Last%TimeCh4%-%Today Ch%5people.physics.tamu.edu/tyana/PHYS218/files/Lect_10_10-1-14_post.pdf · Last%TimeCh4%-%Today Ch%5 Today& • Chapter&5& o Using&Newton’s&1stforbodiesin

Last Time Ch 4 -‐ Today Ch 5

Today •  Chapter 5

o  Using Newton’s 1st for bodies in equilibrium

o  Using Newton’s 2nd Law for accelera@ng bodies

o  Sta@c and kine@c fric@on o  Problems involving circular mo@on

Last Week •  Chapter 4

o  Forces o  Newton’s 1st Law o  Newton’s 2nd Law o  Newton’s 3rd Law o  Free body diagrams

T. S@egler 10/01/2013 Texas A&M University

Newton’s 1st : If there is no net force on an object then there is no change in the object’s mo@on

Newton’s 2nd : If you apply a force to an object then that object would experience an accelera@on in the direc@on of that force. This accelera@on was inversely propor@onal to the mass being acted on.

Newton’s 3rd : For every applied force (ac@on) there is an equal and opposite force (reac@on)

Newton’s Laws

!Ftot = 0!

!v(t) = !v0 = constant

!F! =m!a

!FAB = !

!FBA

(just in case the exam made you forget)


A toboggan of weight w (including the passengers) slides down a hill of angle α at a constant speed. Which statement about the normal force on the toboggan (magnitude n) is correct?

A. n = w

B. n > w

C. n < w

D. not enough informa@on given to decide

Clicker Ques7on


Bodies in equilibrium – Problem Solving Strategy

•  A body in equilibrium is either at rest or moving with constant velocity. •  What this means in terms of Newton’s Laws:

!F = 0!Fx = 0, Fy = 0!!Problem Solving Strategy

•  Iden@fy the relevant Laws •  Iden@fy the target variables

•  Draw a simple sketch showing dimensions and angles •  Draw free body diagrams for each body (include coordinate axes) •  Draw a force vector for each interac@on with the body (w/ magnitude and

direc:on)

•  Find the components of each force •  Set the sum of all x-‐components equal to zero •  Set the sum of all y-‐components equal to zero •  Solve those equa@ons for the target variables


Example : Equilibrium

Draw a free body diagram showing all the forces on the knot. Find the tension in each cord in the figure if the weight of the suspended object is w.


Example : Equilibrium

The weight w = 60.0N (a)  What is the tension in the diagonal string? (b)  Find the magnitudes of the horizontal forces F1 and F2 that must be applied to hold the system

in the posi@on shown.


A car engine is suspended from a chain linked at O to two other chains. Which of the following forces should be included in the free-‐body diagram for the engine?

A. tension T1

B. tension T2

C. tension T3

D. two of the above

E. T1, T2, and T3

Clicker Ques7on


Example – Non-‐zero accelera@on (Atwood’s Machine)

M2

M1

A load of bricks M1 hangs from one end of the rope that passes over a fric@onless pulley. A counterweight M2 = 1.5 M1 is suspended from the other end of the rope as shown. The system is released from rest. (a) Draw two free body diagrams one for each load. (b) What is the magnitude of the upward accelera@on of the load of bricks? (c) What is the tension in the rope while the load is moving? How does the tension compare the the weight of the bricks and the counterweight?


Fric7on

•  Kine:c fric:on acts when a body slides over a surface.

•  The kine:c fric:on force is fk = µkn.

•  Sta:c fric:on acts when there is no rela@ve mo@on between bodies.

•  The sta:c fric:on force can vary between zero and its maximum value: fs ≤ µsn.


Fric7on – Sta@c and Kine@c



In both cases shown a box is sliding across a floor with the same kine@c coefficient of fric@on and the same ini@al velocity. The only difference between the two cases is the mass of the box. In which case will the box slide the furthest before coming to rest? a) Case 1 b) Case 2 c) Same

Prelecture: Ques7on 1

Prelecture: Ques7on 2 and Clicker Ques7on

A box of mass m sits at rest on an inclined plane that makes and angle θ with the horizontal. It is prevented from sliding by sta@c fric@on. The coefficient of sta@c fric@on between the box and the ramp is μs. What is the magnitude of the sta@c fric@onal force ac@ng on the box? a) mg sinθ b) μs mg cosθ c) μs mg


Prelecture: Ques7on 2

What is the magnitude of the sta@c fric@onal force ac@ng on the box? a) μs mg b) μs mg cosθ c) mg sinθ

(a) (c)


Checkpoint Ques7on 1

A box sits on the horizontal bed of a moving truck. Sta@c fric@on between the box and the truck keeps the box from sliding around as the truck drives.

If the truck moves with constant accelera@on to the lej as shown, which of the following diagrams best describes the sta@c fric@onal force ac@ng on the box:

A B C

µS

a


A box sits on a horizontal table. A string with tension T pulls to the right, but sta@c fric@on between the box and the table prevents the box from moving. What is the magnitude of the sta:c fric:onal force ac@ng on the box? a) 0 b) T c) Mg d) μMg

Check Point Ques7on 2 and Clicker Ques7on

(c) (d) (b) (a)


Checkpoint Ques7on 3

A block slides on a table pulled by a string akached to a hanging weight. In case 1 the block slides without fric@on and in case 2 there is kine@c fric@on between the sliding block and the table. In which case is the tension in the string the biggest? a) Case 1 b) Case 2 c) Same


Checkpoint Ques7on 3 -‐ Math


Example : Fric@on A 10.0 kg box of kikens rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kine@c fric@on is μk = 0.25 and the coefficient of sta@c fric@on is μs = 0.35. a) As the angle α is increased find the minimum angle at which the box has begun to move. b) At this angle find the accelera@on once the box has begun to move. c) At this angle how fast will the box be moving ajer it has slid 5.0 m along the ramp?

General math:

log (x/y) = log (x)! log (y)

log (xy) = log (x) + log (y)

log (xn) = n log (x)

lnx " loge x

x = 10(log10 x)

x = e(ln x)

ha

hho

!

"

ha = h cos ! = h sin"

ho = h sin ! = h cos"

h2 = h2a + h2

o tan ! =ho

ha

#A = Axi+Ay j +Az k

#A · #B = AxBx +AyBy +AzBz = AB cos ! = A!B = AB!

#A# #B = (AyBz!AzBy )i+ (AzBx!AxBz)j + (AxBy!AyBx)k

= AB sin ! = A"B = AB"

df/dt = natn#1

If f(t) = atn, then

!

"

"

"

#

"

"

"

$

% t2t1

f(t)dt = an+1 (t

n+12 ! tn+1

1 ) (for n $= !1)%

f(t)dt = an+1 t

n+1 + C (for n $= !1)

ax2 + bx+ c = 0 % x =!b±

&b2 ! 4ac

2a

"

" + ! = 90!" + ! = 180!

"!

!

"

!

"

!

"

!

"!

" !

Equations of motion:

constant acceleration only:

#r(t) = #r$ + #v$t+12#at

2

#v(t) = #v$ + #at

v2x = v2x,0 + 2ax(x! x0)(and similarly for y and z)

#r(t) = #r$ +12 (#vi + #vf )t

always true:

'#v( = !r2#!r1t2#t1

#v = d!rdt

'#a( = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t

%) dt%

#v(t) = #v$ +% t0 #a(t

%) dt%

Circular motion: |#arad| =v2

RT =

2$R

v

Relative velocity: #vA/C = #vA/B + #vB/C

#vA/B = !#vB/A

Constants/Conversions:

g = 9.80 m/s2 = 32.15 ft/s2 (on Earth’s surface)

1 km = 0.6214 mi 1 mi = 1.609 km1 ft = 0.3048 m 1 m = 3.281 ft1 hr = 3600 s 1 s = 0.0002778 hr

1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

10#9 nano- n10#6 micro- µ10#3 milli- m10#2 centi- c

103 kilo- k106 mega- M109 giga- G

Forces: Newton’s:& #F = m#a, #FB on A = !#FA on B

Phys 218 — Exam I Formulae

α

General math:




lnx " loge x

x = 10(log10 x)

x = e(ln x)

ha

hho

!

"



h2 = h2a + h2

o tan ! =ho

ha

#A = Axi+Ay j +Az k




df

dt= natn#1

If f(t) = atn, then

!

"

"

#

"

"

$

% t2t1

f(t)dt = an+1

&

tn+12 ! tn+1

1

'

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

2a

"

" + ! = 90!" + ! = 180!

"!

!

"

!

"

!

"

!

"!

" !

Equations of motion:translational rotational

constant (linear/angular) acceleration only:

#r(t) = #r$ + #v$t+12#at

2

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

2

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

Forces, Energy and Momenta:translational rotational

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

%

#Fdt = !#p(

#Fext = M#acm = d!pcm

dt(

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

&d!

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&ext = Itot#% = d!Ldt

(

#&int = 0

Krot =12Itot$

2

— Both translational and rotational —

W = !K = Ktrans,f +Krot,f !Ktrans,i !Krot,i

Etot,f = Etot,i +Wother ( Kf + Uf = Ki + Ui +Wother

U = !)

#F · d#r ; Ugrav = Mgycm ; Uelas =12k(r ! requil)

2

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

+

Circular motion: arad =v2

RT =

2'R

v

s = R! vtan = R$ atan = R%

Relative velocity:#vA/C = #vA/B + #vB/C

#vA/B = !#vB/A



G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg


1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

1 J = 1 N·m 1 W = 1 J/s1 rev = 360$ = 2' radians 1 hp = 745.7 W



Forces: Newton’s:, #F = m#a, #FB on A = !#FA on B

Hooke’s: #Felas = !k(#r ! #requil)

friction: |#fs| * µs|#n|, |#fk| = µk|#n|

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

(and similarly for #v and #a)

Gravity:

Fgrav = GM1M2

R212

Ugrav = !GM1M2

R12T =

2'a3/2%GM

Phys 218 — Exam III Formulae


Example – Fric@on cont. b) At this angle find the accelera@on once the box has begun to move. c) At this angle how fast will the box be moving ajer it has slid 5.0 m along the ramp?

General math:




lnx " loge x

x = 10(log10 x)

x = e(ln x)

ha

hho

!

"



h2 = h2a + h2

o tan ! =ho

ha

#A = Axi+Ay j +Az k




df/dt = natn#1

If f(t) = atn, then

!

"

"

"

#

"

"

"

$

% t2t1

f(t)dt = an+1 (t

n+12 ! tn+1

1 ) (for n $= !1)%

f(t)dt = an+1 t

n+1 + C (for n $= !1)

ax2 + bx+ c = 0 % x =!b±

&b2 ! 4ac

2a

"

" + ! = 90!" + ! = 180!

"!

!

"

!

"

!

"

!

"!

" !

Equations of motion:

constant acceleration only:

#r(t) = #r$ + #v$t+12#at

2

#v(t) = #v$ + #at

v2x = v2x,0 + 2ax(x! x0)(and similarly for y and z)

#r(t) = #r$ +12 (#vi + #vf )t

always true:

'#v( = !r2#!r1t2#t1

#v = d!rdt

'#a( = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t

%) dt%

#v(t) = #v$ +% t0 #a(t

%) dt%

Circular motion: |#arad| =v2

RT =

2$R

v

Relative velocity: #vA/C = #vA/B + #vB/C

#vA/B = !#vB/A




1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N



Forces: Newton’s:& #F = m#a, #FB on A = !#FA on B

Phys 218 — Exam I Formulae

α

General math:




lnx " loge x

x = 10(log10 x)

x = e(ln x)

ha

hho

!

"



h2 = h2a + h2

o tan ! =ho

ha

#A = Axi+Ay j +Az k




df

dt= natn#1

If f(t) = atn, then

!

"

"

#

"

"

$

% t2t1

f(t)dt = an+1

&

tn+12 ! tn+1

1

'

ax2 + bx+ c = 0 $ x =!b±

%b2 ! 4ac

2a

"

" + ! = 90!" + ! = 180!

"!

!

"

!

"

!

"

!

"!

" !

Equations of motion:translational rotational

constant (linear/angular) acceleration only:

#r(t) = #r$ + #v$t+12#at

2

#v(t) = #v$ + #at

v2f = v2$ + 2a(r ! r$)

#r(t) = #r$ +12 (#vi + #vf )t

!(t) = !$ + $$t+12%t

2

$(t) = $$ + %t

$2f = $2

$ + 2%(! ! !$)

!(t) = !$ +12 ($i + $f )t

always true:

&#v' = !r2#!r1t2#t1

#v = d!rdt

&#a' = !v2#!v1

t2#t1#a= d!v

dt =d2!rdt2

#r(t) = #r$ +% t0 #v(t) dt

#v(t) = #v$ +% t0 #a(t) dt

&$' = "2#"1t2#t1

$ = d"dt

&%' = #2##1

t2#t1%= d#

dt =d2"dt2

!(t) = !$ +% t0 #$(t) dt

$(t) = $$ +% t0 #%(t) dt

Forces, Energy and Momenta:translational rotational

W = #F ·!#r =%

#F · d#r

P = dWdt = #F · #v

#pcm = m1#v1 +m2#v2 + . . .

= M#vcm#J =

%

#Fdt = !#p(

#Fext = M#acm = d!pcm

dt(

#Fint = 0

Ktrans =12Mv2cm

#& = #r # #F and |#& | = F"r

W = & !! =%

&d!

P = dWdt = #& · #$

#L = I1#$1 + I2#$2 + . . .

= Itot#$

= #r # #p(

#&ext = Itot#% = d!Ldt

(

#&int = 0

Krot =12Itot$

2

— Both translational and rotational —

W = !K = Ktrans,f +Krot,f !Ktrans,i !Krot,i

Etot,f = Etot,i +Wother ( Kf + Uf = Ki + Ui +Wother

U = !)

#F · d#r ; Ugrav = Mgycm ; Uelas =12k(r ! requil)

2

Fx(x) = !dU(x)/dx #F = !#)U = !*

$U$x i+

$U$y j +

$U$z k

+

Circular motion: arad =v2

RT =

2'R

v

s = R! vtan = R$ atan = R%

Relative velocity:#vA/C = #vA/B + #vB/C

#vA/B = !#vB/A



G = 6.674# 10#11 N ·m2/kg2

R% = 6.38# 106 m M% = 5.98# 1024 kgR& = 6.96# 108 m M& = 1.99# 1030 kg


1 kgms2 = 1 N = 0.2248 lb 1 lb = 4.448 N

1 J = 1 N·m 1 W = 1 J/s1 rev = 360$ = 2' radians 1 hp = 745.7 W



Forces: Newton’s:, #F = m#a, #FB on A = !#FA on B

Hooke’s: #Felas = !k(#r ! #requil)

friction: |#fs| * µs|#n|, |#fk| = µk|#n|

Centre-of-mass:

#rcm =m1#r1 +m2#r2 + . . .+mn#rn

m1 +m2 + . . .+mn

(and similarly for #v and #a)

Gravity:

Fgrav = GM1M2

R212

Ugrav = !GM1M2

R12T =

2'a3/2%GM

Phys 218 — Exam III Formulae


When released, the cart accelerates up the ramp.

Which of the following is a correct free-‐body diagram for the cart?

A. B. C. D.

m1a m1a w1 w1 w1 w1

T T T T n n n n

A cart (weight w1) is akached by a lightweight cable to a bucket (weight w2) as shown. The ramp is fric@onless.

Clicker Ques7on


Fluid Resistance and terminal speed

•  The fluid resistance on a body depends on the speed of the body.

•  A falling body reaches its terminal speed when the resis@ng force equals the weight of the body.

•  The figures at the right illustrate the effects of air drag.


For a human body falling through air in a spread-‐eagle posi@on the numerical value of the constant D in the equa@on below is about 0.25 kg/m. Find the terminal speed for a lightweight 50 kg skydiver.

Example -‐ Fluid Resistance

fFR = Dv2 (High Speed)


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Last%TimeCh4%-%Today Ch%5people.physics.tamu.edu/tyana/PHYS218/files/Lect_10_10-1-14_post.pdf · Last%TimeCh4%-%Today Ch%5 Today& • Chapter&5& o Using&Newton’s&1stforbodiesin