Physics 201: Lecture 9, Pg 1 Lecture 8 l Goals: Solve 1D & 2D problems introducing forces...

Physics 201: Lecture 9, Pg 1

Lecture 8

Goals:Goals:

Solve 1D & 2D problems introducing forces with/without friction

Utilize Newton’s 1st & 2nd Laws

Begin to use Newton’s 3rd Law in problem solving


Newton’s Laws

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.)

Read: Force on A by B


Exercise Newton’s 3rd Law

A. 2

B. 4

C. 6

D. Something else

a b

Two blocks are being pushed by a massless finger on a horizontal frictionless floor.

How many action-reaction force pairs are present in this exercise?


Newton’s LawThe acceleration of an object is directly proportional to the

net external force acting upon it.

The constant of proportionality is the mass.

This is a vector expression

kji zyx FFFF

zz

yy

xx

maF

maF

maF

NET

NET

NET


Analyzing Forces: Free Body Diagram

A heavy sign is hung between two poles by a rope at each corner extending to the poles.

Eat at Bucky’s

A hanging sign is an example of static equilibrium (depends on observer)

What are the forces on the sign and how are they related if the sign is stationary (or moving with constant velocity) in an inertial reference frame ?


Free Body Diagram

Step two: Sketch in force vectorsStep three: Apply Newton’s 1st & 2nd Laws (Resolve vectors into appropriate components)

mg

T1

T2Eat at Bucky’s

T1

T2

mg

Step one: Define the system


Free Body Diagram

Eat at Bucky’s

T1 T2

mg

Vertical : y-direction 0 = -mg + T1 sin1 + T2

sin2

Horizontal : x-direction 0 = -T1 cos1 + T2 cos2


Pushing and Pulling Forces

String or ropes are examples of things that can pull

You arm is an example of an object that can push or push


Scale Problem

You are given a 5.0 kg mass and you hang it directly on a fish scale and it reads 50 N (g is 10 m/s2).

Now you use this mass in a second experiment in which the 5.0 kg mass hangs from a massless string passing over a massless, frictionless pulley and is anchored to the floor. The pulley is attached to the fish scale.

What force does the fish scale now read?

5.0 kg

50 N

?

5.0 kg


Scale Problem

Step 1: Identify the system(s).

In this case it is probably best to treat each object as a distinct element and draw three force body diagrams. One around the scale One around the massless pulley (even

though massless we can treat is as an “object”)

One around the hanging mass

Step 2: Draw the three FBGs. (Because this is a now a one-dimensional problem we need only consider forces in the y-direction.)

?

5.0 kg


Scale Problem

Fy = 0 in all cases

1: 0 = -2T + T ’

2: 0 = T – mg T = mg

3: 0 = T” – W – T ’ (not useful here) Substituting 2 into 1 yields T ’ = 2mg = 100 N

(We start with 50 N but end with 100 N)

?

5.0 kg

? 1.0 kg

-mg

T

-T -T

-T ’

T”

W

1:2:3:T ’


Newton’s 2nd Law, Forces are conditional

A. P + C < W

B. P + C > W

C. P = C

D. P + C = W

A woman is straining to lift a large crate, without success. It is too heavy. We denote the forces on the crate as follows: P is the upward force being exerted on the crate by the personC is the contact or normal force on the crate by the floor, and W is the weight (force of the earth on the crate). Which of following relationships between these forces is true, while the person is trying unsuccessfully to lift the crate? (Note: force up is positive & down is negative)


Another experiment

A block is connected by a horizontal massless string. There is a known mass 2.0 kg and you apply a constant force of 10 N. What is the acceleration of the block if the table top is frictionless?

Fx = max = -T

Fy = 0 = N - mg

ax = -T/m = 5 m/s2

m

1T

mg

FBD N


Version 2 A block is connected by a horizontal massless string. There

is a known mass 2.0 kg and you apply a constant horizontal force of 10 N. The surface is frictionless and it is inclined 30° from horizontal.

What is the acceleration of the block down the slide?

Fx = max = Tx+Wx

Fy = 0 = N + Ty+Wy

max = -T cos 30°+W sin 30°

ax = -T/m 3½/2 - g / 2

ax = -5(1.732)/2- 5 = -7.2 m/s2

What is the apparent weight of the block on the slide?

N = Ty+Wy = T sin 30° -W cos 30°

N = -10(0.5) + 20(0.866) = 12.3 N

mT

mg = W

FBDN

i j

30° 30°

30°


Version 3 A block is connected by a horizontal massless string. There

is a known mass 2.0 kg and you apply a constant horizontal force of 34.6 N. The surface is frictionless and it is inclined 30° from horizontal.

What is the acceleration of the block down the slide?

Fx = max = Tx+Wx

Fy = 0 = N + Ty+Wy

max = -T cos 30°+W sin 30°

ax = -T/m 3½/2 - g / 2

ax = -17.3(1.732)/ 2- 5 = -35 m/s2

What is the apparent weight of the block on the slide?

N = Ty+Wy = T sin 30° -W cos 30°

N = -34.6(0.5) + 20(0.866) = 0.0 N !!!

mT

mg = W

FBDN

i j

30° 30°

30°


Another experiment: A modified Atwood’s machine

Two blocks, m1 & m2, are connected by a massless frictionless string/pulley on the table as shown. The table surface is frictionless and little g acts downward.

What is the acceleration of the horizontal block?

Requires two FBDs and

Newton’s 3rd Law.

m1

m2

m2g

N

m1g

T

T

Mass 2

Fx = m2a2x = -T

Fy = 0 = N – m2g

Mass 1

Fy = m1a1y= T – m1g

Correlated motion:

|a1y| = | a2x| ≡ a

If m1 moves up moves m2 right


Another experiment: A modified Atwood’s machine

Two blocks, m1 & m2, are connected by a massless frictionless string/pulley on the table as shown. The table surface is frictionless and little g acts downward.

What is the acceleration of the horizontal block.

Requires two FBDs and

Newton’s 3rd Law.

m1

m2

m2g

N

m1g

T

T

Mass 2

B. Fx = m2 a = -T

Fy = 0 = N – m2g

Mass 1

A. Fy = m1 a= T – m1gSubbing in T from B into A

m1 a= - m2 a – m1g

m1 a + m2 a = m1g

a = m1g / (m1 + m2 )


A “special” contact force: Friction

What does it do? It opposes motion (velocity, actual or that which

would occur if friction were absent!) How do we characterize this in terms we have learned?

Friction results in a force in a direction opposite to the direction of motion (actual or, if static, then “inferred”)!

maFFAPPLIED

ffFRICTION mgg

NN

i

j


If no acceleration

No net force So frictional force just equals applied force Key point: It is conditional!

FFAPPLIED

ffFRICTION mgg

NN

ii

j j


Friction...

Friction is caused by the “microscopic” interactions between the two surfaces:


Friction: Static frictionStatic equilibrium: A block with a horizontal force F applied,

As F increases so does fs

Fm

1

FBD

fs

N

mg

Fx = 0 = -F + fs fs = F

Fy = 0 = - N + mg N = mg


Static friction, at maximum (just before slipping)

Equilibrium: A block, mass m, with a horizontal force F applied,

Direction: A force vector to the normal force vector N N and the vector is opposite to the direction of acceleration if were 0.

Magnitude: fS is proportional to the magnitude of N

fs = s N

Fm fs

N

mg


Kinetic or Sliding friction (fk < fs)

Dynamic equilibrium, moving but acceleration is still zero

As F increases fk remains nearly constant

(but now there acceleration is acceleration)

Fm

1

FBD

fk

N

mg

Fx = 0 = -F + fk fk = F

Fy = 0 = - N + mg N = mg v

fk = k N


Model of Static Friction (simple case)

Magnitude:

f is proportional to the applied forces such that

fs ≤ s N s called the “coefficient of static friction”

Direction: Opposite to the direction of system acceleration if were 0


Sliding Friction

Direction: A force vector to the normal force vector N N and the vector is opposite to the velocity.

Magnitude: ffk is proportional to the magnitude of N N

ffk = k N N ( = Kmg g in the previous example)

The constant k is called the “coefficient of kinetic friction”

Logic dictates that S > K for any system


Coefficients of Friction

Material on Material s = static friction k = kinetic friction

steel / steel 0.6 0.4

add grease to steel 0.1 0.05

metal / ice 0.022 0.02

brake lining / iron 0.4 0.3

tire / dry pavement 0.9 0.8

tire / wet pavement 0.8 0.7


Other Forces are Conditional

Notice what happens if we change the direction of the applied force

The normal force can increase or decrease

Let a=0

F

fF mg

Nii

j j


An experiment

Two blocks are connected on the table as shown. The

table has unknown static and kinetic friction coefficients.

Design an experiment to find S

Static equilibrium: Set m2 and add mass to m1 to reach the breaking point.

Requires two FBDsm1

m2

m2g

N

m1g

T

T

Mass 2

Fx = 0 = -T + fs = -T + S N

Fy = 0 = N – m2g

fS

Mass 1

Fy = 0 = T – m1g

T = m1g = S m2g S = m1/m2


Static Friction with a bicycle wheel (not simple)

You are pedaling hard and the bicycle is speeding up.

What is the direction of the frictional force?

You are breaking and the bicycle is slowing down

What is the direction of the frictional force?


For Thursday

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Physics 201: Lecture 9, Pg 1 Lecture 8 l Goals: Solve 1D & 2D problems introducing forces...

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