Chapter 4Forces and Newton’s Laws of

Motion F=ma; gravity

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0) Background

• Galileo – inertia (horizontal motion) – constant acceleration (vertical motion)

• Descartes & Huygens – Conservation of momentum: mass x velocity = constant

• Kepler & Braha – laws of planetary motion (kinematics only) !

Question of the day: !

Explain planetary motion

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1) Newton’s first law: the law of inertia

Natural state is motion with constant velocity - Aristotle: rest is natural state - Galileo: circular motion (orbits) is natural state

A free object moves with constant velocity

No forces at rest or motion in a straight line

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Inertial reference frame !

- A reference frame in which the law of inertia holds !

- Requires ability to identify a free object: If no force acts on a body, a reference frame in which it has no acceleration is an inertial frame. !

- does not hold on a carousal, or an accelerating car

a

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...have the ship proceed with any speed you like, so long as the motion is uniform and not fluctuating this way and that. You will discover not the least change in all the effects named, nor could you tell from any of them whether the ship was moving or standing still. …from Dialogue Concerning the Two Chief World Systems, 1632.

Velocity is relative

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Galileo's Principle of Relativity !

The mechanical laws of physics are the same for every inertial observer. By observing the outcome of mechanical experiments,

one cannot distinguish a state of rest from a state of constant velocity. Absolute rest cannot be defined.

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2) Newton’s second law: F=ma

a) Mass - quantity of matter (determined with a balance) - quantity that resists acceleration (inertial mass) (i) Define 1 kg as mass of a standard cylinder (ii) Addition of masses (scalar): m = m1 + m2

- in particular two identical masses have twice the mass, to satisfy quantity of matter definition

(iii) Observe acceleration vs mass for a given force: mass acceleration 1 kg 1 m/s2

2 kg 1/2 m/s2

3 kg 1/3 m/s2

mass is inversely proportional to acceleration

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b) Force - push or pull - disturbs “natural” state: causes acceleration (i) Define 1 N (newton) as force required to accelerate 1 kg by 1 m/s2 (ii) Addition of forces (vector):

!Fnet =

!F1 +!F2 + ...

Identical forces in opposite direction produce no acceleration

Two identical forces at 60º produce the same acceleration as a third identical force at 0º (cos(60º)=1/2)

Two identical parallel forces corresponds to twice the force.

!Fnet =

!F∑

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(iii) Observe acceleration vs. force for a given object !

Force Acceleration 1 N 1 m/s2

2 N 2 m/s2

3 N 3 m/s2

! Force is proportional to acceleration !

(iv) Types of force: - gravity - electromagnetic - weak nuclear -strong nuclear

electroweak

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c) Second Law

Define const = 1

!F = m!a

For m = 1 kg, and a = 1 m/s2, F = 1 N by definition, so

1 N = 1 kg m/s2

(F is net force)

F ∝ a (const mass)

m ∝ 1a

(const force)} F ∝ma

F = (const)ma

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!F =

Δ(m!v)Δt

• Special case:

• F = ma can be used as the defining equation for force and inertial mass, but only because of the physical observation that force is proportional to acceleration (for a given mass), and mass is inversely proportional to acceleration (for a given force).

• Inertia is the tendency of an object not to accelerate • Newton’s second law formally refers to the rate of change of

momentum:

For constant mass,

!F = m Δ!v

Δt= m!a

!F = 0 ⇒ 0 = m!a ⇒ !a = 0 ⇒ velocity is constant (1st law)

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d) Free-body diagrams and net force Replace object(s) by dot(s). Represent all forces from the dot. Solve F=ma for each object

F1

F2FN

mg

F1

F2

m

!Fnet =

!F∑ Net force is vector

sum of all forces

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e) Components of force

!F = m!a means Fx = max & Fy∑ = may∑

sum of all forces

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Example:

m

F1 = 15 N

F2 = 17 Nθ = 67º

m = 1300 kg Find acceleration.

x

y

θ

F1

F2

Fx = F2 + F1 cosθ∑ = 23 N

Fy = F1 sinθ∑ = 14 Nax =

Fx∑m

= 0.018 m/s2

ay =Fy∑m

= 0.011 m/s2

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Two forces FA and FB are applied to an object whose mass is 8.0 kg. The larger force is FA. When both forces point due east, the object’s acceleration has a magnitude of 0.50 m/s2. However, when FA points due east and FB points due west, the acceleration is 0.40 m/s2, due east. Find (a) the magnitude of FA and (b) the magnitude of FB.

Example

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Problem solving with F=ma

• Choose an inertial reference frame – earth is close enough – measure v and a relative to this frame

• Draw free-body diagrams for each object

Tension

Weight

Normal

Friction

• Indicate direction of velocity and acceleration

Velocity

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• Resolve forces and add components

Tension

Weight

Normal

Friction Velocity

Fy = FN −W +T sinθ

FN T

fW

Fx = T cosθ − f

θ

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• Apply 2nd law

Fx = maxFy = may

• Use kinematic equations for constant accel. if force is const.

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3) Newton’s third lawFor every action, there is an equal and opposite reaction

A B

FAB

FBA

FAB = -FBA

Conservation of momentum:

0 = Δ(mB

!vB +mA!vA )

ΔtChange in momentum of an isolated system is zero

!FAB +

!FBA =

Δ(mB!vB )

Δt+ Δ(mA

!vA )Δt

!FBA =

Δ(mA!vA )

Δt

!FAB =

Δ(mB!vB )

Δt

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A chain of 3 links, each having a mass of 100 g, is pulled on a surface with an acceleration of 2.5 m/s2.

Find the forces acting between adjacent links.

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4) GravityNewton’s law of universal gravitation:

G = 6.673×10−11Nm2 /kg2

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• Spheres look like points from the outside

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�

F = mia

�

F =GmgMr2

If mi = mg,

�

Fa

= Fr2

GM

�

→ a = GMr2

= gAcceleration independent of mass

Gravitation: A very special force.

gravitational masses inertial mass

equal to at least ppt

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It is, however, clear that science is fully justified in assigning such a numerical equality only after this numerical equality is reduced to an equality of the real nature of the two concepts.” -- Einstein

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The principle of equivalenceThe happiest thought of my life… !

The gravitational field has only a relative existence... Because for an observer freely falling from the roof of a house - at least in his immediate surroundings - there exists no gravitational field. -- Einstein

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The principle of equivalence A uniform gravitational field is completely equivalent to

a uniformly accelerated reference frame. !

• No local experiment can distinguish them • Concept of inertial frame no longer useful • Gravity is geometrical

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• Weight, W – force due to earth’s (or a planet’s) gravity

�

rE = 6.38×106m

�

M e = 5.98×1024 kg

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– Near the surface, r = RE

or,

W = mg

where

g = GME

RE2 = 9.8m/s2

W = G ME

RE2 m

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Homework• C&J 4.1

C&J 4-1 An airplane has a mass of 3.1×104 kg and takes off under the influence of a constant net force of 3.7 ×104 N. What is the net force that acts on the plane’s 78 kg pilot?

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Homework• C&J 4.14

A billiard ball strikes and rebounds from the cushion of a pool table perpendicularly. The mass of the ball is 0.38 kg. The ball approaches the cushion with a velocity of 2.1 m/s and rebounds with a velocity of 2.0 m/s. The ball remains in contact with the cushion for a time of 3.3 ms. What is the average net force (magnitude and direction) exerted on the ball by the cushion?

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• C&J 4.33 (8e) Some people are riding in a hot air balloon. The combined mass of the people + balloon is 310 kg. The balloon is motionless in the air because the downward acting weight of the people + balloon is balanced by an upward acting “buoyant” force. If the buoyant force remains constant, how much mass should be dropped overboard so that the balloon acquires an upward acceleration of 0.15 m/s2?

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5) Normal Force• Perpendicular force of surface on object

– constraint

Fy = 0∑FN −W = 0FN =W

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Fy = 0∑

FN −11N −W = 0

FN =W +11N = 26N

Normal force increased

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Fy = 0∑

FN +11N −W = 0

FN =W −11N = 4N

Normal force decreased

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– Perpendicular to surface (not necessarily up)

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At rest or moving with constant velocity

FN

W

FN −W = ma = 0FN =W = 700N

Apparent weight (measured by scale) is the normal force

• Apparent weight

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Accelerating up

FN

W

FN −W = may = maFN =W + ma = 1000N

a = FN −Wm

=FN −WW

g = FNW

−1⎛⎝⎜

⎞⎠⎟g

If FN = 2W , a = g

For FN = 1000N, and W = 700N, a = 37g

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Accelerating down

FN

W

FN −W = may = −maFN =W − ma = 400N

a = W − FNm

=W − FNW

g = 1− FNW

⎛⎝⎜

⎞⎠⎟g

If FN = 0, a = g

For FN = 400N, and W = 700N, a = 37g

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Free fall

FN = 0

W

FN −W = may = −ma = −mgFN =W − mg = 0

weightlessness

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6) Friction• Parallel force of surface on

object – (between surfaces)

• Resists motion – opposite to direction of motion or

applied force – (<= applied force; does not cause

motion)

• Material dependent

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• Kinetic friction – opposes sliding motion fk = µkFN

coefficient of kinetic friction

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• Static friction – constraint

fs ≤ µsFN

fsMAX = µsFN

coefficient of static friction

fs = F if F ≤ fsMAX

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Homework • C&J 4.99 A student presses a book between his hands, as the drawing

indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs 31 N. The coefficient of static friction between his hands and the book is 0.40. To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

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7) Tension and pulleys• Tension: force exerted by rope or cable

– For an ideal line, the same force is exerted at both ends

– objects connected by a taut line have the same acceleration

• Pulley: changes direction of force – For an ideal pulley, the magnitude of the tension is

the same on both sides – magnitude of acceleration of connected objects is the

same

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• T1 = T2 = T

• a1 = a2 = a • For the example,

a1y = -a2y

• Simplify problem, by choosing sign for a sense of the motion

m1

m2

T1

T2

+a

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m1

m2

T1

T2

+a

T

m1g

T

m2g

Equations of motion:T − m1g = m1am2g − T = m2a

Solve for a (eliminate T ):

a = m2 − m1

m1 + m2

⎛⎝⎜

⎞⎠⎟g

Solve for T (eliminate a):

T =2m1m2

m1 + m2

⎛⎝⎜

⎞⎠⎟g

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m1

m2

+a

Acceleration can be determined by considering external forces (tension is an internal force holding objects together)

m2g

m1g

Fext (+ ) = m∑( )∑ a+

m2g − m1g = (m1 + m2 )a

a = m2 − m1

m1 + m2

⎛⎝⎜

⎞⎠⎟g

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8) Equilibrium applications• Equilibrium means zero acceleration • Balance forces in x and y directions

Fx = 0∑Fy∑ = 0

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Example: Find tension on leg (F)Free body diagram for pulley:

F = T1 cos35º+T2 cos35º

T

T

mg

Free body diagram for weight:

T=mg

Since T = T1 = T2 , F = 2T cos 35º= 36 N

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9) Non-equilibrium applications

• Non-equilibrium means non-zero acceleration • Determine acceleration from 2nd law:

Fx = max∑Fy = may∑

• Solve kinematic equations

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Homework• C&J 4.87 • The alarm at a fire station rings and an 86-kg fireman,

starting from rest, slides down a pole to the floor below (a distance of 4.0 m). Just before landing, his speed is 1.4 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

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A sphere of mass 3.0x10-4 kg is suspended from a cord. A steady horizontal breeze pushes the sphere so that the cord makes a constant angle of 37° with the vertical. Find (a) the push magnitude (b) the tension in the cord.

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Final Exam Dec. 2005 Q24 A block is at rest on a rough inclined plane and is connected to an object with the same mass as shown. The rope is massless and the pulley frictionless. The coefficient of static friction between the block and the plane is μs. What is the magnitude of the static frictional force acting on the block?

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C&J 4-98 Two forces, and act on the 7.00-kg block shown in the drawing. The magnitudes of the forces are F1 = 59.0 N and F2 = 33.0 N. What is the horizontal acceleration (magnitude and direction) of the block?

�

! F 1

�

! F 2

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The drawing shows a large cube (mass = 25 kg) being accelerated across a horizontal frictionless surface by a horizontal force P. A small cube (mass = 4.0 kg) is in contact with the front surface of the large cube and will slide downward unless P is sufficiently large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that P can have in order to keep the small cube from sliding downward?

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The drawing shows a large cube (mass = 25 kg) being accelerated across a horizontal frictionless surface by a horizontal force P. A small cube (mass = 4.0 kg) is in contact with the front surface of the large cube and will slide downward unless P is sufficiently large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that P can have in order to keep the small cube from sliding downward?

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Oct. 2005 Midterm test Q7 A 1.0 kg mass is released at the top of an inclined plane that is 5.0 m long and makes an angle of 32° above the horizontal. If the coefficient of friction is 0.20, what is the speed of the object when it reaches the bottom?

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