Aug+23+2009+Solutions+FyANVC08++Ch+3,+4,+8 4,+9 2 5,+10!1!6+Vectors,+Newton's+Laws,+Torque,+Pressure

8/3/2019 Aug+23+2009+Solutions+FyANVC08++Ch+3,+4,+8 4,+9 2 5,+10!1!6+Vectors,+Newton's+Laws,+Torque,+Pres

1/19

Solutions V1 FyANVC08 Ch 3, 4, 8.4, 9.2-5, 10.1-6 Vectors, Newton's laws, Torque, Pressure NV-College

Directions Please Write your name on all of the papers on your table, NOW!

Test time 8:10-10:40

Resources Calculators, and Formulas for the National Test in Mathematics Courses A &

B, and the formula sheet for Physics A, FYANV-College. You may also use

one page of your own formula sheet. The page must have your name and nocalculations or solution to any problem are supposed to be on the paper.

The test: The test consists of a total of18problems.

For the problemsshort answers are not enough. They require:

that you write down what you do, that you explain your train of thought, that you, when necessary, draw figures. When you solve problems graphically/numerically please indicate how

you have used your resources.

Problem 17 and 18 are larger problems which may take up to 60 minutes to

solve completely. These problems are of the greatest importance for the

highest grade, MVG. It is important that you try to solve them. A description

of what I will consider when evaluating your work, is attached to the problem.Try all of the problems.It can be relatively easy, even towards the end of the

test, to receive some points for partial solutions. A positive evaluation can be

given even for unfinished solutions.

Score The maximum score is 87 points 4 o6f them VG points and 3 Problems.

mark levels The maximum number of points you can receive for each solution is indicated

after each problem. If a problem can give 2 Pass-points and 1 Pass with

distinction- point this is written (2/1). Some problems are marked with,

which means that they more than other problems offer opportunities to show

knowledge that can be related to the criteria for Pass with Special Distinction inAssessment Criteria 2000.

Lower limit for the mark on the test

Pass (G): 29 points

Pass with distinction (VG) : 58 points of which at least 15 Pass with

distinction points (VG).

Pass with special distinction (MVG): 60 points of which at least 46 VG Pass

with distinction points. The requirements for Pass with distinction must be

well satisfied. Your teacher will also consider how well you solve the -problems. Have Fun Behzad

P 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 SumG 2 1 1 1 1 1 1 1 1 1 2 1 2 2 18

VG 2 1 1 1 1 2 1 1 2 2 1 1 2 2 1 21

G

VG

P 16a 16b 17 18a 18b 19 20a 20b 21a 21b 21c 21d 22 Sum 87 Grade

G 2 2 2 2 2 2 2 2 2 2 1 0 2 23 41

VG 1 1 2 1 4 2 4 4 2 4 25 46

MVG 10

G

VG

MVG

Not for sale!Free to use for educational purposes! [email protected] 1


2/19



In the multiple choice problems below, circle the correct alternative and write clearly the

correct answer in the space provided as Alternative:

1. A jumbo jet cruises at a constant velocity of hkm /00012 . In such situations theengine of the jumbo jet exerts a thrusting force of kN.120 . The acceleration of the

jet, and the force of air resistance on the jet are:

a) 0=a ,b) hkma /00012= ,c) 2/82.9 sma = ,d) 2/00012 hkma = ,e) kNF ceresis .120tan = in the opposite direction of its motion.f) kNF

ceresis.120tan = in the same direction of its motion.

g) kNF ceresis .120tan = perpendicular direction of its motion.h) kNF ceresis .120tan = up in the opposite direction of its weight.

Answer: Alternatives: ______________________ [2/0]

Why? Explain in the space provided below! [0/2]

Suggested Explanation: Answer: Alternative (a), and (e):Alternative (a):The velocity of the jumbo jet is constant. Therefore, itsacceleration must be zero: 0=a . [0/1]

Alternative (e)Due to the fact that 0=a , the total force acting on the

jumbo jet must be zero, i.e. 0=netFr

. Therefore, the air resistance must

be exactly equal to the thrusting force of its engine. Therefore, kNF ceresis .120tan = in the opposite direction of its motion.

2. Martina walks 9 blocks north, and then 12 blocks west. Her displacement isa) 15 blocks north of west.

b) 21 blocks north of west.c) 3 blocks north of west.d) 9 blocks north.e) 12 blocks west.

Answer: Alternative _______________ [1/0]


Suggested solutions: Answer: Alternative (a): 15 blocks north of west.

Displacement: blocks15129 22 =+ . [0/1]


3/19



3. An object is thrown upwards at sm /25 . Its acceleration at the top of its motion is:a) 2/82.9 sma = downwards.

b) 2/82.9 sma = upwards.c) 0=a .d) sma /25= downwards.

Alternative: ______________ [1/0]


Suggested explanation:

Alternative (a): 2/82.9 sma = downwards. The gravitational

acceleration of the Earth at sea level is constant and is 2/82.9 sma =

downwards. All free objects projected to the air, if the air resistance

is ignored, accelerate at the rate of 2/82.9 sma = downwards at all

points of their motion, including at the top. no matter where theobject is on the air. [0/1]

Base your answers to the questions 4 to 6on the information a

the diagram below. Note that the VG-points are given to your

explanations that give you the opportunity to demonstrate the

depth of your understandings of the physics of the problem. The

explanations

nd

must be written clearly in the space provided.

4. The applied force on the object is represented by the vector:[

a) 1/0]

Cb) D c) B d) A Why? Explain. [0/1]

nswer: Alternative _______________

Explanations:

A

Suggested solutions: Answer: Alternative (c): The applied force on

the object is represented by the vector B . This is due to the factthat friction force is always smaller than he applied force. Therefot reB can not be the friction force, and it must therefore be the external

rce. Therefore the object must be accelerating up the incline, andthe applied force on the object is represented by the vectorfo

B . [0/1]

A

B

C

D


4/19





. The object is [1/0]a) at rest.

b) moving up at constant velocity.c) moving down at constant velocity.d) accelerating up the incline in the direction

5

B .

e) accelerating down the incline in the direction A .f) depending on the magnitude of the friction force it could be at rest, moving downor going up.

Answer: Alternatives: ______________________ [1/0]


Suggested solutions: Answer: Alternative d: The object isaccelerating up the incline in the directionB . It is due to the factthat friction force is always smaller than the applied force. B is theapplied force and it is larger than the force A which is the sum ofthe friction force and the component of the weight parallel to thesurface of the incline:

cossin += mgmgA

amFnetr

r

= maAB =

Therefore, there is the net force on the object and accordingNewtons second law of motion the object, therefore, will accelerateupwards. [0/1]

6. The angle of the incline is increased. The friction force on the objecta) increases as the angle of the incline increases.

b) is constant and is independent of the angle of the incline.c) decreases as the angle of the incline increases.d) first increases for a while, and then decreases.

Answer: Alternative(s): ______________________ [1/0]


Suggested solutions: Answer: Alternative (c) The friction force onthe object decreases as the angle of the incline increases. The

friction force is part of the vector A , and due to the fact that theonly normal force is that of cosmgFN = and the friction force is

always proportional to the normal force, i.e. cosmgFf Nf == ,

and due to the fact that the cosine fu nof the angle in the first quadrant,

nction is a decreasing functio 900 , the friction force

decreases as the angle of the incline increases. The friction force isnegative for a vertical wall, i.e. = 90 . [0/2]


5/19



Base your answers to t e on the information given below. Note that theh questions 7 to 9

iven to your explanations that give you the opportunity to demonstrate the

ics of the problem. The explanations must be

VG-points are g

depth of your understandings of the phys

written clearly in the space provided.

An object suspended from a dynamometer is lowered into a container filled with water.When the object is completely immersed into the water, volume of the water in the

container increases by3

.250 cm . The dynamometers reading on the air is N805.21 .

You may need the following information: 2/8.9 smg = , and density of water33 /100.1 mkg= w

. The object is made of

wer: Alternative(s): ______________________ [1/0]

Why? Explain your reasoning and necessary calculations in the space provided below!

[0/1]

7

a) gold.b) leadc) cupperd) irone) wood

Ans

Suggested solutions: Answer: Alternative (c) cupper

Due to the fact that 3.250 cm of water is spilled over, the volume of

the object m aust lso be 325 cm , and ts mass i.0 i s

kgkgm 225.28.9

805.21==

34363 1050.210250.250 mmcmV ===

The density of the object is: CumkgV 4

89001050.2

=m 2 3/

52.2

==

[0/1]

8. If the water container was on an scale, and it was showing g.450 before immersingthe object into it, after the object is immersed into the water, the scale will read

a) g.700 b) g.450 c) g.250 d) N.805.21 e) N055.22 Answer: Alternative(s): ___________ [1

Why? Draw a free-body diagram and explain your reasoning

in the space provided below! [0/1]

Suggested solutions: Alternative (a).

/0]

gm

FN 700=

The reading of the scale is increased by theweight of the displaced water, i.e. by kgm 25.0= . The scale will

therefore, read NF

( ) NNgmmgF waterdispN 8.9700.08.9250.0450.0. =+=+=

M

Mg

BW FgmWw ==

NF


6/19



We may converting the apparent weight NFN 8.9700.0 = to its

equivalent mass gkgg

Fm N 700700.0 === [0/1]

Therefore, the scale reads gm 700= .

nally stop

9. As the object is immersed into the water the reading of dynamometera) increases and fi s at N805.21 when the object is complete

into the water.

ly immersed

b) decreases and finally stops at N355.19 when the object is completely immersedinto the water.

c) N5 80.21d) increases and finally stops at N255.24 when the object is completely immersed

water.

ases and finally stops at

into the

e) decre N45.2 when the object is completely immersedater.into the w

Answer: Alternative(s): ____________________ [1/0]

Why? Draw a free-body diagram and explain your reasoning and solutions in the space provided below!

Suggested solutions: Answer: Alternative bThe mass of the displaced water is kgm 25.0= :

kgmVmm

25.0.21000 3 ==== V

WW 10504

Buoyant force on the object in the water is the weight of the

displaced water:NgmF

WB

Therefore, reading of the dyn

45.28.925.0 ===

amometer, TF is:

NFgmF BT 355.1945.2805.21 === [0/1]

Therefore: As the object is immersed into the water thereading of dynamometer decreases and finally stops at

N.355 tely immersed into the

t r [0/1]

19 when the object is comple

wa e .

TF

Mg

M

BF


7/19



10.Philip is floating in fresh water. His density is 3/900 percentbody is above the water surface.

a)mkg How many of his

%10 .

b) %20 .c) %30 d) %50 .e) %80 .f) %90


ody diagram and explain your reasoning and solutions in the space

vide e [0/2]

rats on the water, the Buoyant

Why? Draw a free-b

pro d b low!

Suggested solutions: Answe : Alternative (a) [1/0]Due to the fact that Philip floforce must be exactly equal to his weight mg . The Buoyant

force is equal to the weight of the displaced water. Letsassume his total volume is V and the part of his body that is

under water surface is displacedV

gVgV displacedw /=/

wV

displacedV =

%909.01000

900===

V

Vdisplaced 0/1]

m

[

[Free-body-diagra

Answer:

:0/1]

%90=V

Vdisplacedof Philips volume is inside the water, and the

w rest, i.e. 10% is above the surface of the ater.

Mg

M

BF


8/19



Base your answers to the questions 11 to 12 on the rmation and the diagram below.Note that the VG-points are given to

st depth of your understandi

infoyour explanations that give you the opportunity to

ngs of the physics of the problem. Theust be written clearly in the space provided.

demon rate theexplanations m

throom scale in an elevator. The elevator does not haved is in general very quiet and is very difficult without hearing the bell and

-number to know if it is moving or not. Maximilian has designed ane ich by u

e elevator, but you

Ma

Maximilian stands on a digital baiany w ndow an

h o rseeing t e fl oexperim nt wh sing it, you can calculate not only the magnitude of the acceleration

can say if it is accelerating up or down.of th

ximilian weights kg0.70 .

The scale reads11. kg0.80 .

a) The elevator accelerates upwards at the rate of 2/8.9 sma = b) The elevator accelerates downwards at the rate of 2/8.9 sma = c) The elevator accelerates upwards at the rate of 2/4.1 sma = d) The elevator accelerates downwards at the rate of 2/4.1 sma =


Why? Draw a free-body diagram and explain your reasoning and solutions in the space provided below! [0/1]

Suggested solutions: AThe object is made of cupper:

nswer: Alternative (c) [1/0]

MaMgFN =

2/4.198

8 smMF

a N ==

= [0/1]70

.970

7080

M

g

=

2.The scale reads1 kg0.70 .

a) The elevator accelerates upwards at the rate of 2/8.9 sma = b) The elevator accelerates downwards at the rate of 2/8.9 sma = c) The elevator is stationary.d) The elevator is moving up at a constant velocitye) The elevator is moving down at a constant velocityAnswer: Alternative(s): ______________________ [2/0]Why? Explain your reasoning and necessary calculations in the space below! [0/1]

Suggested solutions: Answer: Alternative (c, d, e) The object is notaccelerating. It can be going up or down at a constant velocity, or itis stationary and therefore not moving. According to Newtons lawsof motion neither the scale nor us may notice if it is stationary ormoving at a constant velocity. Newtons first, second and third lawcombined:

amFnet

r

r

= maMgFN

=

Using MggFN == 70 maMgFN == 0 [0/1]2/0.0 sma =

NF

Mg

Ma

+


9/19



13.As illustrated below, two weights of kg00.6 and kg00.4 are suspended from thtwo ends of a

e

cm.150 long light rod. The rod is uniform and its mass is kg00.1 .

may be assumed weightless. At what point should the beam be picked up if it igoing to have any tendency to rotate?

s not

a) cm4.61 from the end kg00.6 is suspended from.b) cm0.60 from the end kg00.6 is suspended from.c) cm4.61 from the end kg00.4 is suspended from.d) From the center of mass of the rod.

Answer: Alternative _______________ [1/0]

ram showing clearly all the forces and the pivot point.he sp

Why? Draw a free body diagExplain and show the details of your calculations in t ace provided below.

[0/2]

Suggested solutions: Answer: Alternative a: The rod may be picked upfrom a point cmcm 4.6136.61 from the end kg0.6 is suspended from.

The equilibrium condition will require a hts suchhat the torque causing the rotation to the right is equal to the torque

strated below if this point is at

point between the weigtcausing rotation to the left. As illu mx

from the kg0.6 weight, then mx5.1 from the kg0.4 object, and

mx75.0 from the center of mass of the rod ich iswh kg0.1 .

Static Equilibrium( ) ( )xgxgxg /+/=/ 7.015.146 5 [0/1]

xxx += 75.0466

xx 575.66 = 75.656 =+ xx 75.611 =x

mm 4.616136.011

== [0/1]x75.6

=

mx mx5.1

kg0.1

NF

kg0.6 kg0.4

x75.0


10/19



14.Three physics books each kg10.2 lie on top of eachother on a table as shown below. Draw the figure in your

pap

mi e the

tio

er. Calculate and clearly show forces applied on the

ddle book on the figure. You may assum

gravita nal acceleration2/10 smg = . Note that the

forces ust be proportional. [1/2]m

The weight of each book is

Suggested solution:

NmgW 211010.2 ===

The upper book is pushing the middle bookdown with its weight NW 211 = . [1/0]

n the other hand the lowest book is pusheddown with the weight of the books above, i.e.O

NmgW 421010.2222 ===

Therefore, the middle book is pushed downback with the same force NFN 422 = . [0/1]

[0/1]

15.Suggest an experimental method to measure the static and kinetic coefficient of friction.

alternative method also.

SuWekinWe the agradually unt e incli d angleis that of the static friction:

Explain in sufficient detail why the method suggested may give reasonably correct result.

Can you suggest another alternative method to measure the kinetic coefficient of friction?If so explain the [2/1]

ggested Answer:may use an incline to obtain experimentally the static as well as the

etic energy of the surfaces under investigation:may raise ngle of the incline (surface where the object is)

il the object starts moving. The tangent of th ne

Cs tan=

CC

CsCsC gmgm

cos

sin0cossin ===//// [0/1]

The kinetic friction can be measured in a plenty ways. For example we usethe same incline as above but this time we lower the angle until the objectstops. The tangent of the angle is now the coefficient of kinetic friction.Another alternative would be giving the object an initial velocity onhorizontal surface, and lettingdistance traveled may then be measured which in turn gives informationregarding the deceleration of the object. The deceleration is proportionalto that of the friction force and t

tan

athe object slide until it comes to a stop. The

herefore the kinetic coefficient of friction:

x

vaaxvaxvv

2202

2

02

0

2

0

2 =+=+=

On the other hand, ff maf = and mgFfkNkf == which yields

fk amgm /=/ .Therefore:

NW 211

=

NFN 422 =


11/19



xg

v

g

afk

2

2

0==

Answer: On a horizontal surfacexg

vk

2

2

0= where 0v is the initial velocity of

the object and x is the stopping distance.s use your imagination. There are a lot of wayJu s to determine the

coe ion.We may use a dynamometer to measure instantly the coefficient of staticanWe pull the object by a dynamometer on a horizontal surface until it startsmo ter is exactly equal to that

of

tfficient of kinetic frict

d kinetic friction.

ving. The force measured by the dynamome

the friction force: == gff mF sNs mg

e coefficient of the static friction is just the forc

ffs =

Th e measured by the

dynamometer divided by the weight of the object. [2/0]

16.A Saturn V rocket has a mass of kg61075.2 and exerts a force of N61033 on the

of the rocket for the first few kilometers of the elevation

gases it expels. Assuming that the gravitationalacceleration is constant and independent of the elevation

and the mass of the ejected gas is negligible compared to

the mass of the rocket itself, calculate

a. the initial vertical acceleration of the rocket. [2/0]b. the force exerted on kg62 Astronaut Pannos.[2/0]

Suggested solutions: Answer: 2/2.2 sma = ;

NFNP 744=

Data: kgm 61075.2 = ,S NF6

1033= ,

Problem: ?=a , kgmPannos 62= , ?=NPannosF

NgmS766 101095.268.91075.2 == N 7.2

NgmS7107.2

If the direction of the acceleration is

taken to be positive, Newtons secondlaw: amF

r

r

= implies:

amamgF == 666 1075.21095.261033

a= 75.295.2633 a= 75.205.6

2/

75.2

05.6sma = Answer: 2/2.2 sma =

The force on Pannos is the normal force applied by the seats on Pannos.Free-body-diagram illustrates the forces on Pannos: Newtons second law:

( ) ( ) NagmamgmFamgmF PPNPPP NP P 7442.28.962 =+=+=+==

Answer: NFNP 744=

a

NF61033=

kgmS6

1075.2 =

NgmS7

107.2

NFNP 744=

2/2.2 sma =kgmPannos 62=

NgmPannos 808


12/19



36

BA

17.The force of gravity on a kg10 metal block is ten times as great as that on a kg1 wooden block of the same shape and design. Why then doesnt the heavier block fall

faster? Explain. [2/0]

Because the gravitational acceleration is the same for all freely-falling objects, regardless of

in a given time of the free fall. Even inserved

at of the gravitational force, i.e. their weight:

Suggested Explanation:

their masses, they fall as fast and gain identical speedthe air, due to the fact that their shape and design are identical no difference may be ob

in the resisting force for the objects.

Newtons second law dictates that for a freely falling object (ignoring the resistance of the air)the only applied force is th

gaamgmamFnet =/=/=r

r

independent from the mass of the object!

tgvv += 0 : velocity of a freely falling object , t: falling time

xgvv += 2202 : velocity of a freely falling object, , x : falling distance

Due to the fact that none of the relationships above is a function of the mass of the object, the

speed of any free-falling object is independent from their respective masses.

18.A kg0.3 object hangs from a rope which passes over a frictionless and very light pulleyas shown below. The pulley is sup ed by another

rope connected to dynamometer as shown.Ignoring the mass of the pulley, calculate:

a. the value which the dynamometerA shows.[2/1]

b. the value which the dynamometerB shows.

port

a

[2/1]

gested solution:SugData: kgm 0.3=

The object weights NmgW 0.3 3010 ===

Due to the fact that the system islibrium, the total force onin equi

the system is zero:

= 0F

==

=

000

BAF

mgA

xx

yy

NAABx

36tan3036sin36cos

36sin =====

8.2130

NN

AmgAAy 1.3736cos

3036cos ====

Answer: N,A 1.37= NB 8.21=

NA 1.37=

NB 8.21=

NmgW 30==

36

x

y


13/19



19. A diving board is a cantilever. A cantilever is a beam that projects beyond its support.

Find the forces exerted by the two supports of the

m0.5 cantilever diving beam that has a mass of

kg0.8 when Per who is kg0.90 stands at its end.

[2/2]

Suggested solutions: Answer: kNF 7.31 ,

kNF 6.42

Data: kgmbeam 0.8= , kgmPer 0.90= , kggmPer 882=

The condition of static equilibrium requires that = 0Fr

and = 0r

. i.e.:

= downup FF = clockwunterClockwise

Lets name the forces

iseco

as 1F, 2F , gm , andbeam gm as illustrated in thePer

figure below, and then apply the conditions of static equilibrium to theituation, i.e.s

The sum of all forces up is equal to the sum of all forces down. The sum of all clock-wise torques is equal to the total torque counter-

clockwise about a given point, for example about point A :

++= 12 gmgmFF Perbeam

=+ 10.15.14 Fgmgm beamPer

[0/2]

by substitution technique: FindWe may solve this simultaneous equation

1F from the second equation and substitute it into the first one to find 2F kNNbeamPer 7.36.364581 gmgmF .985.18.99045.14 =+=+= [1/0]

kNNgmgmFFPerbeam12 6.48.460782.9908.986.3645 =++ [1/0]=++=

Answer: kNF 7.31 , kNF 6.42

m0.1m5.1

m4

m0.1m5.1

m4

gmbeam

A

kNF 7.31

kNF 6.42

kggmPer 882=


14/19



When assessing your work I am going to consider:

How well you have presented your work. How systematic and general your presentation, your solution and

your reasoning are.

If your calculations are correct.Your analysis of the results and your conclusions.

and physical language in

igure below, the heavier mass is

How well you are using the mathematicalyour presentation.

20.In the f g750 , and the other two identical mass are eachg200 a. The string m y be assumed weightless, and the pulley

a) Draw free-body diagram for each mass. [2/1]b) Calculate the acceleration of the system after it is released

frictionless.

from rest. [2/4/]

c) Calculate the tensions in each string. [2/2]Suggested solution: Answers:

( )mMmgMg

a2

2

+

= ;

( )

( )

+=

mM

MmgFT

2

41

Data:

+

=mM

MmgFT

2

22

kggM 750.0750 == , kggm 20.0200 ==

Problem: ?=a ?1 =TF , ?2 =TF

[1/0]

ote that due to the fact that the strings are inelastic andnstretchable, and the pulley is frictionless, the tension in

a given string is the same, and all masses move at thesame acceleration.The free-body-diagrams illustrated for all three massesinvolved in the system are illustrated in the figures to theleft. In the free-body-diagram all forces involved as well as the mass andacceleration of the mass must beplotted. Using Free-body-diagram, Newtons second law [0/1]of motion can be written easily. Traditionally direction of acceleration forthe mass involved is taken as positive:

Nu

==

=

mamgFmamgFF

MaFMg

T

TT

T

2

21

1

[0/2/] mg

m ?=a

?2 =TF

+

Mg

M

?=F1T

?=a+

mg

m

?1 =TF

?=a

?2 =TF

+


15/19



[1/0]

( ) == agMMaMgFT1

( )

+=+= agmmamgFTT 21

T2

ther to eliminate two of the three

= mamgFF

We may add all three equations togeunknowns.

( )

= mamgF

F

g

addT

T

2

1Answer:

=

=

mamgF

MaFM

T

T

2

1

+= amMmgMg 22

( )

( )( )

gmM

mM

mM

mgMga

2

2

2

2

+

=

+

= [0/1/]

Su ns systems above,nsions in the string

bstitute the value of the acceleration in the equatioand find the te

( ) ( )

( )( )

+

+=+=

mM

mgMggmagmFT

2

22

==mgMg

gMagMFT

2

1 + mM 2

( )

( )

++=

M

MmgMgmFT

22

+/

+

+

m

gmg

mM

mgMg

T

2

2

2

2

1

[1/1]

+

=mgMg

MF

2

( )

( ) + mM 2

=

=

MmgF

F

T

T

22

1+ mM 2

[]

Mmg4

( )

( )

2/98.28.9

400.0750.0

400.0750.0sma =

+

= , 2/0.3 sma [1/0]

( ) ( )N

mM

MmgFT 11.58.9

400.0750.0

2.075.04

2

41 =

+

=

+= ,

( )N

mM

MmgFT 56.2

2

22 =

+=

=FT 22

=

N

NFT

56.

11.51 [2/2]

Alternatively, we may calculate the tensions directly as:

( )

=

===

mF

NagMF

T

T 1.5)38.9(75.0

2

1 ( ) =+=+ Nag 56.2)38.9(2.0

Note that according to ( )( )

gmM

mMa2

2

+= it is the difference betw

weight of the larger block and the sum of the other two

een the

masses, i.e.( )gmM 2 that accelerates the total mass of the system ( )mM 2+

asses together isr[]

Note also that the tension on the string connecting the mlarger (twice) than the tension on the string connecting the two smallemasses.


16/19



M

m

21.A kgM 00.1= wooden block on a plain inclined at = 54 is connected to an emptygm 565= bucket by a light cord running over a frictionless pulley. The coefficient of

static and friction between the block and the surface of the block are 50.0=s and

30.0=k respectively.

Sand is gradually added to the bucket until the system just begins to move.

a) Assume a proper scale for the forces. Draw free-body diagram for the massesinvolved. [2/2]

te the acceleration of the system. [1/4/]

d) Calculate the tension in the cord. [0/2]

Suggested solution

Data:

b) Calculate the mass of sand added to the bucket. [2/4]c) Calcula

kgM 00.1= , kggm 565.0565 == , = 54 , 40.0=s ,

Problem: ?=sandm , ?=a , ?=TF

As the sand is added to the bucket, it gets heavier and heavier. The friction force, ff

gradually increases to its maximum value ( ) cosMgs on of theintended motion. Just before, the last grain of the sand is added to the bucket, the total force

on the wooden block, as well as the total force on the bucket containing sand is zero,

according to the Newtons first law:

in the opposite directi

( )

= 0g

( ) ( )

=+

+

0sincos Ts

sandT

FMgMg

mmF

( )

( ) ( ) ( )

+=+

+

gmmMgMg

gmm

sands

sand

sincos

=FT

( ) ( )( ) ( )

//sands

+=+gmmgM sincos ( ) ( )( )

+=+ sincosMmm ssand( ) ( )( ) mMm = ssand + sin cos

( ) ( )( ) 565.054sin54cos40.000.1 +=sandm

g480kg0.479 =sandm and therefore, the tension in the string is:

( ) ( N10.2N30.5658.9 ) 10.20.479 =+=+= gmmF sandT , ( ) N10.2+= gmmF sandT

?=TF

sandmm +

( )gmm sand+

kgM 00.1=

( )sinMg

( ) cosMgs

?=TF


17/19



Immediately after the last grain of the sand is added to the bucket, the kinetic friction takes

educ e

acceler

aking the direction of the acceleration positive, the second law of Newton,

over, r ing the resistance force in the opposite direction of the motion. This results in th

ation of the system. The situation will be as in the figure below

T amFnetr

r

= may

be written as:

( ) ( =+m )( ( )) =+ aMggammFgm

kT

T

sincos

Adding the equations together results in the cancellation of the tension:

)( + MMF

sandsand

( ) ( )

( ) ( )( )( ) ( ) ( )( ) ( )

++=++

=+/

+=/+

MaammMgMggmm

MaMgMgF

ammFgmm

sandksand

kT

sandTsand

sincos

sincos

( ) ( ) ( )( )

( )sand

ksand

mmM

Mggmma

++

++=

sincos

( ) ( ) ( )( )agmmammgmmF sandsandsandT +=++= ( )( agmmF sandT )+= This means that the difference between the weight of the bucket including the sand na d that of

the parallel component of the weight of the ck and opposing kinetic frictionwooden blo force,

i.e. the force ( ) ( ) ( )( ) sincos ++= ksandnet MggmmF accelerates the total mass of the

system ( )sand

mmM ++ .

( ) ( ) ( ))( )

( 222 /108.5/0575.0479.0565.000.1

54sin54smsm =

++

cos2.08.90.1479.0565.08.9a

++=

( )( ) ( )( ) NagmmMF sandT 17.10057.08.9479.0565.000.1 =++=++= 22 /108.5 sma , NF

T17.10=

( )gmm sand+

( )sandmm + a

TF

( ) cosMgk

( )sinMg

TF

M

a


18/19



22.A g0.128 of necklace is suspected of being gold-plated lead instead of pure gold. If itdropped into a full glass of water

is

g5.10 of water spills over. What proportion of the

necklace is pure gold? Density of pure gold is 3/0.19 cmg=gold , and that of lead is3

lead /0.11 cmg=

[2/4/]

Suggested solutions: Answer: Only %5.36 of the bracelet is made of the

pure gold. The rest, i.e. %5.63 is made of lead.

ata:D gm 0.128necklace = , gmw 5.10= ,35.10 cmV =

irst methodF :

3

necklace

necklacenecklace /19.12

0.128cmg

V

m===

5.10

gmm goldlead 0.128=+ gmmgoldlead

= 0.128 [0/1]35.10 cmVV goldlead =+

35.10 cmmm gold

goldead

lead =+

[0/1]

Using 3gold /0.19 cmg= ,3

lead /0.11 cmg= , and gmm goldlead =128 :

5.101911

128=+

goldgold mm [0/1]

5.10

1911

11

11

128

19

19=+

goldgold mm [0/1]

19115.10110.12819 =+ goldgold mm

5.219411192432 =+ goldgold mm

5.2375.219424328 =goldm = g69.9 [1/gmgold 28

5.237== 0]

%2323.0128

69.29

necklace

===g

g

m

mgold [1/0]

Answer: Only %23 of the bracelet is made of the pure gold. The rest, i.e.

%77 is made of lead.

econd methodS :

ets assume :L goldVx , and leadVy . Therefore, we may construct a two

equation two unknown equation system as:

=+

=+

5.10

1281119

yx

yx [0/2]

To solve this equation system, we may multiply both sides of the secondequation by 11 and subtract the results from the first one so y is

eliminated:

=+=+

5.11511111281119

yxyx


19/19


5.125.1151281119 == xx [1/0]5.128 =x

356.15.12

cmx == [0/18

]

ggold 69.2956.1 = [1/0]xm 1919 ==

%2323.0128necklacem

69.===gold [0/1]

29m

Answer: Only %23 of the elet is made of thebrac pure gold. The rest, i.e.

%77 is made of lead.

ote that this is the percentage weight, while the percentage volume is:N

%1414.056.1

==goldV

5.10=

necklaceV

It is customary to use the mass percentage rather than volume

percentage of gold in jewelry. Why?

Documents

Aug+23+2009+Solutions+FyANVC08++Ch+3,+4,+8 4,+9 2 5,+10!1!6+Vectors,+Newton's+Laws,+Torque,+Pressure