Newton’s Laws

• Objects change velocity only when they are accelerated

• Acceleration happens when an object is acted on by a force

• The acceleration is proportional to the force, and inversely proportional to the mass (F=ma)

• Forces act in equal and opposite pairs

The dark side of the force(s)……

A book is held stationary in place. Which of the following statements is most correct?

1. The book is stationary, there are no forces acting on the book.

2. The book exerts a downward force on my hand.

3. My hand exerts an upward force on the book.

4. The force of the book on my hand is equal and opposite to the force of my hand on the book.

5. Number 1 is false; 2,3 and 4 are all true.

Forces on a Bicycle

• Bicycle accelerating to the right.• Wheels are turning clockwise.• What is the force of the road on the bicycle?

? ?

What is the force of the road on the bicycle?

1. To the left.2. To the right.3. There is no left or right

force, only up.

? ?

Examples of Action-Reaction Force Pairs

FORCE PAIRS are also called CONTACT FORCES.

The total force is the same. Box 1 is twice as heavy as Box 2. Does the contact force change?

?

?

Compare the contact forces in the two cases.

1. The total force is the same. By symmetry, the contact forces are the same.

2. The contact forces are higher in the top diagram.

3. The contact forces are higher in the bottom diagram.

The contact forces matter!

• Same force, different result!• Bottom case breaks the egg, due to high contact forces.

Free body diagrams

• Two forces act on box 1 (neglecting friction)• One force acts on box 2

What are the forces?

• First step: Use F=Ma for TOTAL MASS to find acceleration, a.• Second step: Use F=Ma for MASS #1 to find Force on Box #1.• Third step: Use “equal and opposite” to find Force on Box #2.• Fourth step: Add up forces on Box #2.

Finding the forces.

• First step: Use F=Ma for TOTAL MASS to find acceleration, a.

• Second step: Use F=Ma for MASS #1 to find Force on Box #1.

• Third step: Use “equal and opposite” to find Force on Box #2.

• Fourth step: Add up forces on Box #2.

NFFF

NF

NsmkgaMF

smkg

N

M

Fa

kgMMM

TOTBox

TOT

TOT

TOT

67.63.130.20

3.13

3.13/33.10.10

/33.10.15

0.20

15

22

2

211

2

21

Reaction Forces & Acceleration

Compare the ACCELERATION of the Right Side boat to the Left.

1. They are equal because of “equal and opposite forces”.

2. The right boat has higher acceleration.

3. The right boat has lower acceleration.

Force of gravity.• Gravity (approximately) is a constant

acceleration, g, independent of mass.• Therefore, the FORCE of gravity is proportional

to the mass (Galileo). Fgravity = -Mg

F = -mgF = -mg

F = +mg

Tower of Babel

• All of the blocks have the same mass, M. They sit on top of a table.

• What is the total force on the stack of blocks?• What is the force of the table on the stack of blocks?• What are the contact forces?

What is the total force on the Tower?

1. 0Mg2. 1Mg3. 2Mg4. 3Mg

Each block has mass M.

What are the contact forces?

1. 1Mg, 2Mg, 3Mg2. 3Mg, 2Mg, 1Mg3. 2Mg, 1Mg, 0Mg4. 0Mg, 1Mg, 2Mg

F_a, F_b, and F_c are equal to:

F_c

F_b

F_a

Problem 65: Addition of Forces

• When two people push in the same direction on mass m, they cause an acceleration a1. When they push in opposite directions, the acceleration is a2.

• What is the force of each person in terms of the given information?

•Step 1: DRAW A PICTURE!•Step 2: “Givens” and “asks”•Step 3: Relationships (F=ma)

Problem 65: solution

M

F1

F2

a1

M

F1

F2

a2

221

121

MaFF

MaFF

How do you complete the problem?

Q: Find F1, F2.

Hint: Add and subtract the equations.

Problem 65: Given a1 and a2, the ratio F2/F1 is:

1. a2/a1

2. a1/a2

3. (a1-a2)/(a1+a2)

4. (a1+a2)/(a1-a2)

Problem 63: Hot air balloon

• A balloon with some passengers hovers motionless at a total mass of 1220kg. A last passenger climbs aboard, and the balloon sinks at 0.56 m/s2.

• What was the mass of the last passenger?

•Step 1: DRAW A PICTURE!•Step 2: “Givens” and “asks”•Step 3: Relationships (F=ma)

Prob. 63: Setting it up.

• Givens: M1 and a

• Asks: Mp

F1

Flift Flift

F2

a = -0.56 m/s2

M1=1220kgM2=M1+Mp

Prob. 63: Solution

F1

Flift Flift

F2

a = -0.56 m/s2

M1=1220kgM2=M1+Mp

Initially, acceleration is zero, so force of gravity is equal to force of lift.

LIFTFF 1

We know the force of gravity initially, since the mass is given.

gMFF LIFT 11

After the last passenger loads, the difference between the force of gravity F2 and the lifting force leads to an acceleration, a. aMgMgM

gMF

aMFF LIFT

212

22

22

Getting hammered?

1. Left side picture is better.2. Right side picture is

better.3. Neither one has an

advantage, the forces are the same.

Which picture describes the better approach for tightening a loose hammer-head?