Physics 111: Lecture 8, Pg 1 Physics 111: Lecture 8 Today’s Agenda l Friction Recap l Drag Forces çTerminal speed l Dynamics of many-body systems çAtwood’s

Physics 111: Lecture 8, Pg 1

Physics 111: Lecture 8

Today’s Agenda

Friction Recap Drag Forces

Terminal speed Dynamics of many-body systems

Atwood’s machineGeneral case of two attached blocks on inclined planesSome interesting problems


Friction Review:

Friction is caused by the “microscopic” interactions between the two surfaces:See discussion in text


Model for Friction

Arah vektor gaya gesek fF adalah tegak lurus dengan vektor gaya normal N, dan berlawanan arah dengan gaya yang bekerja pada sistem

Gesekan Kinetic (sliding): Besarnya vektor gaya gesek berbanding lurus dengan besarnya gaya normal N.

fF = KN

Gesekan Static: Gaya gesekan menyeimbangkan total yang yang bekerja pada sistem sehingga sistem tidak bergerak. Besarnya gaya gesek statis maksimum berbanding lurus dengan gaya normal N.

fF SN


Kinetic Friction:

K koefisien gesek kinetik.

i : F KN = ma

j : N = mg

so F Kmg = ma

maF

mg

N

i

j

KN


Static Friction:

Koefisien gesek statis S, menentukan gaya gesek statis maksimum, SN, yang timbul antara persentuhan dua buah benda.

S dapat ditentukan dengan cara memperbesar F sampai benda mulai bergeser:

FMAX - SN = 0 N = mg

FMAX = S mg S FMAX / mg

FMAX

mg

N

i

j

S N


Lecture 8, Act 1Two-body dynamics

Sebuah balok bermassa m, jika diletakkan pada permukaan bidang miring yang kasar (m > 0) dan diberikan gaya sesaat, akan bergerak turun pada bidang miring dengan kecepatan konstan. Jika balok yang sejenis(same m) denganmassa 2m diletakkan pada bidang miring yang sama kemudian diberikan gaya sesaat, maka balok tersebut akan:

(a) berhenti

(b) dipercepat

(c) Bergerak dengan kecepatan tetapm


Lecture 8, Act 1Solution

Gambarkan diagram benda bebas dan tentukan gaya total dalam arah X

q

i

j

mg

N

q

mKN

FNET,X = mg sin - q mKmg cos q

= ma = 0 (kasus balok pertama)

Jika massa digandakanmaka semua komponen menjadi dua kaliLebih besar. Tetapi percepatan tetap nol

Speed will still be constant!


Drag Forces:

When an object moves through a viscous medium, like air or water, the medium exerts a “drag” or “retarding” force that opposes the motion of the object.

Fg = mg

FDRAG

j

v


Drag Forces:

This drag force is typically proportional to the speed v of the object raised to some power. This will result in a maximum (terminal) speed.

Fg = mg

FD = bvn

j

v feels like n=2

Parachute


Terminal Speed:

Suppose FD = bv2. Sally jumps out of a plane and after falling for a while her downward speed is a constant v.What is FD after she reaches this terminal speed?What is the terminal speed v?

FTOT = FD - mg = ma = 0.

FD = mg

Since FD = bv2

bv2 = mgFg = mg

FD = bv2

j

v

bmg

v =


Many-body Dynamics

Systems made up of more than one object

Objects are typically connected:

By ropes & pulleys today

By rods, springs, etc. later on


Atwood’s Machine:

Find the accelerations, a1 and a2, of the masses.

What is the tension in the string T ?

Masses m1 and m2 are attached to an ideal massless string and hung as shown around an ideal massless pulley.

Fixed Pulley

m1

m2

j

a1

a2

T1T2


Atwood’s Machine... Draw free body diagrams for each object Applying Newton’s Second Law: ( j -components)

T1 - m1g = m1a1

T2 - m2g = m2a2

But T1 = T2 = T since pulley is ideal

and a1 = -a2 = -a.since the masses are connected by the string

m2gm1g

Free Body Diagrams

T1 T2

ja1 a2


T - m1g = -m1 a (a)

T - m2g = m2 a (b) Two equations & two unknowns

we can solve for both unknowns (T and a).

subtract (b) - (a): g(m1 - m2 ) = a(m1+ m2 )

a =

add (b) + (a): 2T - g(m1 + m2 ) = -a(m1 - m2 ) =T = 2gm1m2 / (m1 + m2 )

Atwood’s Machine...

21

221

mm

)mm(g

+

--

g)mm(

)mm(

21

21

+-


Atwood’s Machine...

m1

m2

j

a

a

TT

So we find:

am m

m mg

( )

( )1 2

1 2

g)mm(

mm2T

21

21

+=

Atwood’s Machine


Is the result reasonable? Check limiting cases!

Special cases:

i.) m1 = m2 = m a = 0 and T = mg. OK!

ii.) m2 or m1 = 0 |a| = g and T= 0. OK!

Atwood’s machine can be used to determine g (by measuring the acceleration a for given masses).

a)mm(

)mm(g

12

12 +=

-

am m

m mg

( )

( )1 2

1 2

gmm

mm2T

21

21

)(


Attached bodies on two inclined planes

All surfaces frictionless

m1

m2

smooth peg

1 2


How will the bodies move?

From the free body diagrams for each body, and the chosen coordinate system for each block, we can apply Newton’s Second Law:

Taking “x” components:

1) T1 - m1g sin 1 = m1 a1X

2) T2 - m2g sin 2 = m2 a2X

But T1 = T2 = T

and a1X = -a2X = a

(constraints)

m1

yx

T1

N

m1g

m2

m2g

T2

Nx y

1

2


Solving the equations

Using the constraints, solve the equations.

T - m1gsin 1 = -m1 a (a)

T - m2gsin 2 = m2 a (b)

Subtracting (a) from (b) gives:

m1gsin 1 - m2gsin 2 = (m1+m2 )a

So: -a

m m

m mg =

+1 1 2 2

1 2

sin sinq q


Special Case 1:

m1m2

1 2

m1 m2

If 1 = 0 and 2 = 0, a = 0.

Boring

-a

m m

m mg=

+1 1 2 2

1 2

sin sinq q


Special Case 2:

If 1 = 90 and 2 = 90, am m

m mg

( )

( )1 2

1 2

m2

TT

m1

Atwood’s Machine

m1m2

1 2

-a

m m

m mg=

+1 1 2 2

1 2

sin sinq q


Special Case 3:

If 1 = 0 and 2 = 90, am

m mg

2

1 2( )

m1

m2

Lab configuration

m1m2

1 2

-a

m m

m mg=

+1 1 2 2

1 2

sin sinq q

-

Air-track



In which case does block m experience a larger acceleration? In (1) there is a 10 kg mass hanging from a rope. In (2) a hand is providing a constant downward force of 98.1 N. In both cases the ropes and pulleys are massless.

(a) Case (1) (b) Case (2) (c) same

m

10kga

m

a

F = 98.1 N

Case (1) Case (2)


Lecture 8, Act 2 Solution

m

10kga

Add (a) and (b):

98.1 N = (m + 10kg)a

kg10m

N198a

.

Note:kg10m

mN198T

.

(a)

(b)

T = ma (a)

(10kg)g -T = (10kg)a (b)

For case (1) draw FBD and write FNET = ma for each block:



The answer is (b) Case (2)

T = 98.1 N = mam

N198a

. For case (2)

m

10kga

m

a

F = 98.1 N

Case (1) Case (2)

kg10m

N198a

.

m

N198a

.


Problem: Two strings & Two Masses onhorizontal frictionless floor:

m2 m1T2 T1

Given T1, m1 and m2, what are a and T2?

T1 - T2 = m1a (a)

T2 = m2a (b)

Add (a) + (b):

T1 = (m1 + m2)a a

a

i

21

1

mm

T

+=

Plugging solution into (b):

21

212 mm

mTT

+=



Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings?

(a) T1 > T2 > T3 (b) T3 > T2 > T1 (c) T1 = T2 = T3

T3 T2 T13m 2m m

a


Lecture 8, Act 3Solution

Draw free body diagrams!!

T33mT3 = 3ma

T3 T22mT2 - T3 = 2ma

T2 = 2ma +T3 > T3

T2 T1m

T1 - T2 = ma

T1 = ma + T2 > T2

T1 > T2 > T3



Alternative solution:

T3 T2 T13m 2m m

a

Consider T1 to be pulling all the boxes

T3 T2 T13m 2m m

a

T2 is pulling only the boxes of mass 3m and 2m

T3 T2 T13m 2m m

a

T3 is pulling only the box of mass 3m

T1 > T2 > T3


Problem: Rotating puck & weight.

A mass m1 slides in a circular path with speed v on a horizontal frictionless table. It is held at a radius R by a string threaded through a frictionless hole at the center of the table. At the other end of the string hangs a second mass m2.What is the tension (T) in the string?What is the speed (v) of the sliding mass?

m1

m2

v

R


Problem: Rotating puck & weight...

Draw FBD of hanging mass:Since R is constant, a = 0.

so T = m2g

m2

m2g

T

m1

m2

v

R

T


Problem: Rotating puck & weight...

Draw FBD of sliding mass:

m1

T = m2g

v gRm

m 2

1

m1g

N

m1

m2

v

R

T

Use F = T = m1a

where a = v2 / R

m2g = m1v2 / R

T = m2g

Puck


Recap of today’s lecture

Friction Recap. (Text: 5-1) Drag Forces. (Text: 5-3)

Terminal speed.

Dynamics of many-body systems. (Text: 4-7)Atwood’s machine.General case of two attached blocks on inclined planes.Some interesting special cases.

Look at Textbook problems Chapter 6: # 3, 7, 21, 75

Documents

Physics 111: Lecture 8, Pg 1 Physics 111: Lecture 8 Today’s Agenda l Friction Recap l Drag Forces çTerminal speed l Dynamics of many-body systems çAtwood’s