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Lecture 8

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- Goals:
- Differentiate between Newton’s 1st, 2nd and 3rd Laws
- Use Newton’s 3rd Law in problem solving

Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday)

Finish Chapter 7

1st Exam Wed., Feb. 18th from 7:15-8:45 PM Chapters 1-7

in room 2103 Chamberlin Hall

Static Equilibrium Case

Dynamic Equilibrium (see 1)

Dynamic case with non-zero acceleration

“Normal” means perpendicular

Normal

Force

Friction Force

f

S F = 0

Fx= 0 = mg sin q – f

Fy= 0 = mg cos q – N

with mg sin q = f ≤ mS N

if mg sin q> mS N, must slide

Critical angle ms = tan qc

mg sin q

q

y

mg cos q

q

q

x

Block weight is mg

Static Equilibrium Case

Dynamic Equilibrium

Friction opposite velocity

(down the incline)

“Normal” means perpendicular

Normal

Force

v

Friction Force

fK

S F = 0

Fx= 0 = mg sin q – fk

Fy= 0 = mg cos q – N

fk= mk N = mk mg cos q

Fx= 0 = mg sin q – mk mg cos q

mk = tan q(only one angle)

mg sin q

q

y

mg cos q

q

q

x

mg

Normal

Force

Friction Force

Sliding Down

v

mg sin q

fk

Sliding

Up

q

q

Weight of block ismg

3. Dynamic case with non-zero acceleration

Result depends on direction of velocity

Fx= max = mg sin q± fk

Fy= 0 = mg cos q – N

fk= mk N = mk mg cos q

Fx= max = mg sin q±mk mg cos q

ax = g sin q±mk g cos q

- With a cross sectional area, A (in m2), coefficient of drag of 1.0 (most objects), sea-level density of air, and velocity, v(m/s), the drag force is:
D = ½ C Av2 c A v2in Newtons

c = ¼ kg/m3

In falling, when D = mg, then at terminal velocity

- Example: Bicycling at 10 m/s (22 m.p.h.), with projected area of 0.5 m2 exerts ~30 Newtons
- Minimizing drag is often important

- Terminal velocity reached when Fdrag = Fgrav (= mg)
- For 75 kg person with a frontal area of 0.5 m2,
vterm 50 m/s, or 110 mph

which is reached in about 5 seconds, over 125 m of fall

Read: Force of B on A

Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.

Law 2: For any object, FNET = F = ma

Law 3: Forces occur in pairs: FA , B = - FB , A

(For every action there is an equal and opposite reaction.)

If object 1 exerts a force on object 2 (F2,1) then object 2 exerts an equal and opposite force on object 1 (F1,2)

F1,2 = -F2,1

For every “action” there is an equal and opposite “reaction”

IMPORTANT:

Newton’s 3rd law concerns force pairs which

act ontwo different objects(not on the same object) !

Newton also recognized that gravity is an attractive, long-range force between any two objects.

When two objects with masses m1and m2 are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:

F = m1 g = G m1 m2 / r2

g = G m2 / r2

If we know big G, little g and r then will can find m2 the mass of the Earth!!!

FB,E = - mB g

FB,E = - mB g

FE,B = mB g

FE,B = mB g

EARTH

Consider the forces on an object undergoing projectile motion

Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi)

Consider the following two cases (a falling ball and ball on table),

Compare and contrast Free Body Diagram

and

Action-Reaction Force Pair sketch

mg

FB,T= N

mg

The Free Body Diagram

Ball Falls

For Static Situation

N = mg

FB,T

FT,B

Certain forces act to keep an object in place.

These have what ever force needed to balance all others (until a breaking point).

Main goal at this point : Identify force pairs and apply Newton’s third law

FB,E = -mg

FB,T= N

FT,B= -N

FB,E = -mg

FE,B = mg

FE,B = mg

First: Free-body diagram

Second: Action/reaction pair forces

- You have a bird in a cage that is resting on your upward turned palm. The cage is completely sealed to the outside (at least while we run the experiment!). The bird is initially sitting at rest on the perch. It decides it needs a bit of exercise and starts to fly. Question: How does the weight of the cage plus bird vary when the bird is flying up, when the bird is flying sideways, when the bird is flying down?
- So, what is holding the airplane up in the sky?

- greater than
- equal to
- less than

A fly is deformed by hitting the windshield of a speeding bus.

v

The force exerted by the bus on the fly is,

that exerted by the fly on the bus.

- greater than
- equal to
- less than

Same scenario but now we examine the accelerations

A fly is deformed by hitting the windshield of a speeding bus.

v

The magnitude of the acceleration, due to this collision, of the bus is

that of the fly.

By Newton’s third law these two forces form an interaction pair which are equal (but in opposing directions).

Thus the forces are the same

However, by Newton’s second law Fnet = ma or a = Fnet/m.

So Fb, f = -Ff, b = F0

but |abus | = |F0 / mbus | << | afly | = | F0/mfly |

Answer for acceleration is (C)

a

b

- 2
- 4
- 6
- Something else

- Two blocks are being pushed by a finger on a horizontal frictionless floor.
- How many action-reaction force pairs are present in this exercise?

Fa,f

Fb,a

Ff,a

Fa,b

FE,a

FE,b

Fg,b

Fg,a

Fb,g

Fa,g

Fa,E

Fb,E

a

b

6

v

- A box of mass m1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction
(mk= 0.5) on top of a second box having mass m2 = 2 kg, which in turn slides on a smooth (frictionless) surface.

- What is the acceleration of the secondbox ?
Central question: What is force on mass 2?

(A) a = 0 N (B) a = 5 N (C) a = 20 N(D) can’t tell

- What is the acceleration of the secondbox ?

slides with friction (mk=0.5)

T

m1

a = ?

m2

slides without friction

v

- First draw FBD of the top box:

N1

m1

fk = mKN1 = mKm1g

T

m1g

- As we just saw, this force is due to friction:

- Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2.

Action

Reaction

f2,1= -f1,2

f1,2 = mKm1g = 5 N

m1

m2

(A) a = 0 N(B) a = 5 N (C) a = 20 N(D) can’t tell

- Now consider the FBD of box 2:

N2

f2,1 = mkm1g

m2

m1g

m2g

- Finally, solve Fx = ma in the horizontal direction:

mK m1g = m2a

= 2.5 m/s2

f2,1 = mKm1g

m2

- A box of mass m1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and mk= 0.5.The second box can slide freely (frictionless) on an smooth surface.
Compare the acceleration of box 1 to the acceleration of box 2 ?

a1

friction coefficients ms=1.5andmk=0.5

T

m1

a2

slides without friction

m2

a1

friction coefficients ms=1.5andmk=0.5

T

m1

a2

slides without friction

m2

- A box of mass m1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and mk= 0.5.The second box can slide freely on an smooth surface (frictionless).
If no slippage then the maximum frictional force between 1 & 2 is

(A) 20 N (B) 15 N (C) 5 N (D) depends on T

fs = 10 N and the acceleration of box 1 is

Acceleration of box 2 equals that of box 1, with |a| = |T| / (m1+m2) and the frictional force f ism2a

(Notice that if T were in excess of 15 N then it would break free)

N

fSS N = S m1 g = 1.5 x 1 kg x 10 m/s2

which is 15 N (so m2 can’t break free)

fS

T

m1 g

a1

friction coefficients ms=1.5andmk=0.5

T

m1

a2

slides without friction

m2

Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday)

Finish Chapter 7

1st Exam Wed., Feb. 18th from 7:15-8:45 PM Chapters 1-7

in room 2103 Chamberlin Hall