Newton’s Law(s) and Order

Chapter 5.6-5.7

Important Vocabulary:Normal ForceContact Force

“It’s not just an equation, it’s the Law!”

Which string breaks first?

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?

Problem 5-9: force and acceleration.

• M=950kg. dT=1.20s. V1=16.0m/s V2=9.5m/s

• What is the average Force?

• How far did the car travel?

F=Ma The mass is given, so need to find a.a is equal to change in velocity over time (definition).Once have “a”, get distance from master equation.

m

aTTVx

NsmkgMaF

smT

VVa

3.15

)2.1)(42.5(2

120.1*0.16

2

1

5150)/42.5)(950(

/42.5

221

2

22.10.165.912

M=950kg, V1, V2 and T=1.2s are given.What is the average Force?How far did the car travel?

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

M1g

Flift Flift

M2g

a = -0.56 m/s2

M1=1220kgM2=M1+mp

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

LIFTFgM 1

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

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

ag

aMm

aMgMgM

aMFgM

p

LIFT

1

212

22

Mr. Ed the talking horse.

?

Paradox? According to Newtons’ laws, Mr Ed says, I can’t move the cart. If I pull with force F, the wagon pulls back with an equal and opposite force F. The net force is ZERO, so the cart doesn’t move.

F-F

Mr. Ed Paradox.

1. Trust the horse, the cart doesn’t move.2. From the perspective of the cart, there is a

net force.3. From the perspective of the horse, there is

a net force.4. Both 3 and 4 are correct.

The Normal Force• “Normal” refers to the direction of the force• The Normal Force is the contact force due to gravity, acting in the

direction opposite to gravity.• When an object is moving with constant speed under the influence

of gravity, the normal force equals the force of gravity—the “weight.”

F=-Mg

N

Normal Force is not always equal to the weight

Equal

Not equal

Forces are vectors!From problem 5-23.

Fx1

Fx2

Fy1

Fy2

Fxtot=Fx1 + Fx1

Fytot=Fy1-Fy2

(NOTE: adding magnitudes)

Forces are vectors, Prob. 5-25.

Given the mass of the skier, and neglecting friction, what are the forces on the skier (direction and magnitude)?

Mg

N

Fnet

HINTS ON FINDING EQUAL ANGLES:

1. Parallel line rule.2. Perpendicular line rule.