the normal force).

Fnet = Sum of the forces (∑ F) X axis & y axis are treated separately

Fnet = Fg + FN Fg is in the negative direction

Fnet = (-Fg) + FN

By the 1st Law, Fnet = 0 at constant speed

(-Fg) + FN = 0

FN = Fg Fg = Weight (w) w= m• g g is constant, 10 N/kg or 10 m/s2

85kg • 10 N/kg = 850N FN = Fg = w= 850N

Newton’s 2nd Law (Fnet = ma) An elevator accelerates downward at 3.0 m/s2 A. Construct a force diagram for the 85kg passenger.

B. Write an equation for the vertical forces on the passenger. Fnet = FN + (- Fg )

Fnet = ma

FN + Fg = ma

FN + (- Fg ) = ma FN = ma – (- Fg) C. Calculate the force the floor exerts on the passenger.

FN = m(a) – (- Fg)

85kg (-𝟑𝒎

𝒔𝟐 ) – (-850N) = 595N

Fg = w = mg = 85kg (10 N/kg)= 850N

=Fnet ‘a’ is downward = -3 m/s2, m = 85kg

x

y

Fg

FN

Fnet = ma +

FN

Fnet = ma

Fg -y

Fg-x

y

x

Fpull = 0.5 N

Fg

The acceleration for cart A is _____ (>,<,=) than the acceleration of cart B.

Fnet = Sum of the forces (∑ F) X axis & y axis are treated separately For A: Fnet-x = Fpull + Fg ║

For B Fnet-x = Fpull Fnet A is > Fnet B 2nd law: 𝒂 ∝ 𝑭𝒏𝒆𝒕 Acceleration of A > B

A: B:

A:

B: Fg Fpull = 0.5 N

FN

Fg -y

Fg-x

Fg

Fpull = 0.5 N

FN

Fp -y

Fp-x

Fg

Fpull = 0.5 N

FN

FN

Newton’s 3rd Law 1. Students are sitting in identical office chairs.

Student “A" then suddenly pushes outward on student “B” with his feet, causing both chairs to move.

a. See force diagrams on next slide. b. True of false: each student exerts a force on the other, but "A" exerts the larger force. Explain you answer.

False. Each student exerts the same amount of force on the other because the “pushes” are 3rd Law pairs.

To be a 3rd law pair the forces must be the same force (push), but in opposite direction, and involve the same two objects (“A” & “B”). • 3rd law force pairs are always equal in magnitude (absolute value) • Newton’s 3rd Law states that all forces come in pairs; paired forces are

equal in magnitude, but opposite in direction. FA, B = -F B, A

IB Physics, Bell Work, Tuesday, April 14

A: FN, B, chair FN, A, chair

Fg, B, E,

FPush, B, A, Ff, B, floor,

B:

FPush, A, B,

Fg, A, E,

Ff, A, floor,

The 3rd law force pairs are circled. 3rd law pairs have the same force, but

in opposite directions ,with the same two objects, and are always on

different force diagrams. The force pairs are numerically equal (same

magnitude) to each other.

*

□

Fnet

you

FN, y, f

=

=

Fg, y,

FN, y, c Ffs, y, f

U4, Review Sheet #4

4. You accelerate a grocery cart along a level floor in the presence of friction effects between the cart and the floor. Draw force diagrams for you, and the cart. Fully label all vectors. Show the force pairs.

you

Fp, y, f

E

Friction Objective: • For a given pair of surfaces, the friction force is generally some

fraction of the normal force, where that fraction is called the coefficient of friction, m (greek letter mu).

• Therefore, Ffriction = mFN. Every pair of surfaces has its own coefficient of friction.

Two types of friction Kinetic Friction is a type of friction that acts on moving objects when one or both surfaces are moving. • This force is exerted on one surface by another when the two

surfaces rub against each other because one or both of them are moving.

Static friction is the force exerted on one surface by another when there is no motion between the two surfaces.

Friction The relationship between friction and the normal force:

Friction, f ∝ 𝑭𝑵

– if Normal force doubles, the friction doubles.

– The harder one object is pushed against the other, the

greater the force of friction that results.

The coefficient of friction.

– Frictional force also depends on the materials that the

surfaces are made of.

– If one graphs friction vs. normal force, the slope of the line

is the coefficient of friction. Thus,

Static Friction Force

Kinetic Friction Force

Friction Problem from Worksheet 5 3. A horizontal 100 N force is applied to a 50 kg classmate resting on a level tile floor. The coefficient of kinetic friction is 0.15. What is the acceleration of the classmate?

FN = Fg Fg = w w = mg FN = mg FN = 50kg(10N/Kg) = 500N

Ffk = μk FN

Fnet = 100N + (- 75 N) = ma

= 0.15• 500N = 75N

= 500N

= 75N

4. Suppose the classmate in was resting on a carpet where the coefficient of static friction is 0.25. Is the horizontal 100 N force sufficient to cause the classmate to accelerate? Ffs = μs FN = (0.25)(500N) = 125N FA < the max static Ff , classmate will not move.