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Last Time: Chapter 4 Today: More Chapter 4

Today •  More on free body diagrams •  Chapter 4 examples

Monday •  Forces •  Newton’s 1st Law •  Newton’s 2nd Law •  Free body diagrams •  Examples (if we get that far)

T. SCegler 09/24/2014 Texas A&M University

Prelecture: Ques3on 1 and Clicker Ques3on

A block slides down a fricConless inclined plane. Which of the following diagrams most closely resembles the free body diagram of this block?


Checkpoint: Ques3on 1

A box of mass m hangs by a string from the ceiling of an elevator that is acceleraCng upward.

Which of the following best describes the tension T in the string? a) T < mg b) T = mg c) T > mg


Checkpoint: Ques3on 2

A block sits at rest on a horizontal fricConless surface. Which of the follwoing sketches most closely resembles the correct free body diagram for all forces acCng on the block. Each red arrow represents a force.



Example Constant force (eoc 4.7)

A 68.5 kg skater moving iniCally 2.40 m/s on rough horizontal ice comes to rest uniformly in 3.52 s due to fricCon. What force does fricCon exert on the skater?

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

You are standing at rest and begin to walk forward. What force pushes you forward?

A. the force of your feet on the ground

B. the force of your acceleraCon

C. the force of your velocity

D. the force of your momentum

E. the force of the ground on your feet

Clicker Ques3on



Example Non-‐constant force (eoc 4.60)

An object with mass m iniCally at rest is acted on by a force where k1 and k2 are constants. Calculate the velocity of the object as a funcCon of Cme.

!F = k1i + k2t

3 j

vx (t) = ax (t)dtt1

t2!x(t) = vx (t)dtt1

t2!

Remember !F =m!a

Clicker Ques3on

The graph to the right shows the velocity of an object as a funcCon of Cme.

Which of the graphs below best shows the net force versus Cme for this object?

t

vx

0

A. B. C. D. E.

t

ΣFx

0 t

ΣFx

0 t

ΣFx

0 t

ΣFx

0 t

ΣFx

0


Example eoc 4.31

A chair of mass 12.0 kg is si`ng on a horizontal floor that is not fricConless. You push the chair with a force 40.0 N that is directed at an angle of 37° below the horizontal and the chair slides along the floor. a) Draw a free body diagram for the chair. b) Use the diagram and Newton’s laws to calculate the normal force the floor exerts on the chair.


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!

"!

!

"

!

"

!

"

!

"!

" !



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

2

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


#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%


RT =

2$R

v


#vA/B = !#vB/A




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





Example eoc 4.31 cont.

A chair of mass 12.0 kg is si`ng on a horizontal floor that is not fricConless. You push the chair with a force 40.0 N that is directed at an angle of 37° below the horizontal and the chair slides along the floor. a) Draw a free body diagram for the chair. b) Use the diagram and Newton’s laws to calculate the normal force the floor exerts on the chair.


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!

"!

!

"

!

"

!

"

!

"!

" !



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

2

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


#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%


RT =

2$R

v


#vA/B = !#vB/A




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





Example -‐ Person in an elevator

A 160 lbs (72.6 kg) person is on a scale in a moving elevator. What does the scale read in each of the following scenarios? a) The elevator is going up at a constant velocity of 3 m/s? down at 3 m/s?

b) The elevator is acceleraCng up at 2 m/s2 ? Down at 2 m/s2 ?

S=scale P=person E=elevator


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!

"!

!

"

!

"

!

"

!

"!

" !



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

2

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


#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%


RT =

2$R

v


#vA/B = !#vB/A




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





Example -‐ Person in an elevator cont.

b) The elevator is acceleraCng up at 2 m/s2 ? Down at 2 m/s2 ? S=scale P=person E=elevator


You are traveling on an elevator up the Sears tower, and you are standing on a bathroom scale.

As you near the top floor and are slowing down, the scale reads

A) More than your usual weight

B) Less than your usual weight

C) Your usual weight

Clicker Ques3on


Example – Box in a rocket

A 6.50 kg instrument is hanging by a verCcal wire inside a space ship that is blasCng off the at the surface of the earth. The ship starts from rest and reaches an alCtude of 276 m in 15.0 s with constant acceleraCon. a) Draw a free body diagram for the instrument at this Cme. b) Find the force that the wire exerts on the instrument.

x = x0 + vx,0t +12axt

2Remember

vx = v0 x + axt!F =m!aT. SCegler 09/24/2014 Texas A&M University

A cart with mass m2 is connected to a mass m1 using a string that passes over a fricConless pulley, as shown below. The cart is held moConless.

The tension in the string is A) m1g B) m2g C) 0

m2

m1

g

a

Clicker Ques3on


Example -‐ mass vs weight (eoc 4.19)

At the surface of Jupiter's moon Io, the acceleraCon due to gravity is g = 1.81 m/s2. A watermelon weighs 44.0 N at the surface of the earth. a) What is the watermelon’s mass at the earth’s surface? b) What are it’s mass and weight at the surface of Io?



For the situaCon depicted, what is the correct free body diagram for Box A?

A B F

(rough)

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

Clicker Ques3on

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