Solutions to Problems in Leighton and Vogt's Exercises in Introductory Physics

(1)

1 1

"Exercises in

"Exercises in Introductor

Introductory

y Physics" (Addison-Wesley, 1969)

Physics" (Addison-Wesley, 1969)

Pier F. Nali Pier F. Nali (Revised April 10, 2016) (Revised April 10, 2016)

CHAPTER 1

CHAPTER 1 Atoms in Motion

Atoms in Motion

B-3

(2)

a) a) 23 23 19 19 33 0 0 3 3 33 6

6..002255 1100 mmoolleeccuullees / ms / moollee 10

10 mmololececululeses/c/cmm 2 222..4411 1100 ccmm // mmoollee G G N N n n Molar_vol._at_STP Molar_vol._at_STP × × × × ∼ ∼ ∼ ∼ ∼∼ _.. 3 3 3 3 2222 33 3 3 _ _ __ 11..0 0 g g ccmm 1 100 110 0 mmoolleeccuulleess//ccmm _ _ 00..00001 1 g g ccmm L L L L G G n

n lliiqquuiidd aaiirr ddeennssiittyy

n n n

n aaiirr ddeennssiittyy

− − − −

⇒

∼ ∼ ∼∼ ∼∼ ∼∼ _.. b) b) 3 3 23 23 19 19 33 _ _ 00..00001 1 g g ccmm 1 10 0 gg//mmoolleeccuullee 2 2..77 1100 mmoolleeccuullees s / / ccmm G G a

aiirr ddeennssiittyy m m n n − − − − × × ∼ ∼ ∼ ∼ ∼∼ _.. c) Said

c) Said σ σ the (approx.) diameter of an air molecule the inverse of the mean free path is the (approx.) diameter of an air molecule the inverse of the mean free path is

2 2 1 1 G G n n σ σ ⋅⋅ ∼ ∼   , where, where 7 7 3 3 11 1100 ccmm L L n n σ

σ ∼ ∼ ∼∼ −− _{, assuming for a rough estimate that the molecules are condensed}_{, assuming for a rough estimate that the molecules are condensed}

in liquid air in such a way that each molecule occupies a cubic cell of side

in liquid air in such a way that each molecule occupies a cubic cell of side σ σ . Thus. Thus

( (

77

))

22 1199 33 55 11 1 1 1 100 ccm− − m ××1010 mmoolleeccuulleess//ccmm 1010 ccmm−− ∼ ∼ ∼∼ 

 or, calculating the reciprocal, the mean free path or, calculating the reciprocal, the mean free path 5 5 1 100 ccm−− m   ∼∼ _.. d) d) FromFrom P PP::P((STSTPP )) ∼∼nn_G_G ::nn_G_G((SSTTPP )) ∼∼((STSTPP )) ::_,, ( ( )) ( ( )) 55 1 atm, 1 atm, 1 10 0 ccmm,, 1 1 m,m, STP STP STP STP P P − − ∼ ∼   ∼∼   ∼∼ one gets

one gets PP ∼∼110 0 aattm−−77 m..

B-4

Said

Said PP the pressure of argon (A) gas spread in a l the pressure of argon (A) gas spread in a layer of volumeayer of volume V V (at constant temperature), we have (at constant temperature), we have

( ( )) ( ( )) 33 33 4 4 10 10 33 7 7660 0 mmm m HHgg 2222..4411 1100 ccmm //mmoollee 6 6..00 1100 mmm m HHgg 2 2..883399 1100 ccmm //mmoollee STP STP STP STP P

P MMoollaarr__vvooll..__aatt__SSTTPP PP MMoollaarr__vvooll..__aatt__SSTTPP V V V V PP P P −− × × × × ×× = =

⇒

= = = = == × × = = ××

(3)

and and

( (

33

))

00 2323 1133 33 10 10 33 6 6..002255 1100 aattoommss//mmoollee _ _ AA __ __ __ 22..112233 1100 aattoommss // ccmm 2 2..883399 1100 ccmm //mmoollee N N n

nuummbbeerr ddeennssiittyy aattoommss ppeerr ccmm

V V × × = = == == ×× × × .. Said

Said A A a constant of a constant of proportionality representing the effective target area per A proportionality representing the effective target area per A atom and expressing (onatom and expressing (on experimental basis) the percent intensity reduction of the atomic beam of K atoms as experimental basis) the percent intensity reduction of the atomic beam of K atoms as

_

_ 33..00 %%

A

A number× × number densitydensity thickness× × thickness == one gets one gets

2 2 14 14 22 1 133 33 11 3 3..00 %% 33..00 1100 1 1..44 1100 ccmm //aattoomm _ _ 22..11223 13 10 a0 attoommss//ccmm 1100 ccmm A A n

nuummbbeerr ddeennssiittyy tthhiicckknneessss

− − − − − − × × = = == ≈≈ ×× × × × × ×× ..

B-5

The cubic lattice has

The cubic lattice has N N× × ×NN×N N points and points and

(

N N− × − 11

)

) (

× −

(

NN − × 11

)

) (

× −

(

N N −11

))

spacings between nearest spacings between nearest neighbours (Cl and Na ions in alternance). Then

neighbours (Cl and Na ions in alternance). Then llaattttiiccee vvo__ oll..= − ⋅ =

( (

NN − ⋅ ≈11

))

33 ss33 ≈ NN ss3 33 3 (for large N (for large N ) and) and

3 3 33 3 3 3 3 3 3 33 33 2244 1 1 11 11 _ _ // AA ccmm 2 222..44225588 1100 2.820A 2.820A N N n

nuummbbeerr ddeennssiittyy aattoommss

N N ss ss °° − − − − °°





≈ ≈ == == ==





× ×

















..

Since the NaCl molecule is diatomic Since the NaCl molecule is diatomic

1 1 2 2 0 0 _ _ __ __ __ NNaaCCll _

_ __ NNaaCCll nnuummbbeerr ddeennssiittyy mmoolleeccuullaarr wweeiigghhtt oof f d deennssiittyy oof f N N × × = = ,, Cl Cl Na Na Cl Cl Na Na Cl… Cl….. Na Na….…. 2 x interatomic spacing 2 x interatomic spacing ss Cl Cl Na Na

(4)

where: where: 3 3 _ _ _ N_ NaaCCll 22..11665 5 gg//ccmm ,, _ _ __ __ NNaaCCll 5588..44554 4 gg//mmoollee,, d deennssiittyy oof f m

moolleeccuullaarr wweeiigghhtt oof f

= = = = and hence and hence 1 1 23 23 2 2 0 0 _ _ __ __ __ NNaaCCll 6

6..0022 1100 mmoolleeccuulleess/m/moollee _

_ __ NNaaCCll n

nuummbbeerr ddeennssiittyy mmoolleeccuullaarr wweeiigghhtt oof f N N d deennssiittyy oof f × × = = = = ×× ..

B-6

The number of helium atoms expelled per second by one gram of radium in equilibrium with its The number of helium atoms expelled per second by one gram of radium in equilibrium with its disintegration products is

disintegration products is 1313..66 1× × 100 aatto1100 omms s g g ss−−11 −−11 = = 111177..550044 1××100 aatto1144 omms s g g d−−11 daayy −−11. The corresponding. The corresponding volume of helium produced per day by one gram of radium is volume of helium produced per day by one gram of radium is

3 3 33 11 4 4 33 11 11 3 3 0 0..00882244 1100 ccm m ddaayy 4 4..229911 1100 ccm m gg ddaayy 1 19922 1100 gg − − −− − − − − −− − − × × = = ×× ×

× and thus the number density of and thus the number density of helium atoms at STPhelium atoms at STP

is is a) a) ( ( )) 1 144 11 11 19 19 33 4 4 33 11 11 1 11177..550044 1100 aattoomms s g g ddaayy 2 2..77 1100 aattoommss//ccmm 4 4..229911 1100 ccm m g g ddaayy STP STP He He n n − − −− − − − − −− × × = = ≈ ≈ ×× × ×

and the calculated Avogadro’s number and the calculated Avogadro’s number b)

b)

( ( )) 1199 33 33 33 2233 0

0 22..77 1100 aattoomms s ccmm 2222..4411 1100 ccmm //mmoollee 66..11 1100 aattoommss//mmoollee

STP STP He He N N ==nn ××Molar_vol._aMolar_vol._at_STPt_STP ≈≈ ×× −− ×× ×× ≈≈ ×× ..

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C-1

The molecular weight of H(CH

The molecular weight of H(CH22))1818COOH is appCOOH is approximately 298 roximately 298 g/mole g/mole and the molar vand the molar volume isolume is 1 1 3 3 3 3 _ _ 22998 8 g g mmoollee 37 372.2.5 c5 cmm /m/mololee 0.8 g cm 0.8 g cm m

moolleeccuullaarr weweigighht t density density − − − − ≈ ≈ ≈≈ ..

The volume of oil is deduced by dividing 0.81 mg by its density, i.e. The volume of oil is deduced by dividing 0.81 mg by its density, i.e.

3 3 3 3 33 3 3 0 0..8811 1100 gg 1 1..00 1100 ccmm 0.8 g cm 0.8 g cm − − − − − − × × ≈ ≈ ×× ..

The surface area over which this volume of oil is spread is

The surface area over which this volume of oil is spread is 11 22 22 33 22 4

4π × π × 8844 ccmm ≈ ≈ 55..554422 1××100 ccmm , so the, so the

thickness of the oil film (calculated as if its density were the same as in non spread state) is thickness of the oil film (calculated as if its density were the same as in non spread state) is

3 3 33 33 33 6 6 2 2 22 33 22 1 1 4 4 0 0..8811 1100 gg 00..8 8 g g ccmm 11..00 1100 ccmm 0 0..22 1100 ccmm 8 84 c4 cmm 55..554422 110 c0 cmm ll π π − − − − −− − − × × ×× ≈ ≈ ≈≈ ≈≈ ×× × × ×× ..

If the oil molecules were arranged in linear chains 20 atoms high (

If the oil molecules were arranged in linear chains 20 atoms high ( ≈≈ thicknessthickness llof the oil film)of the oil film) ×× 44 atoms wide (

atoms wide ( 11 5 5 ll

≈

≈ ×× ) so that a molecule occupies a volume roughly estimated (assuming in liquid state a) so that a molecule occupies a volume roughly estimated (assuming in liquid state a rotational degree of motion around the main axis of the chain) as rotational degree of motion around the main axis of the chain) as

( (

66

))

33 3 3 22 22 33 1 1 11 5 5 55 0 0..22 1100 ccmm 3 3..22 1100 ccmm 2 255 2255 ll ll ll ll − − − − × × × × ×× == ≈≈ ≈≈ ×× ..

The reciprocal of the latter (

The reciprocal of the latter (00..33 1××100 m2222moolleeccuulleess//ccmm33) is the number density) is the number density 00

_ _ .. N N Molar

Molar volvol . So. So

2

222 33 2222 33 33 2244

0

0 00..33 1100 mmoolleeccuulleess//ccmm __ .. 00..33 1100 mmooll..//ccmm 337722..5 5 ccmm //mmoollee 1100 mmoolleeccuulleess//mmoollee..

N

(6)

C-2

..

Said

Said mm the mass of a gas molecule the gas density is the mass of a gas molecule the gas density is ρ ρ

( (

gasgas

))

= ⋅⋅= mm N N _g_g. If one assume for sake of. If one assume for sake of simplicity that in a solid the molecules are tightly packed in cells of side

simplicity that in a solid the molecules are tightly packed in cells of side σ σ then then ρ ρ

( (

solidsolid

))

≈≈mm σ σ 33 (1)(1).. Thus Thus

( (

))

( (

))

( (

))

( (

))

( (

))

( (

))

2 2 3 3 3 3 22 ₃₃ 3 3 g

gaass 11 ggaass ggaass

ssoolliidd N Ngg N N _g_g ssoolliidd NNgg N N gg ssoolliidd

ρ ρ ρ ρ ρ ρ σ σ σ σ σ σ ρ ρ ρ ρ ρ ρ





≈ ≈ ⋅⋅

⇒

≈ ≈ ⋅⋅

⇒

≈ ≈ ⋅⋅













.. (1) (1)

For our purposes here

For our purposes here we may ignore that in closed packaging configuration we may ignore that in closed packaging configuration the proportion of space occupied by sphericalthe proportion of space occupied by spherical particles is particles is ( ( )) ( ( )) _ _ __ __ __ 74,048% 74,048% 3 3 22 particle particle unitcell unitcell n

nrr ooff ppaarrttiicclleess ppeerr cceellll V V V V π π × × ∆

∆== == ≈≈ and therefore the number of particles per and therefore the number of particles per

unit cell is unit cell is ( ( )) ( ( )) 3 3 3 3 2 2 particles/celparticles/celll 3 3 22 ₆₆ unitcell unitcell particle particle V V V V π π σ σ π π σ σ ∆

∆ ⋅⋅ == ⋅⋅ == rather than 1 particle/cell,rather than 1 particle/cell, ρ ρ

( (

solidsolid

))

isis 22⋅⋅mm σ σ 33rather thanrather than

3 3

m

m σ σ and so on, so the subsequent results should containand so on, so the subsequent results should contain 2 2

N N gg

in locus of

in locus of N N _g_g. This is the densest packing of equal spheres. This is the densest packing of equal spheres in three

in three dimensions, as dimensions, as proved bproved by Gauss y Gauss in in 1831. 1831. See Gauss, See Gauss, C. C. F. "F. "Besprechung des Besprechung des Buchs von Buchs von L. L. A. A. Seeber:Seeber: Intersuchungen über die Eigenschaften der positiven ternären quadratischen Formen usw." Göttingsche Gelehrte Anzeigen Intersuchungen über die Eigenschaften der positiven ternären quadratischen Formen usw." Göttingsche Gelehrte Anzeigen (1831, July 9) 2, 188-196, 1876 = Werke, II (1876), 188-196. For a proof of this result see, e.g., J. W. S. Cassels, —An (1831, July 9) 2, 188-196, 1876 = Werke, II (1876), 188-196. For a proof of this result see, e.g., J. W. S. Cassels, —An Introduction to the Geometry of Numbers, Springer-Verlag, 1971 [Chap. II,

(7)

Comparing

Comparing ==11

( (

22π σ π N N _g_gσ 22

))

_and_and 11

( (

))

3

3 gasgas vv η

η = = ρ ρ ⋅ ⋅ ⋅⋅_{one also gets}_{one also gets}

( (

))

( (

))

( (

))

2 2 1 1 22 ggaass 22 33 g g N N σσ == ππ == ρρ ⋅⋅vv ππ ⋅⋅ ηη _..

Comparing again the right-hand sides of latter two expressions for

Comparing again the right-hand sides of latter two expressions for N N _g_gσ σ 22 and cubing the second and cubing the second members one gets

members one gets

( (

))

( (

))

( (

))

( (

))

2 2 3 3 ₃₃ ₃₃ ₃₃ gas gas g gaass 5544 22 solid solid g g N N ρ ρ ρ ρ vv π π η η ρ ρ









⋅ ⋅





≈≈

_

_

⋅⋅ ⋅⋅









and finally – assuming on

and finally – assuming on experimental basisexperimental basis(2)(2) ρ ρ

( )

( ) (

ssoolliidd ≈≈ ρ ρ

(

lliiqquuiidd

))

-

-(

( )

) (

(

))

22 33 19 19 33 3 3 33 g

gaass lliiqquuiidd

0 0..77 1100 mmoolleeccuulleess//ccmm 54 54 22 g g vv N N ρ ρ ρ ρ π π η η





_⋅⋅





≈ ≈ ×× ⋅ ⋅ ⋅⋅ ∼∼ .. (2) (2)

This assumption is valid where we are dealing only whit orders of magnitude and is based on the observed condensation This assumption is valid where we are dealing only whit orders of magnitude and is based on the observed condensation behaviour of many substances. Unfortunately, at the times of Loschimdt’s work (1865) was not yet known how to liquefy behaviour of many substances. Unfortunately, at the times of Loschimdt’s work (1865) was not yet known how to liquefy the air, so he could not directly observe condensation behaviour for air. However, the density of a substance in the the air, so he could not directly observe condensation behaviour for air. However, the density of a substance in the condensed state can be estimated by its chemical composition, as Loschmidt did. See Porterfield, William W., and Walter condensed state can be estimated by its chemical composition, as Loschmidt did. See Porterfield, William W., and Walter Kruse. "Loschmidt and the Discovery of the Small.",

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1 1

"Exercises in

"Exercises in Introductor

Introductory

y Physics" (Addison-Wesley, 1969)

Physics" (Addison-Wesley, 1969)

Pier F. Nali Pier F. Nali (Revised April 10, 2016) (Revised April 10, 2016)

CHAPTER 2

CHAPTER 2 Conservation of Energy, Statics

Conservation of Energy, Statics

1

According to the principle of virtual work (PVW) “the sum of the heights times the weights does not According to the principle of virtual work (PVW) “the sum of the heights times the weights does not change”, said

change”, said hh₁₁ and and hh₂₂ the heights of the weights the heights of the weights W W ₁₁ and and W W ₂₂ respectively, then respectively, then

( (

WW h11 1h1 W hW22 2h2

))

W hW h11 11 W hW22 h22 00 ∆

∆ + + = = ∆ ∆ + + ∆ ∆ == .. If we assume that the weight

If we assume that the weight W W ₁₁ goes down goes down ∆∆hh₁₁ as the weight as the weight W W ₂₂ rises rises ∆∆hh₂₂, then, then

1 1 1 1 22 11 11 22 22 2 2 h h hh WW WW ∆ ∆ = = − − ∆ ∆

⇒

  ==    ..

(9)

2

T

Thhe “e “ssuum om of tf thhe he heeigighhts ts titimmees ts thhe we weeigighhtsts” f” foor ar an an arrbbiitrtraary ry nnuummbbeer or of wf weeigighhtts bs beeccoommeess _ii _ii 00

ii

W

W h∆ ∆ =h =

∑

and considering the distances positive or negative depending on the side of the fulcrum, so that and considering the distances positive or negative depending on the side of the fulcrum, so that

ii ii jj j j h h hh ∆ ∆ = =  ∆∆ 

 (or (or ∆ ∆ ∝hhii ∝ii for any for any ii,, beingbeing j j j j h h ∆ ∆ 

 independent of j independent of j), ), it it becomes becomes thus thus

∑

_ii W W iiii ==00..

3

The

The PVW PVW can can be be expressed expressed in in a a more more general general form form as as 00

N N nn ij ij ii ii jj





∆ ∆ ==









∑

∑ ∑

∑

F F _ii rr _{, where the sum over}_{, where the sum over j}_{j is}_is

extended to all the

extended to all the nn forces acting on the mass forces acting on the mass ii and the sum over and the sum over ii is extended to all the is extended to all the N N masses.masses. In

In the the present present case case (a (a single single body) body) N N = =11, , so so the the above above formula formula reduces reduces to to 00

n n ii ii





∆ ∆ ==







∑



F F _ii rr _.. a)

(10)

b)

b) For For nn = =22 isis

( (

F F ₁₁+ + F F ₂₂

))

ii∆ ∆ =rr =00_{and, again for the arbitrariness of the choice of the vector}_{and, again for the arbitrariness of the choice of the vector} _∆_∆rr_{, is}_{, is}

1 1+ + 22 ==00 F F FF _{, whence}_{, whence} 1 1 = −= − 22 F

F FF _{; so, the two forces are equal in magnitude, opposite in direction and}_{; so, the two forces are equal in magnitude, opposite in direction and}

collinear. collinear. c)

c) For For nn = =33 is, again, is, again, F F ₁₁+ + F F ₂₂+ + FF₃₃ ==00_{. If we consider the plane on which lie the two vectors}_{. If we consider the plane on which lie the two vectors}

1 1 F F_and_and 2 2 F

F _{(apart from the trivial case in}_{(apart from the trivial case in which the two vectors are parallel, which degenerates to the case of}_{which the two vectors are parallel, which degenerates to the case of}

2 2 n

n = = ),), FF₃₃_{also lies in the same pane, being expressible as a linear combination of the other two}_{also lies in the same pane, being expressible as a linear combination of the other two}

vectors. Let

vectors. Let XX00 the vector joining the origin to the point of intersection of the lines of action of the the vector joining the origin to the point of intersection of the lines of action of the two vectors

two vectors FF₁₁_and_and

2 2

F

F _and_and XX_{the vector joining the origin to an arbitrary point of one or the other}_{the vector joining the origin to an arbitrary point of one or the other}

line of action. Then the two lines of action are defined by the system of equations line of action. Then the two lines of action are defined by the system of equations

( (

))

( (

))

0 0 11 0 0 22 0 0 0 0



₋₋ _{∧ =}_∧ ₌





− − ∧ ∧ ==



X X X X FF X X X X FF ..

By adding the latter two equation, immediately follows that

By adding the latter two equation, immediately follows that XX00 also lies on the line of action of also lies on the line of action of FF₃₃.. 0

0

X

X _{defines a single point}_{defines a single point (through which the three lines pass) as}_{(through which the three lines pass) as each line of action cannot intersect}_{each line of action cannot intersect}

each other in more than one point (provided the lines are not p

each other in more than one point (provided the lines are not p arallel).arallel). d)

d) FrFrom om F F _iiii∆ ∆ = rr = ∆ F F _ii∆ ⋅ rr ccos⋅ ∆os∆_ii_{it follows that}_{it follows that} 1 1 11 ccooss 00 n n nn ii ii ii ii ii F F r r = = ==









∆ ∆ = = ∆ ∆ ⋅⋅ ∆ ∆ ==











∑

F F



ii rr



∑



, and – again for the, and – again for the

arbitrariness of the choice of arbitrariness of the choice of ∆∆r r -

-1 1 ccooss 00 n n ii ii ii F F = = ∆ ∆ ==

∑

..

(11)

A-1

Suppose the weight W goes down

Suppose the weight W goes down x x as the centre of mass of as the centre of mass of the plank risesthe plank rises y y. Therefore (refer to the. Therefore (refer to the above figure)

above figure) _y_y ₌₌ L L₂₂sinsin_∆_∆_θ_θ_,,_x_x_{= −}_{= − −}

_{( (}

_{cc cc}₋ _′′

₎₎

_and_and

( (

))

2 2 22 ssiinn ,, 1 1 ccooss ,, .. a a aa LL b b bb LL cc aa bb θ θ θ θ ′′ = = − − ∆∆ ′′ = = + + − − ∆∆ ′ ′ = = ′ ′ ++ ′′ So – assuming

So – assuming ∆∆θ θ small small -- 22 ( ( 22 ))

( (

22

))

2 2

1

1 22 aLaL ssiinn 44 L L LL bb ssiinn 11 aaLLssiinn

cc cc cc

cc_{′′ =}₌cc ₋₋ _∆_∆_θ_θ₊₊ ++ ∆∆θ θ _≅_≅cc ₋₋ _∆_∆_θ_θ _and_and aLaL_sin_sin cc

x

x≅ − ≅ − ∆∆θ θ ,, where we have used the trigonometric double-angle formula

where we have used the trigonometric double-angle formula 11 ccooss 22ssiinn22 2 2 α α α α −

− == and the approx. and the approx. 1 1 11 2 2 x x x x +

+ ≅ ≅ ++ for for x x small. small. aa W W W W 45° 45° a’ a’ b b b’ b’ cc c’ c’ x x L L y y _∆_∆_θ_θ

(12)

Being

Being also also cc = = 22⋅⋅aa it follows it follows

2

2ssiinn 22

L L

x

x ≅ ≅ − − ∆ ∆ = θ θ = − − ⋅⋅ yy. From PVW then we have. From PVW then we have

( ( ))

_{( (}_plank_plank₎₎ W W ⋅ ⋅ − − = xx = WW ⋅⋅yy, whence, whence _{( (} ₎₎ ( ( )) 33 kg-wtskg-wts 2 2 22 plank plank plank plank W W y y W W W W x x = =−− == == ..

(note: the length of the plank doesn't affect this derivation). (note: the length of the plank doesn't affect this derivation).

A-2

Let

Let W W the weight of the ball, the weight of the ball, F F and_P_P and F F the forces exerted on the ball by the plane and by the wall_W_W the forces exerted on the ball by the plane and by the wall respectively and made use of PVW in the form

respectively and made use of PVW in the form

1 1 ccooss 00 n n ii ii ii F F = = ∆ ∆ ==

∑

..

Thus, with respect to an horizontal line joining the point in which the ball touches the wall with the Thus, with respect to an horizontal line joining the point in which the ball touches the wall with the centre of mass of the ball,

centre of mass of the ball, FF_w_wccooss 0

( ( ))

0 ₊₊F F _P_Pccooss

( (

π π ₂₂ ₊₊_α_α

))

₌₌00_,_{, tth}_haat_{t iis}_s _ssiin_n ₀₀

w w PP

F

F − − F F α α == ; and, with respect to; and, with respect to a vertical line lying in the plane of the sheet and passing through the centre of mass of the ball, a vertical line lying in the plane of the sheet and passing through the centre of mass of the ball,

2 2

ccooss ccooss ccooss 00

W

W PP

F

F π π ₊₊WW _π_π₊₊F F _α_α₌₌ _,_{, tth}_haat_{t iis}_s _cco_oss ₀₀

P P W W F F α α − − + + == .. So So ,, cos cos ttaann ,, P P W W W W F F F F W W α α α α = = = =

and the forces on the

and the forces on the planes areplanes are 1 1 kg-wt, kg-wt, cos cos F F ttaan n kkgg--wwtt.. P P W W F F α α α α = = = =

(13)

A-3

Again make use of PVW, this time i

Again make use of PVW, this time in the formn the form

1 1 0 0 n n ii ii ii W W hh = = = =

∑

..

-- Without the loads:Without the loads:

( (

WW_C_C ++WW_A_AA_{A C}_′_′_CD_D

))

⋅⋅AAPP ==WW_BB_B_{B G}_′′_GH_H ⋅⋅PBPB, , that that is is WW_C_C + + WW_AA_A_{A C}_′_′_CD_D == 22W W _BB_B_{B G}_′′_GH_H ..

-- With the With the loads, assume the weightloads, assume the weight W W goes down₁₁ goes down x x as the weight as the weight W W rises₂₂ rises y y, being, being 11 2 2 x

x = −= − yy .. Thus

Thus

(

WW_C_C + + WW_A_AA_{A C}_′_′_CD_D + + ⋅ WW₁₁

)

⋅

(

− − = xx

)

=

(

WW_BB_B_{B G}_′′_GH_H + ⋅⋅+ WW ₂₂

))

yy, that is, that is WW_C_C ++WW_A_AA_{A C}_′_′_CD_D ++WW₁₁ ==22

( (

WW_B_BB_{B G}_′′_GH_H ++WW₂₂

))

, and, and subtracting to this equation the one without the loads above one gets

subtracting to this equation the one without the loads above one gets WW₁₁ ==22W W ₂₂, so, so

1 1 2 2 22 11 0.25 kg-wts0.25 kg-wts W W = = W W == ..

A-4

ll11 _ll₂₂ L L x x A A y y A A

A A y y −∆ −∆ ϕ ϕ

(14)

The weight A can move

The weight A can move both vertically and horizontally.both vertically and horizontally. -

- Vertical Vertical displacement displacement of of A A (assumed (assumed downwards):downwards):

( (

))

1 kg 1 kg 1 kg 1 kg A A AA BB BB W W ⋅ −⋅ −∆ ∆ = yy = WW ⋅⋅ ∆ ∆ + yy + ⋅⋅ ∆WW ∆yy .. -

- Horizontal Horizontal displacement displacement of of A A (assuming (assuming B B risesrises ∆∆ y y_B_B as A moves to the left): as A moves to the left):

( (

1 kg1 kg

))

1 kg

1 kg B B BB

W

W ⋅ ⋅ −−∆ ∆ yy = = ⋅⋅ ∆WW ∆yy

Case 1 – vertical displacement Case 1 – vertical displacement For the flattened triangle

For the flattened triangle  _((ll_1,_1, A A

y y

∆

∆ , , ll₁₁++ ∆∆ll₁₁), using Carnot’s theorem and assuming), using Carnot’s theorem and assuming ∆∆ y y_A_A small,small,

( (

))

( (

))

( ( ))

2222 1 1 ₁₁ 2 2 22 22 22 1

1 11 11 2211 ccooss 22 11 2211 ssiinn 11 11 22 ssiinn

A A AA y y yy A A AA AA AA ll _ll ll _{+ ∆}₊_{∆ =}ll _{= +}ll _{+ ∆}ll yy_∆ π π ₊_{+ +}_ϕ_ϕ _{+ ∆}_{∆ =}yy _{= +}ll _{+ ∆}ll yy_{∆ −}₋ _ϕ_ϕ ₊_{+ ∆}_{∆ =}yy _{= −}ll ₋ ∆ ∆ _ϕ_ϕ₊₊ ∆∆ _.. Neglecting the term of order

Neglecting the term of order ∆∆ y y_A_A22 and and using using the the approx.approx. 11 2 2 1 1+ + ≅ x x ≅ +11+ xx, , thenthen

(

( )

11

)

(

11

))

1 1 11 11 11 22 ssiinn 11 11 ssiinn A A AA y y yy ll ll ll ++ ∆ ∆ ≅ ll ≅ − ll − ∆ ∆ ϕ ϕ ≅ −≅ ll − ∆∆ ϕ ϕ andand

( (

))

1 1 1 1 11 11 ssiinn 11 ssiinn A A y y A A ll ll ll ∆∆ ϕ ϕ ll yy ϕ ϕ ∆ ∆ ≅ ≅ − − − − ≅ ≅ −−∆∆ , , wherewhere 1 1 11 11 sin

sin y y A A yyAA

ll ll ll ϕ ϕ = = ≅≅ + + ∆∆ . So. So 11 ₁₁ A A A A y y ll yy ll ∆ ∆ ≅ ≅ −−∆∆ ..

Similarly, for the triangle (not shown)

Similarly, for the triangle (not shown)  _((ll_2,_2, A A y y ∆ ∆ , , ll₂₂++ ∆∆ll₂₂),), ₂₂ 2 2 A A A A y y ll yy ll ∆ ∆ ≅ ≅ −−∆∆ . But. But 1 kg 1 kg 1 1 ll yy ∆ ∆ = ∆= ∆ andand 2 2 B B ll yy ∆ ∆ = ∆= ∆ , whence, whence

(

)

₁₁_kg_k_g

(

)

(

))

₁₁_k_kg_g 1 1 22 11 22 1 1 11 11 11 1 1kkgg ssiinn 3300 ssiinn 4455 22 11kkgg ₂₂ 22 ₂₂ .. A A AA AA AA A A AA AA BB AA AA BB B B BB BB y y yy yy yy W W yy WW yy WW yy WW WW WW ll ll ll ll W W WW WW WW WW ⋅ ⋅ −−∆ ∆ ≅ ≅ ⋅ ⋅ −−∆ ∆ + ⋅ −+ ⋅ −∆ ∆ → → ≅ ≅ ⋅ + ⋅ + ⋅ ⋅ == = = ⋅ ⋅ °° + + ⋅ ⋅ °° = = + + = = ++

Case 2 – horizontal displacement Case 2 – horizontal displacement

In the same way, for the triangle (not even shown as well)

In the same way, for the triangle (not even shown as well)  ₍₍_ll₁₁₋_{− ∆}_∆_ll₁₁ _,_, A A

x x

∆

∆ , , ll ) with₁₁) with ∆∆ x x_A_A small,small,

( ( ))

22 2 2 1 1 11 2 2 22 1

1 11 11 2211 ccooss 3300 11 1 21 2 ccooss 3300

A A AA x x xx A A AA ll ll

ll −− ∆ ∆ = ll = ll + ∆ + ll xx∆ °° ++ ∆ ∆ = xx = ll + + ∆ ∆ °° ++ ∆∆ , and, with the above approximations,, and, with the above approximations,

(

11

)

(

11

)

(

11

))

1

1 11 11 11 22 ccooss 3300 11 11 ccooss 3300 11 11 11 11 ccooss 3300 ccooss 3300

A A AA AA x x xx xx A A ll ll ll ll −− ∆ ∆ ≅ ll ≅ ll + + ∆ ∆ ° ° ≅ + ≅ ll + ∆ ∆ ° → ° → ∆ ≅ ∆ ll ≅ − + ll − + ll ∆∆ ° ° ≅ ≅ −−∆ ∆ xx °° .. Substituting Substituting 1 1 cos30

cos30 x x A A

ll

°° == in the above expressions one gets in the above expressions one gets ₁₁

1 1 A A A A x x ll xx ll ∆

∆ ≅ ≅ −−∆∆ , and similarly, for triangle, and similarly, for triangle



 ( (ll₂₂ + ∆+∆ll₂₂ , , ∆∆ x x_A_A , , ll ):₂₂): ₂₂

2 2

co

coss 4545 A A A A AA L L xx ll xx xx ll − − ∆ ∆ ≅ ≅ ∆ ∆ °° = = ∆∆ .. So from PVW So from PVW

(15)

( (

_{1 kg}_{1 kg}

))

( (

))

1 1kkgg 11kkgg 11 22 11kkgg 1 1 22 3 3 22 1 1kkgg 11kkgg 11kkgg 22 22 1 1 22 3 3 1 kg 1 kg 2 2

ccooss 3300 ccooss 4455 .. A A AA B B BB BB AA BB AA A A AA B B BB BB B B x x LL xx W W yy WW yy WW ll WW ll WW xx WW xx ll ll x x LL xx W W WW WW WW WW WW ll ll W W W W − − ⋅ ⋅ −−∆ ∆ = = ⋅⋅ ∆ ∆ → → ⋅ ⋅ −−∆ ∆ = = ⋅⋅ ∆ ∆ → → ⋅⋅ ∆ ∆ = = ⋅⋅ ∆∆ − − → → ⋅ ⋅ = = ⋅ ⋅ → → ⋅ ⋅ °° = = ⋅ ⋅ °° → → ⋅ ⋅ = = ⋅⋅ → → = = ⋅⋅ Whence finally Whence finally 11 11 11 33 1 kg 1 kg 2 2 ₂₂ 22 22 .. A A BB W

W ≅ ≅ + + WW = = ++ W W In summary: In summary:

( (

11 33

))

2 2 22 3 3 2 2 kg-wts, kg-wts, kg-wts. kg-wts. A A B B W W W W = = ++ = =

A-5

From the PVW (in the form From the PVW (in the form

∑

):): -

- Case Case 1 1 – – vertical vertical fixed fixed line:line:

(

)

(

))

( (

))

( ( ))

( (

))

2 2 1 1 22 11 22 11 22 2 2 AB AB 2 2 11 1 1 22 ACAC ₂₂ 11 22

ccooss ccooss ccooss 00 ccooss 11 ssiinn 1 1 .. T T T T WW TT TT WW TT TT WW T T TT WW TT TT WW ϕ ϕ++ ϕϕ++ ππ == → → ++ ϕϕ == →→ ++ −− ϕϕ == →→ + + −− == →→ ++ == -

- Case Case 2 2 – – horizontal horizontal fixed fixed line:line:

(

( )

)

(

( ))

1

1ccooss 22 22ccooss 22 ccooss 22 00 11 22

T T π π _ϕ_ϕ TT π π _ϕ_ϕ WW π π TT TT − − − − + + − − + + = = → → == .. Thus Thus ₁₁ ₂₂ 5050 lb-wts lb-wts 2 2 22 W W T T = = T T = = == ϕ ϕ

(16)

A-7

Let V the reaction force of the edge in contact with the wheel and assume null the reaction force of the Let V the reaction force of the edge in contact with the wheel and assume null the reaction force of the wall at the ground soon as the force F is

wall at the ground soon as the force F is applied to the axle. Then from PVWapplied to the axle. Then from PVW

∑

:: -

- CCaasse e 1 1 – – vveerrttiiccaal l ffiixxeed d lliinnee: : ccoVV oss WWccooss 00 VVccooss WW VV R h R h WW R

R

ϕ

ϕ ++ π π == → → ϕ ϕ == →→ −− == .. -

- Case 2 Case 2 – – horizontal horizontal fixed fixed line:line: FFccooss 0

( ( ))

0 VV ccooss

( (

₂₂

))

WWccooss ₂₂ 00 FF VVssiinn VV x x R R π π _ϕ_ϕ π π _ϕ_ϕ + + ++ ++ == → → == == ,, where where x x= − − = − RR22

(

R hR− =

)

h

)

22 = hh

(

22RR h−−h

))

.. Thus Thus FF WW x x W W hh

( (

22RR hh))

))

R R hh RR hh − − = = == − − −− .. V V x x ϕ ϕ V V x x ϕ ϕ

(17)

A-9

Using Using 1 1 ccooss 00 n n ii ii ii F F = = ∆ ∆ ==

∑

we have: we have: -

- horizontal horizontal lineline AB AB ::

(

)

₂₂

(

))

ccooss 4455 ccooss ccooss 00 ccooss 00 ccooss 4455 ccooss 00

O

O PP QQ xx xx

F

F _π_π _{+ °}₊_°₊₊FF π π ₊₊FF _{°° +}₊FF _π_π _{+ =}₊_ϕ_ϕ ₌

⇒

₋₋FF _{°° +}_{+ −}FF ₋FF _ϕ_ϕ ₌₌ _,, -

- vertical vertical lineline OPOP ::

(

₂₂

)

₂₂

(

₂₂

))

ccooss 4545 ccooss 00 ccooss ccooss 00 ssiinn 4455 ssiinn 00

O O PP QQ xx xx F F π π + ° + ° + + FF °° + + FF π π + + FF π π + =+ ϕ ϕ =

⇒

− − FF °° + + − FF − FF ϕ ϕ == ;; and confronting: and confronting:

ccoossϕ ϕ ==ssiinnϕ ϕ , , whence whence ϕ = ϕ = 4545°° or or ϕ ϕ = = 225225°°..

x x

F

F is is positive positive when when ϕ ϕ = = 4545°° therefore it is parallel to, therefore it is parallel to, and has the same direction of,and has the same direction of, F F . Its magnitude_O_O. Its magnitude is

is FF_x_x = = F F

( (

2 12− =− 1

))

= 2200..7 7 NN, 45° down from, 45° down from AB AB .. When the plate is in equilibrium the t

When the plate is in equilibrium the torque, with respect to the cornerorque, with respect to the corner OO, is, is

( (

))

ssiinn 00 00 00 ssiinn 00 5 50 0 NN 00,,11000 0 mm 0,34 m, 0,34 m, ssiin 4n 455 2200..7 7 NN 00..77007711 x x QQ PP OO xx x x F F xx OOPP FF FF FF FF xx FF OOPP F F OPOP x x F F π π ϕ ϕ ϕ ϕ ⋅ ⋅ ⋅ ⋅ − + − + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ ⋅ ==

⇒

⋅ ⋅ ⋅ ⋅ + + ⋅ ⋅ ==

⇒

⋅ ⋅ ×× = = −− == −− ==−− ° ° ×× i.e. .34 m left of O. i.e. .34 m left of O. x x φ φ F F F F F F F Fxx

(18)

A-10

Assume

Assume WW₁₁ >>W W ₂₂. Then as. Then as W W goes ₁₁ goes down down −− D Dsinsinθ θ the weight the weight W W rises ₂₂ rises ++ D Dsinsinθ θ . From the. From the conservation of energy conservation of energy 11 11 22 22

( (

))

1 1 22 2 2 ssiinn ssiinn 00 W W W W vv WW DD WW DD g g θ θ θ θ



₊₊



⋅ ⋅ + + ⋅ ⋅ − − + + ⋅ ⋅ ==









.. Thus Thus 11 22 1 1 22 2 2 WW W W ssiinn vv gDgD W W W W θ θ − − = = + + ..

A-11

See the previous problem. In the

See the previous problem. In the present case,present case,

(

)

(

))

2 2 1 1 2 2 2 2

ssiinn ssiinn 00 ssiinn ssiinn W W vv WW DD WW DD vv ggDD g g θθ φφ φφ ϑϑ ⋅ ⋅ ⋅ ⋅ + + ⋅ ⋅ + + ⋅ ⋅ − − ==

⇒

= = −− ..

(19)

B-1

At At t t = =00

(

( )

.. ..

) (

₁₁

( )

.. ..

)

₂₂ 00,,

(

( )

.. ..

) (

₁₁

( ))

.. .. ₂₂ 2 2 H H K K EE == KK EE == PP EE == P EPE == MgMg , being , being M M₁₁ = = MM₂₂ == M M At At t t > >00

(

( )

) (

( )

)

11 22

(

( )

)

(

( ))

2 2 1 1 22 11 22 .. .. .. .. ,, .. .. ,, .. .. K K EE == K EK E == MMvv PP EE ==MMggy Py P EE ==MgMgxx x x y y

(20)

Length of the cord:

( (

11 22

))

2 2 H H + +

Length of the cord on the side of M

Length of the cord on the side of M22:: H H −− xx

Length of the cord on the side of M

Length of the cord on the side of M11::

( (

22 11

))

2 2 H H x x − − ++ Height of M Height of M22:: x x Height of M Height of M11:: 1 1 1 1 2 2 22 22 H H xx y y= =



_

− −



_

−−





Then Then at at t t > >00

( (

.. ..

))

₁₁ 11 11 2 2 22 22 H H xx P P EE ==MMggyy ==MMgg



_



_

_



_

−− −



_

−

_











and from t and from the conservation of energyhe conservation of energy

(

( )

) (

₁_1,, ₀₀

( )

) (

_2,,₂ ₀₀

( )

) (

₁_1,, ₀₀

( )

)

₂_2,, ₀₀

(

( )

) (

_1,,₁ ₀₀

( )

)

₂_2,, ₀₀

(

( )

) (

₁_1,, ₀₀

( ))

₂_2,, ₀₀ 2 2 22 2 2 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. 1 1 11 ₁₁ 0 0 00 11 2 2 2 2 22 22 22 22 22 1 1 11 1 1 11 .. 2 2 22 22 tt tt tt tt tt tt tt tt K K EE KK EE PP EE PP EE K EK E KK EE PP EE PP EE H H HH HH xx Mg Mg MgMg MvMv MvMv MgMg MgxMgx H H MgH MgH MvMv MgxMgx MgMg = = ++ == ++ == ++ == == >> ++ >> ++ >> ++ >>

⇒



_

_



+ + ++ ++ == ++ ++

_



−−



−− +

_

+

⇒

























= = ++



−−



++



−−



















Taking the derivative with respect to

Taking the derivative with respect to t t of each member of the above expression of each member of the above expression 1 1 0 0 22 11 00 2 2 Mva Mva MgvMgv





= = ++



−−



++





, where we used, where we used

dx dx vv dt dt = = and and 2 2 2 2 22 d dvv ddvv vv vava d dtt = = ddt t == .. a)

a) So the vertical acceleration of MSo the vertical acceleration of M22 at at t t > > 00 is is

1 1 11 1 1 2 2 22 a a = − = −



_

−−



_

gg





.. b)

b) Being Being aa < <00 the mass M the mass M22 will move downward. will move downward. The eight of The eight of MM22 is is

2 2 H H x

x = = and its velocityand its velocity

( ( ))

00 00

vv == at at t t = =00 whereas whereas 11 22 2

2 22

H H x

x = = at at ++ at at t t > >00; ; ththeen n it it wwilill l ststririkke e ththe e ggrorouund nd (( x x = = 00) at) at

time time ₁₁ 22 1 1 1 1 2 2 H H H H t t _a_a g g − − = = ==





− −









.. c) c) BeingBeing 11 11 2 2 22 22 H H xx y y= =



_

− −



_

−−





the height of Mthe height of M11, , then then for for x x = =00 is is

1 1 1 1 2 2 22 H H y y= =



_

− −



_

<<H H





; so the; so the other mass (M

(21)

B-2

When the boom rotates an angle

When the boom rotates an angle ∆∆θ θ (assumed small, i.e. neglecting terms of the order if (assumed small, i.e. neglecting terms of the order if

( (

∆∆θ θ

))

22oror higher) around the pivot the length

higher) around the pivot the length y y of of the the horizontal horizontal cable cable will will change change of of ∆ ∆ y y xxcoscosθ θ θ ∆∆θ _{. This can be}_{. This can be}

seen using Carnot’s theorem and the approx.

seen using Carnot’s theorem and the approx. 11 2 2

1

1+ + ≅ x x ≅ +11+ xx to find the length of the side opposite to to find the length of the side opposite to the angle the angle θ θ ::

( (

))

( (

))

2 2 22 22 22

ccooss 22 ccoos cs cooss ssiinn ssiinn 1 21 2 ccttgg 11 ssiinn ccttgg ccooss .. y y yy yy xx xx xx xx x x xx xx θ θ θθ θθ θθ θθ θ θ θθ θθ θθ θθ θθ θθ θθ ′′ ∆ ∆ = = − − = = + + − − ++ ∆ ∆ − − ≅≅ + + ∆∆ − −  ⋅⋅ ∆∆ = = ∆∆

Or, in a more straightforward way,

( (

))

ssiinn ssiinn ccooss ccoos ss siinn ssiinn ccooss ccooss .. y y xx xx xx xx xx y y yy yy xx θ θ θθ θθ θθ θθ θθ θθ θθ θθ θ θ ′′ ≅≅ ++ ∆∆ == ∆∆ + + ∆∆ ≅ ≅ ++ ∆∆

⇒

′′ ∆ ∆ = = − ≅− ≅

The height of the weight W changes of

The height of the weight W changes of ∆ ∆ = Z Z = LLccooss

( (

θθ + ∆ +∆ − θθ

))

− LLccoossθθ ≅ ≅ − − LLssiinnθθ∆∆θθ and the weight w of and the weight w of

the boom (assumed applied to the centre of mass of the boom) moves of the boom (assumed applied to the centre of mass of the boom) moves of

( (

))

2

2ccooss 22ccooss 22ssiinn L L LL LL z z θθ θθ θθ θθ θθ ∆ ∆ = = ++ ∆ ∆ − − ≅ ≅ − − ∆∆ .. Then from PVW: Then from PVW: 2 2 0

0 ccooss ssiinn ssiinn 00

ttaann .. 2 2 L L T T yy WW ZZ ww zz TT xx WW LL ww L L ww T T W W x x θ θ θ θ θ θ θ θ θ θ θ θ θ θ ∆ ∆ + + ∆ + ∆ + ∆ ∆ ==

⇒

⋅ ⋅ ∆ − ∆ − ⋅ ⋅ ∆ ∆ − − ⋅ ⋅ ∆ ∆ ≅≅

⇒





= =

_

++

_





y y

(22)

B-3

Let

Let L L = 10 ft = 10 ft the length of the length of the ladder,the ladder, r r = = 8 ft the initial height of the ro8 ft the initial height of the rollers at the top end allers at the top end and supposend suppose that, slightly tilting the ladder leaning against the vertical wall, the

that, slightly tilting the ladder leaning against the vertical wall, the rollers would go downrollers would go down ∆∆r r .. Then (assuming

Then (assuming ∆∆r r small) small)

(

)

) (

22

(

))

22

( (

))

(

)

(

))

2 2 22 22 22 22 22 22 2 2 22 22 22 0. 0. L L rr rr ss ss LL rr rr rr ss ss ss rr ss rr rr ss ss LL rr rr ss ss rr rr ss ss = = ++ ∆ ∆ + ++ + ∆∆

⇒

≅ ≅ + + ∆ ∆ + + + ∆ + ∆ = = + + + + ∆ ∆ + + ∆ ∆ = = + + ∆ ∆ + + ∆∆

⇒

∆ + ∆ = ∆ + ∆ = Thus Thus ss r r r r ss ∆ = ∆ = − − ∆∆ .. Furthermore, said

Furthermore, said hh_W_{W w}_,,_w the heights of the weights W the heights of the weights W and w, hangings from rungs atand w, hangings from rungs at L L_W_{W w}_,,_w feet from the feet from the bottom

bottom end, end, thenthen hh_{W w}_W_,,_w L LW wW,,w r r L L = = and and hh_{W w}_W_,,_w L LW wW,,w r r L L ∆ ∆ = = ∆∆ .. W

W is is hung hung ll = = 2.5 2.5 feetfeetfrom the top end and thereforefrom the top end and therefore L L_W_W = = −LL ll− whereas whereas

2 2 w w L L L L == ..

Let R, H and V, respectively, the force of the

Let R, H and V, respectively, the force of the roller on the wall, the horizontal roller on the wall, the horizontal and vertical forces of theand vertical forces of the ladder on the ground, then, from PVW,

ladder on the ground, then, from PVW,

(

)

(

))

0 0 00 0 0 .. 2 2 W W ww W W ww W W ww W W ww L L LL r r H H ss WW hh ww hh HH rr WW rr ww r r ss LL LL L L LL rr ss ss LL H H WW ww HH WLWL wLwL WW LL ll ww ss LL LL LLrr LLrr





∆ ∆ + + ∆ ∆ + + ∆ ∆ ==

⇒

_

− − ∆ ∆ +

_

+ ∆ + ∆ + ∆ ∆ ==

⇒









− − ++ ++ ==

⇒

== ++ ==

_

−− ++

_





s s r r L L

(23)

Now, from PVW in the form Now, from PVW in the form

∑

one gets one gets R R== H H and and VV = = WW ++ww, whence, whence a) a) R R = =45 45 lb-wtslb-wts b b) ) H H = = 445 5 llbb--wwttss,, V V ==90 90 llbb--wwttss

B-4

Let 3

Let L L == 3RR the length of the plank. Then we have the length of the plank. Then we havessiinn 33 6600 2 2 22 L L R R α α = = ==

⇒

α α = = °°..

For the heights of the weights W and w

For the heights of the weights W and w one gets thatone gets that

( ( ))

2 2 2 2 cos cos 2 2 W W L L h h = = − RR− RR −− θ θ ,,

( (

))

cos cos w w h

h = = − RR− RR α θ α −−θ and (for a small tilt and (for a small tilt ∆∆θ θ ))

(

( )

)

22

(

( ))

22

2 2

ssiinn 11 ssiinn ccooss ssiinn ssiinn 2 2 22 ₂₂ W W R R L L LL h h RR RR RR R R θ θ θθ θθ θθ αα θθ θθ θθ θθ ∆ ∆ ≅≅ −− ∆∆ == −− ∆∆ == ∆∆ == ∆∆ ,, being being 11 2 2 cos

cosα α = = ,, ∆ ≅ ∆ hh_w_w ≅ RRsinsin

( (

α α θ − − ∆θ

))

∆θ θ .. From PVW, remembering that

From PVW, remembering that∆∆hh_W_Wandand ∆∆hh_w_w lie in opposite directions, we have lie in opposite directions, we haveW hW∆ ∆ h_W_W = = ∆ww h∆h_w_w, whence, whence

(

)

(

)

(

))

ssiinn ssiinn ssiinn ssiinn ssiinn 3300 ..

2 2 22 22 R R W W W W θ θ θ ∆ = ∆ θ = wwRR α α θ − ∆ − θ θ ∆ = θ = RR α θ α − ∆− θ θ ∆θ

⇒

θ θ = = α θ θ α −−θ θ α

⇒

= = −α θ −θ θ

⇒

θ = = α α = = °° α α α α θ θ α α θ −−θ

(24)

B-5

F

Frroom m PPVVW W wwe e hhaavveeW xxW∆ ∆ ssiinn 3300°° + + ∆ WW yy∆ ssiinn 6600°° ==00 .. F

Frroom m tthhe e rreeccttaanngguullaar r ttrriiaanngglle e sshhoowwn n ((sseee e pprroobblleem m BB--33) ) x x xx yy yy 00 yy x x xx ttaann xx y y α α ∆ + ∆ = ∆ + ∆ =

⇒

∆ ∆ = = − − ∆ ∆ = = − − ∆∆ .. Then Then sin30 sin30 cos30 cos30

ssiinn 3300 ssiinn 6600 00 ssiinn 3300 ssiinn 6600 00 ssiinn 3300 ttaann ssiinn 6600 00 ssiinn 3300 ttaann ssiinn 6600 00 ssiinn 3300 ttaann ccooss 3300 00 ttaann ttaann 330 ,0 ,

W W xx WW yy xx yy xx α α xx α α α α α α °° °° ∆ ∆ °° + + ∆ ∆ °° ==

⇒

∆ ∆ °° ++ ∆ ∆ °° ==

⇒

∆ ∆ °° − − ∆ ∆ °° ==

⇒

°° − − °° ==

⇒

°° − − °° ==

⇒

= = = = °° and

and therefore therefore α α = = 3030°°..

B-6

Being

Being L L the length of the string and the length of the string and r r the radius of the sphere, then, from Carnot’s theorem, the radius of the sphere, then, from Carnot’s theorem,

(

)

22

(

)

22

(

)

(

))

2 2 ccooss L L rr+ ++ + ∆ ∆ = LL= RR + + − R rrR− − − R RR R r − − r α α ++ ∆∆α α .. x x y y R R R R α α L L r r h h

(25)

If

If ∆∆α α is small is small

(

L rr L+ ++

)

+ ∆ ∆ ≅ LL≅ RR22 + + −

(

R rrR− −

)

22 − 22

(

R RR

(

R rr− −

)

ccoossα α + + 22R RR

(

R r − −

))

r ssiinnα α α ∆∆α .. But

But R R22+ − +

( )

(

R rrR− −

)

22− 22R RR

(

( )

R rr− −

)

ccoossα α = = +

(

( ))

L r L+r 22, thus, thus

(

)

(

)

(

)

(

))

( (

))

( (

))

( (

))

( (

))

( (

))

( (

))

( (

))

2 2 2 2 2 2 2 2 ssiinn 11 ssiinn ssiinn ssiinn .. R R RR r r L L rr LL LL rr RR RR rr LL r r L L r r R R RR rr RR RR r r L L rr LL L L rr LL r r α α αα αα αα α α αα αα αα − − + + ++ ∆ ∆ ≅ ≅ + + + + − − ∆ ∆ ≅ ≅ + + + + ∆∆ + + − − −− + + + + ∆∆

⇒

∆ ∆ ∆∆ + + ++     Furthermore, being

Furthermore, being hh the the height height of of the the centre centre of of the the sphere,sphere,

( )

( ) (

ssiinn

(

)

) ( )

( )

ssiinn

( )

ccooss h

h R rrR αα αα R rrR αα R rR r αα αα ∆

∆ = = − − ++ ∆ ∆ − − − − ≅ ≅ − − ∆∆ .. Let

Let F F the the strength strength of of the the string, string, from from PVW: PVW: FF LL W hW h FF WW hh WW L r L r cotcot L L RR α α ∆ ∆ ++ ∆ ∆ = = ∆∆

⇒

= = ≅≅ ∆

∆ . One also gets. One also gets

49 49 4.4.5 5 ccmm==444.4.5c5cmm 4 400 44..5 c5 cmm==4444..55ccmm R R r r L L r r − − = = −− + + = = ++

So the triangle is isosceles so that

So the triangle is isosceles so that cotcot 11 2 2 ₂₂ L L r r F F W W h h α

α ==

⇒

≅≅ ++ . Let. Let hh the height of the isosceles, then the height of the isosceles, then

( (

))

22

( ( ))

22 ₄₄₄_4..5₅22 ₂₂₄_4..5₅22 ₁₁₃₃₈₈₀₀ 2 2 R R h h == +LL r +r −− == −− == and and 4444..55 2222..2255 0.60.6 2 2 22 11338800 11338800 L L r r F F WW WW WW WW h h + + ≅ ≅ = = == × ×  ..

B-7

(26)

The centres of the spheres lie on the vertices of a regular tetrahedron of edge

The centres of the spheres lie on the vertices of a regular tetrahedron of edge aa and height and height 66 3 3 a a , then , then cos cos 66 3 3 θ θ = = .. Let

Let ww the component of the weight_a_a the component of the weight ww along each edge along each edge aa. From symmetry considerations. From symmetry considerations 3

3× × ww_a_accoossθ θ ==ww, , whence whence 2 2 ton-wtston-wts 3cos 3cos 66 a a w w ww w w θ θ = = = = == ..

Along the horizontal surface of the tetrahedron

Along the horizontal surface of the tetrahedron ww has a component_a_a has a component sinsin 33 3 3

a

a aa

w

w θ θ = = ww ⋅⋅ and the two and the two welds at the contact points of each sphere with the two other exert along the same direction a force welds at the contact points of each sphere with the two other exert along the same direction a force

2

2 ccoTT oss 3300°° ==T T 33 , being, being T T the tension of each weld. the tension of each weld. Then from PVW (in the form

Then from PVW (in the form

∑

) and fixing the horizontal direction, one gets) and fixing the horizontal direction, one gets 3 3 3 3 3 3 33 a a a a w w T

T = = ww ⋅⋅

⇒

T T == , and thus, for a factor of , and thus, for a factor of safety of 3,safety of 3, TT_{( ( ))}₃₃ = = × 33× = TT = =ww_a_a =2 2 ttoonn--wwttss..

B-8

θ θ θ θ 30° 30°

(27)

Let

Let ll the lengthe length of th of the cthe cord,ord, hh₁₁,, h the heights of the two masses when they lie at h₂₂ the heights of the two masses when they lie at distancesdistancescc₁₁,, cc from₂₂from the top in downwards direction along the two legs of the

the top in downwards direction along the two legs of the right triangle.right triangle. Then from PVW

Then from PVW TT ll∆ = ∆ = ∆ mm h₁₁∆ + h₁₁+ ∆mm₂₂∆hh₂₂ where: where:

1 1 1 1 11 22 11 3 3 2 2 22 22 22 sin30 sin30 si sinn 6060 h h cc cc h h cc cc ∆ ∆ = = ∆ ∆ °° = = ∆∆ ∆ ∆ = = ∆ ∆ °° = = ∆∆ Being Being 1 1 2 2 cos cos sin sin cc ll cc ll α α α α = = = = one gets one gets 11 1

1 ssiinn ccooss 11 22ssiinn 22ccooss

ll cc ll αα αα αα ll hh αα αα αα ll ∆ ∆ = = − − ∆ ∆ + + ∆∆

⇒

∆ ∆ = = − − ∆ ∆ + + ∆∆ ,, 3 3 33 2

2 ccooss ssiinn 22 22 ccooss 22 ssiinn ll

cc ll αα αα αα ll hh ll αα αα αα ∆ ∆ = = ∆ ∆ + + ∆∆

⇒

∆ ∆ = = ∆ ∆ + + ∆∆ Thus Thus

(

11 33

)

(

11 33

))

1

1 11 22 22 22 11ccooss 22 22ssiinn 22 11ssiinn 22 22ccooss

T

T ll∆∆ ==mm h ∆∆h + +mm ∆∆hh = = mm αα ++ mm αα ∆∆ll −− mm αα −− mm αα ll∆∆αα ..

,,

ll α α ∆

∆ ∆∆ being independent being independent of of one one another, another, tacking tacking ∆ ∆ =ll =00 one finds that the system is in static one finds that the system is in static equilibrium when

equilibrium when mm₁₁ssiinnα α − − 33mm₂₂ccoossα α ==00

⇒

ttaannα α ==3 33 3 .. Read

Reading ing the the tabletables s one one gets gets α α = = 79 .179°°.1 and, and, taking taking ∆ ∆ =α α =00,,

3 3 1 1 1 1 22 2

2 ccooss 7799 ..11 22 ssiinn 7799 ..11 22665 5 gg--wwttss

T

(28)

B-9

Being 0

Being ∆ ∆ = L L =0 we we hhaavve e x x xx yy yy 00 yy x x xx ccoott 3300 xx y

y

∆

∆ + + ∆ =∆ =

⇒

∆ ∆ = = − − ∆ ∆ = = − − °°∆∆ .. But

But ∆ = ∆ x x= rr′ ′ ccooss 3300°° ++ ∆ ∆ − ss rr− ccooss 3300°° = = ∆ ∆ − ss− −

( (

rr − r r ′′

))

ccooss 3300°° . . Assuming

Assuming ∆∆ss small small we we have have rr− − ≅ rr′′≅ ∆ ∆ sscos30cos30°°.. Thus

Thus ∆ ≅ ∆ x x≅ ∆ − ∆ − ss

( (

11 ccooss 322300° ° =

))

= ∆ ∆ ssssiinn 322300°°..

a)

a) from PVWfrom PVW

2 2

ccoott 3300 ccoott 3300 ssiinn 3300 0

0

3 3 ccooss 330 s0 siinn 3300 ..

4 4 y y xx ss T T ss WW yy TT WW WW WW ss ss ss T T WW W W





∆ ∆



−− °°∆∆



°°⋅⋅ °°∆∆ ∆ ∆ + + ∆ ∆ ==

⇒

==−− ==−−

_

_

==

_

_

⇒

∆ ∆



∆ ∆



_

∆∆

_

= = ° ° °° == ss ∆ ∆ x x ∆ ∆ y y ∆ ∆ L L L L rr r’ r’

(29)

b) b)

Using the method of the force components (FC) on cart we have: Using the method of the force components (FC) on cart we have:

3 3 ccooss118800 ccooss 330000 00

2 2 T

T °° + + NN °° ==

⇒

TT == NN (horizontal). (horizontal).

Using FC on W:

Using FC on W: ccooss 00 ccooss 224400 00 11 2 2 N N °° + + WW °° ==

⇒

NN == W W .. So So 33 4 4 T T == W W ..

B-10

Assume the centre of the

Assume the centre of the bobbin goes downbobbin goes down H ∆∆ H andand mm risesrises ∆∆hh as the bobbin as the bobbin rolls downwards. Thenrolls downwards. Then from PVW: from PVW: M M H∆ ∆ = H = ∆mm h∆h.. horizontal horizontal N N N N T T normal normal

(30)

For a rolling of the bobbin of

For a rolling of the bobbin of L∆ ∆ = L = ∆RR∆α α then then M M goes goes down down ∆ ∆ = ∆ H H = ∆LLsinsinθ θ , whereas m rises (assuming, whereas m rises (assuming

α α ∆ ∆ small)small) ll rr r r LL R R α α ∆

∆ ≅ ≅ ∆ ∆ = = ∆∆ and goes down H and goes down ∆∆ H ..

Thus Thus

( (

))

ssiinn ssiinn ssiinn

ssiinn ssiinn ssiinn ssiinn 00,, 55

30 30 .. rr r r h h ll HH LL MM LL mm LL R R RR rr rr mm rr M M mm mm MM mm mm R R RR MM mm RR θ θ θ θ θ θ θ θ θ θ θθ θθ θ θ









∆ ∆ = = ∆ ∆ −− ∆ ∆ = = ∆ ∆

_

−−

_

⇒

∆ ∆ = = ∆ ∆

_

−−

_

⇒









= = −−

⇒

+ + ==

⇒

= = ==

⇒

+ + = = °°

B-11

If the chain rises

If the chain rises ∆∆hh resting in a horizontal circle, then it resting in a horizontal circle, then its radius varies of s radius varies of x x r r hh h h

∆

∆ = = ∆∆ and its length and its length of of∆ ∆ = ll = ∆22π π ∆xx.. F Frroom m PPVVWW: : 22 22 2 2 rr WhWh W W hh TT ll WW hh TT xx TT hh TT h h r r π π π π π π ∆ ∆ = = ∆∆

⇒

∆ ∆ = = ⋅ ∆ ⋅ ∆ = = ⋅ ⋅ ∆∆

⇒

== .. x x ∆ ∆ h h ∆ ∆

(31)

B-12

Assuming w rises

Assuming w rises ∆∆hh_w_w whereas W goes down whereas W goes down ∆∆hh_W_W, being:, being:

( (

llWW11 llWW22 llW W 33

))

00 ∆ + ∆ + + + == ,, 1 1 22 11 33 1 1 2 2 W W WW WW WW ll ll ll ll ∆ ∆ = ∆= ∆

⇒

∆ ∆ = = − − ∆∆ ,,

( (

))

1 1 22 00 w w ww ll ll ∆ ∆ + + == ,, 3 3 11 22 W W ww ww ll ll ll ∆ ∆ = = −−∆ ∆ = = ∆∆ ,, then then 2 2 33 22 22 w w ww WW ww h h ll ll ll ∆ ∆ = = ∆ ∆ ++ ∆ ∆ = = ∆∆ ,, 1 1 33 22 1 1 11 11 2 2 22 44

ssiinn ssiinn ssiinn ssiinn

W W WW WW ww ww h h ll θθ ll θθ ll θθ hh θθ ∆ = ∆ = ∆ ∆ = = − − ∆ ∆ = = − − ∆ ∆ = = − − ∆∆ .. From From PVW: PVW: ww h∆ + ∆ ∆ + h_w_w W hW∆ =h_W_W =00. So. So 11 4 4 4 4 ssiinn 00 sin sin w w ww w w w w hh θ θ WW hh WW θ θ ∆ ∆ − − ⋅ ⋅ ∆ ∆ ==

⇒

== ..

B-13

1 1 W W ll 2 2 W W ll 1 1 w w ll 2 2 w w ll 3 3 W W ll y y x x EF EF FG FG EG EG FG FG DF DF