Embed Size (px)
Citation preview
Mechanics - Basic Physical ConceptsMathematics
Quadratic Eq.: a x2 + b x+ c = 0, x = −b±√b2−4 a c2 a
Cartesian and polar coordinates:
x = r cos θ, y = r sin θ, r2 = x2 + y2, tan θ = yx
Trigonometry: cosα cosβ + sinα sinβ = cos(α− β)
sinα+ sinβ = 2 sin α+β2 cos α−β2
cosα+ cosβ = 2 cos α+β2 cos α−β2
sin 2 θ = 2 sin θ cos θ, cos 2 θ = cos2 θ − sin2 θ
1− cos θ = 2 sin2 θ2 , 1 + cos θ = 2 cos2 θ
2Vector algebra: ~A = (Ax, Ay) = Ax ı+Ay
Resultant: ~R = ~A+ ~B = (Ax +Bx, Ay +By)
Dot: ~A · ~B = AB cos θ = AxBx +Ay By +Az Bz
Cross product: ı× = k, × k = ı, k × ı =
~C = ~A× ~B =
∣
∣
∣
∣
∣
∣
ı k
Ax Ay AzBx By Bz
∣
∣
∣
∣
∣
∣
C = AB sin θ = A⊥B = AB⊥, use right hand rule
Calculus: ddx
xn = nxn−1, ddx
lnx = 1x ,
ddθ
sin θ = cos θ, ddθ
cos θ = − sin θ, ddx
const = 0
Measurements
Dimensional analysis: e.g.,
F = ma→ [M ][L][T ]−2, or F = m v2
r → [M ][L][T ]−2
Summation:∑Ni=1(a xi + b) = a
∑Ni=1 xi + bN
Motion
One dimensional motion: v = d sdt
, a = d vdt
Average values: v =sf−si
tf−ti , a =vf−vi
tf−tiOne dimensional motion (constant acceleration):
v(t) : v = v0 + a t
s(t) : s = v t = v0 t+ 12 a t2, v = v0+v
2v(s) : v2 = v2
0 + 2 a s
Nonuniform acceleration: x = x0 + v0 t + 12 a t2 +
16 j t3 + 1
24 s t4 + 1120 k t5+ 1
720 p t6 + . . ., (jerk, snap,. . .)
Projectile motion: trise = tfall =ttrip
2 =v0y
g
h = 12 g t2
fall, R = vox ttrip
Circular: ac =v2
r , v = 2π rT
, f = 1T
(Hertz=s−1)
Curvilinear motion: a =√
a2t + a2
r
Relative velocity: ~v = ~v′ + ~u
Law of Motion and applications
Force: ~F = m~a, Fg = mg, ~F12 = −~F21
Circular motion: ac =v2
r , v = 2π rT
= 2π r f
Friction: Fstatic ≤ µsN Fkinetic = µk N
Equilibrium (concurrent forces):∑
i~Fi = 0
Energy
Work (for all F): ∆W = WA→B = WB − WA =
F s‖ = F‖s = Fs cos θ = ~F · ~s→∫BA
~F · d~s (in Joules)
Effects due to work done: Fext = ma+ Fc + fncWext|A→B = KB −KA + UB − UA +Wdiss|A→B
Kinetic energy: KB−KA =∫BA m~a·d~s, K = 1
2 mv2
K (conservative ~F ): UB − UA = −∫BA
~F · d~s
Ugravity = mg y, Uspring = 12 k x2
From U to ~F : Fx = −∂ U∂x
, Fy = −∂ U∂y
, Fz = −∂ U∂z
Fgravity = −∂ U∂y
= −mg, Fspring = −∂ U∂x
= −k x
Equilibrium: ∂ U∂x
= 0, ∂2 U∂x2 > 0 stable, < 0 unstable
Power: P = dWdt
= F v‖ = F v cos θ = ~F · ~v (Watts)
Collision
Impulse: ~I = ∆~p = ~pf − ~pi →∫ tfti
~F dt
Momentum: ~p = m~v
Two-body: xcm = m1 x1+m2 x2m1+m2
pcm ≡M vcm = p1 + p2 = m1 v1 +m2 v2Fcm ≡ F1 + F2 = m1 a1 +m2 a2 = M acm
K1 +K2 = K∗1 +K∗2 +Kcm
Two-body collision: ~pi = ~pf = (m1 +m2)~vcm
v∗i = vi − vcm , v′i = v∗′i + vcm
Elastic: v1 − v2 = −(v′1 − v′2),v∗′i = −v∗i , v′i = 2 vcm − vi
Many body center of mass: ~rcm =
∑
mi~ri∑
mi=
∫
~r dm∫
mi
Force on cm: ~Fext =d~pdt
= M~acm , ~p =∑
~piRotation of Rigid-Body
Kinematics: θ = sr , ω = v
r , α = atr
Moment of inertia: I =∑
mi r2i =
∫
r2dm
Idisk = 12 M R2, Iring = 1
2 M (R21 +R2
2)
Irod = 112 M `2, Irectangle =
112 M (a2 + b2)
Isphere =25 M R2, Ispherical shell =
23 M R2
I = M (Radius of gyration)2, I = Icm +M D2
Kinetic energies: Krot =12 I ω2, K = Krot +Kcm
Angular momentum: L = rmv = rmω r = I ω
Torque: τ = dLdt
= m d vdt
r = F r = I dωdt
= I α
Wext = ∆K+∆U+Wf , K = Krot+12 mv2, P = τ ω
Rolling, angular momentum and torque
Rolling: K = 12
(
Ic +M R2)
ω2 = 12
(
Ic
R2 +M)
v2
Angular momentum: ~L = ~r × ~p, L = r⊥ p = I ω
Torque: ~τ = d ~Ldt
= ~r × d~pdt
= ~r × ~F , τ = r⊥ F = I α
Gyroscope: ωp = dφdt
= 1LdLdt
= τL
= mg hI ω
Static equilibrium∑ ~Fi = 0, about any point
∑
~τi = 0
Subdivisions: ~rcm = mA~rAcm+mB ~rBcmmA+mB
Elastic modulus = stress/strain
stress: F/A
strain: ∆L/L, θ ≈ ∆x/h, −∆V/V
Oscillation motion
f = 1T, ω = 2π
T
SHM: a = d2 xdt2
= −ω2 x, α = d2 θdt2
= −ω2 θ
x = xmax cos(ω t+ δ), xmax = A
v = −vmax sin(ω t+ δ), vmax = ωA
a = −amax cos(ω t+ δ) = −ω2 x, amax = ω2A
E = K + U = Kmax =12 m (ωA)
2 = Umax =12 k A
2
Spring: ma = −k x
Simple pendulum: maθ = mα` = −mg sin θ
Physical pendulum: τ = I α = −mg d sin θ
Torsion pendulum: τ = I α = −κ θ
Gravity~F21 = −G
m1 m2
r212
r12, for r ≥ R, g(r) = G Mr2
G = 6.67259× 10−11 Nm2/kg2
Rearth = 6370 km, Mearth = 5.98× 1024 kg
Circular orbit: ac =v2
r = ω2 r =(
2πT
)2r = g(r)
U = −G mMr , E = U +K = −GmM
2 r
F = −dUdr= −mG M
r2 = −mv2
r
Kepler’s Laws of planetary motion:
i) elliptical orbit, r = r01−ε cos θ r1 =
r01+ε , r2 =
r01−ε
ii) L = rm ∆r⊥∆t −→
∆A∆t =
12r∆r⊥∆t = L
2m = const.
iii) G Ma2 =
(
2π aT
)2 1a , a = r1+r2
2 , T 2 =(
4π2
GM
)
r3
Escape kinetic energy: E = K + U(R) = 0
Fluid mechanics
Pascal: P = F⊥1
A1= F⊥2
A2, 1 atm = 1.013× 105 N/m2
Archimedes: B =M g, Pascal=N/m2
P = Patm + ρ g h, with P = F⊥Aand ρ = m
V
F =∫
P dA −→ ρ g `∫ h0 (h− y) dy
Continuity equation: Av = constant
Bernoulli: P + 12 ρ v2 + ρ g y = const, P ≥ 0
Wave motion
Traveling waves: y = f(x− v t), y = f(x+ v t)
In the positive x direction: y = A sin(k x− ω t− φ)
T = 1f, ω = 2π
T, k = 2π
λ, v = ω
k= λ
T
Along a string: v =√
Fµ
General: ∆E = ∆K +∆U = ∆Kmax
P = ∆E∆t =
12∆m∆t (ωA)
2
Waves: ∆m∆t =∆m∆x ·
∆x∆t =
∆m∆x · v
P = 12 µ v (ωA)
2, with µ = ∆m∆x
Circular: ∆m∆t =∆m∆A · ∆A∆r ·
∆rdt= ∆m∆A · 2π r v
Spherical: ∆m∆t =∆m∆V · 4π r2 v
Sound
v =√
Bρ , s = smax cos(k x− ω t− φ)
∆P = −B ∆VV= −B ∂s
∂x∆Pmax = B κsmax = ρ v ω smax
Piston: ∆m∆t =∆m∆V · A∆x∆t = ρAv
Intensity: I = PA= 12 ρ v (ω smax)
2
Intensity level: β = 10 log10II0, I0 = 10
−12 W/m2
Plane waves: ψ(x, t) = c sin(k x− ω t)
Circular waves: ψ(r, t) = c√rsin(k r − ω t)
Spherical: ψ(r, t) = cr sin(k r − ω t)
Doppler effect: λ = v T , f0 =1T, f ′ = v′
λ′
Here v′ = vsound ± vobserver, is wave speed relative
to moving observer and λ′ = (vsound ± vsource)/f0,
detected wave length established by moving source of
frequency f0. freceived = freflectedShock waves: Mach Number= vsource
vsound= 1sin θ
Superposition of waves
Phase difference: sin(k x− ωt) + sin(k x− ω t− φ)
Standing waves: sin(k x− ω t) + sin(k x+ ω t)
Beats: sin(kx− ω1 t) + sin(k x− ω2 t)
Others: a cos(k x− ω t) + b sin(k x− ω t)
y = sin(k x− ω t), z = sin(k x− ω t)
Fundamental modes: Sketch wave patterns
String: λ2 = `, Rod clamped middle: λ
2 = `,
Open-open pipe: λ2 = `, Open-closed pipe: λ
4 = `
Temperature and heat
Conversions: F = 95 C + 32
◦, K = C + 273.15◦
Constant volume gas thermometer: T = aP + b
Thermal expansion: α = 1`d `dT, β = 1
Vd VdT
∆` = α `∆T , ∆A = 2αA∆T , ∆V = 3αV ∆T
Ideal gas law: P V = nRT = N k T
R = 8.314510 J/mol/K = 0.0821 L atm/mol/K
k = 1.38× 10−23J/K, NA = 6.02× 1023, 1 cal=4.19 J
Calorimetry: ∆Q = cm∆T, ∆Q = L∆m
First law: ∆U = ∆Q−∆W , W =∫
P dV
Conduction: H = ∆Q∆t = −k A
∆T∆` , ∆Ti =
−HA
`iki
Stefan’s law: P = σ AeT 4, σ = 5.67× 10−8 Wm2 K4
Kinetic theory of gas
Ideal gas: ∆px = 2mvx, F = ∆px∆t =
mv2x
d
Pressure: P = N FA= mN
Vv2x =
mN3V v2
P = 23NVK, Kx =
K3 =
12 k T , T = 273 + Tc,
P V = N k T , n = N/NA, k = 1.38×10−23 J/K,
NA = 6.02214199× 1023 #/kg/mole
Constant V: ∆Q = ∆U = n CV ∆T
Constant P: ∆Q = n CP ∆T
γ =CP
CV
, CP − CV = R
CV =d2 R, for transl.+rot+vib, d = 3 + 2 + 2
Adiabatic expansion: P V γ = constant
Mean free path: ` = vrms t(vrel)rms t
1π d2 n
V
= 1√2π d2 n
V
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 3
This print-out should have 32 questions, checkthat it is complete. Multiple-choice questionsmay continue on the next column or page:find all choices before making your selection.The due time is Central time.
Airplane Momentum
11:03, calculus, numeric, > 1 min.001
An airplane of mass 12000 kg flies level to theground at an altitude of 10 km with a constantspeed of 175 m/s relative to the Earth.What is the magnitude of the airplane’s
angular momentum relative to a ground ob-server directly below the airplane in kg·m2/s?Correct answer: 2.1× 1010 kg ·m2/s.Explanation:
Since the observer is directly below the air-plane,
L = hmv
002
Does this value change as the airplane contin-ues its motion along a straight line?
1. No. L = constant. correct
2. Yes. L increases as the airplane moves.
3. Yes. L decreases as the airplane moves.
4. Yes. L changes in a random pattern asthe airplane moves.
5. Yes. L changes with certain period as theairplane moves.
Explanation:
L = constant since the perpendicular dis-tance from the line of flight to Earth’s surfacedoesn’t change.
Mass on Solid Cylinder
11:03, calculus, numeric, > 1 min.003
Given: g = 9.8 m/s2 .A 4 kg mass is attached to a light cord,
which is wound around a pulley. The pulley
is a uniform solid cylinder of radius 8 cm andmass 1 kg.What is the resultant net torque on the
system about the center of the wheel?Correct answer: 3.136 kg m2/s2.Explanation:
The net torque on the system is the torqueby the external force, which is the weight ofthe mass. So it is given by
τ = r F sinφ = rmg sin 90◦ = rmg
= (0.08 m) (4 kg) (9.8 m/s2)
= 3.136 kg m2/s2 .
004
When the falling mass has a speed of 5 m/s,
the pulley has an angular velocity ofv
r.
Determine the total angular momentum ofthe system about the center of the wheel.Correct answer: 1.8 kgm2/s.Explanation:
The total angular momentum has twoparts, one of the pulley and one of the mass.So it is
‖~L‖ = ‖~r ×m~v + I ~ω‖
= rmv +1
2M r2
(v
r
)
= r
(
m+M
2
)
v
= (8 cm)
[
(4 kg) +(1 kg)
2
]
v
= (0.36 kgm) v
= (0.36 kgm) (5 m/s)
= 1.8 kgm2/s .
005
Using the fact that τ = dL/dt and your resultfrom the previous part, calculate the acceler-ation of the falling mass.Correct answer: 8.71111 m/s2.Explanation:
Use the torque-angular momentum rela-tion, we have
τ =dL
dt=
d
dt(0.36 kgm v) = 0.36 kgm a ,
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 4
Solving for acceleration
a =τ
0.36 kgm
=(3.136 kg m2/s2)
(0.36 kgm)
= 8.71111 m/s2 .
Meter Stick
11:04, calculus, multiple choice, > 1 min.006
A uniform meter-stick with length L pivotsat point O. The meter stick can rotate freelyabout O. We release the stick from the hori-zontal position at t = 0.
OL
Determine the angular acceleration imme-diately after the release of the stick. (Hint:Consider the equation of motion at t = 0.)
1.g
4L
2.g
3L
3.g
2L
4.3g
4L
5.5g
6L
6.g
L
7.5g
4L
8.3g
2Lcorrect
9.7g
4L
10.2g
L
Explanation:
The equation of motion τ = Iα gives:
mgL
2= Iα =
1
3mL2α,
So
α =3g
2L.
007
Assume the stick has a sufficient width, coinsmay be placed on the stick. Put coin 1at P1, where OP1 = L/2, coin 2 at P2,where OP2 = 3L/4 and coin 3 at P3, whereOP3 = L. Now we release the stick from thehorizontal position at the time t = 0.
L/2O
3L/4L
Which coins are expected to stay on thestick immediately after the release of thestick? Assume when ac, the acceleration ofthe coin is greater than or equal to as, the localacceleration of the meter stick, i.e., ac ≥ as,the coin will stay on the stick. But whenac < as, the coin will be detached from (i.e.,will not stay on) the stick. (Hint: Comparethe linear acceleration of the segment of thestick beneath each coin.)
1. None will stay on the stick
2. Only coin 1 will stay on the stick correct
3. Only coin 2 will stay on the stick
4. Only coin 3 will stay on the stick
5. Only coin 1 and coin 2 will stay on thestick
6. Only coin 1 and coin 3 will stay on thestick
7. Only coin 2 and coin 3 will stay on thestick
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 5
8. All will stay on the stick
Explanation:
From part 1, we know the angular accelera-tion of the stick is
α =3g
2L.
The downward linear acceleration of the stickat P1 is
a1 = α(L/2) =3g
4.
At P2, the downward linear acceleration is
a2 = α(3L/4) =9g
8.
At P3, the downward linear acceleration is
a3 = αL =3g
2.
At the moment when the stick is released, theaccelerations of all the coins are g, so at coin1, the corresponding segment of the stick fallsslower than the coin, at coin 2 and coin 3,the corresponding segments of the stick falloff faster than the coins. This implies thatonly coin 1 will stay on the stick.
Rolling Disk
11:04, calculus, multiple choice, > 1 min.008
Consider the descending motion of a uniformdisk with mass m and radius b. It is rollingdown vertically as indicated in the sketch.
Τ
h
aP
vP
b
Determine the downward linear accelerationa.
1. a =2
3g correct
2. a =3
4g
3. a =1
4g
4. a =1
2g
5. a =1
3g
6. a =3
2g
Explanation:
Apply “F = ma” to the center of mass of thedisk,
mg − T = ma. (1)
Apply “τ = Iα” to the disk (where the torqueis calculated about the center),
Tb = Iα
=1
2mb2α
=1
2mba. (2)
=⇒
T =1
2ma. (3)
In the 3rd step we used a = αb. Substitutingequation (3) into (1) leads to,
mg −1
2ma = ma
=⇒
a =2
3g,
and
T =1
2ma =
1
3mg.
009
Determine the tension T .
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 6
1. T =2
5mg
2. T =1
2mg
3. T =2
3mg
4. T =1
3mg correct
5. T = mg
6. T =3
2mg
7. T =3
4mg
Explanation:
010
Determine the descending speed vP of its cen-ter, after the disk has dropped for a heighth.
1. vP =
√
4
3gh correct
2. vP =
√
2
3gh
3. vP =
√
1
2gh
4. vP =
√
1
3gh
5. vP =
√
3
2gh
6. vP =√
2gh
7. vP =√
gh
8. vP =
√
5
3gh
Explanation:
Applying “v2 − v20 = 2as” to the motion of
the center of mass, (v0 = 0),
v2P = 2as = 2
(
2
3g
)
h
=⇒
vP =
√
4
3gh.
Alternative method:
The application of work-energy theoremgives:
mgh =1
2mv2
P +1
2Iω2
P .
But
1
2Iω2
P =1
2
(
1
2mb2
)
ω2P =
1
4mv2
P .
So
mgh =
(
1
2+1
4
)
mv2P
=⇒ vP =
√
4
3gh.
Clay Rotates a Rod
11:05, calculus, multiple choice, > 1 min.011
A uniform rod, supported and pivoted at itsmidpoint, but initially at rest, has a mass 2mand a length l. A piece of clay with mass mand velocity v hits the very top of the rod,gets stuck and causes the rod-clay system tospin about the pivot point O in a horizontalplane. Viewed from above the scheme is
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 7
m, v
l O
2m
With respect to the pivot point O, whatis the magnitude of the initial angular mo-mentum Li of the piece of clay and the finalmoment of inertia If of the clay-rod system?
1. Li = mv l , If =5
12ml2
2. Li = mvl
2, If =
8
12ml2
3. Li = mv l , If =3
12ml2
4. Li = mv l , If =8
12ml2
5. Li = mv l , If =4
12ml2
6. Li = mvl
2, If =
5
12ml2 correct
7. Li = mvl
2, If =
4
12ml2
8. Li = mvl
2, If =
7
12ml2
Explanation:
Basic Concepts: Conservation of angularmomentum ~L
~L = ~r ×~p = m~r × ~v
∑
~τext =d ~L
dt
Lz = I ω .
Therefore, if the net external torque acting ona system is zero, the total angular momentumof that system is constant.Since the total external torque acting on
the system clay-rod is zero, the total angularmomentum is a constant of motion. The totalinitial angular momentum Li is simply the
angular momentum of the clay, since the rodis at rest initially
Li = ‖~r ×~p‖ = mr v
=mv l
2.
The final moment of inertia If of the clay-rodsystem is the moment of inertia of the rodplus the moment of inertia of the clay
If = Irod + Iclay
=1
122ml2 +m
(
l
2
)2
=5
12ml2 .
012
The final angular speed ωf of the rod-claysystem is
1. ωf =6
5v .
2. ωf =12
5
v
l.
3. ωf =6
5
v
l. correct
4. ωf =4
6v .
5. ωf =12
7
v
l.
6. ωf =5
6
v
l.
7. ωf =6
2
v
l.
8. ωf =5
12
v
l.
9. ωf =12
7v .
10. ωf =3
5
v
l.
Explanation:
The total final angular momentum is thesame as the total initial angular momentum.According to L = Iω, we have
Lf = Li
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 8
=mv l
2= If ωf
=5
12ml2 ωf ,
so
ωf =6
5
v
l.
Bohr Model of Hydrogen 02
11:07, calculus, numeric, > 1 min.013
In the Bohr’s model of the hydrogen atom,the electron moves in a circular orbit of radius5.29 × 10−11 m around the proton. Assumethat the orbital angular momentum of theelectron is equal to h.Calculate the orbital speed of the electron.
Correct answer: 1.37503× 107 m/s.Explanation:
The angular momentum of the electron in theground state of the hydrogen atom (this thecase here)in the Bohr’s model is h, therefore:
L = mv r = h
and solving for v,
v =h
mr
=(6.62608× 10−34 J s)
(9.10939× 10−31 kg)(5.29× 10−11 m)
= 1.37503× 107 m/s
014
Calculate the angular frequency of the elec-tron’s motionCorrect answer: 2.5993× 1017 s−1.Explanation:
The angular frequency is given by
ω =v
r
=(1.37503× 107 m/s)
(5.29× 10−11 m)
= 2.5993× 1017 s−1
AP B 1993 MC 57
12:01, trigonometry, multiple choice, < 1 min.015
Two objects, of masses 6 and 8 kilograms,are hung from the ends of a stick that is70 centimeters long and has marks every 10centimeters, as shown.
6 kg
A B CD E
8 kg
If the mass of the stick is negligible, atwhich of the points indicated should a cord beattached if the stick is to remain horizontalwhen suspended from the cord?
1. A
2. B
3. C
4. D correct
5. E
Explanation:
For static equilibrium, τnet = 0.Denote x the distance from the left end
point of the stick to the point where the cordis attached.
0 = τ = (6) g x− (8) g (70− x)
=⇒ 6x− 8 (70− x) = 0
=⇒ 14x = 560
=⇒ x = 40 cm.
Therefore the point should be point D.
Equilibrium of Hinged Lever 01
12:01, calculus, multiple choice, > 1 min.016
A uniform rod pivoted at one end, point O, isfree to swing in a vertical plane in a gravita-tional field. However, it is held in equilibriumby a force F at its other end.
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 9
x
y
`
F
Fx
Fy
WRx
R Ryθ
O
Force vectors are drawn to scale.What is the condition for translational
equilibrium along the horizontal x direction?
1. −Rx + Fx = 0 correct
2. Fx = 0
3. Rx − Fx cos θ = 0
4. Rx − Fx sin θ = 0
5. Fx cos θ −Rx sin θ = 0
Explanation:
Using the distances, angles and forces de-
picted in the figure, the condition∑
Fx = 0
for translational equilibrium in the x directionis
−Rx + Fx = 0 .
017
What is the condition for translational equi-librium along the vertical y direction?
1. Ry + Fy −W = 0 correct
2. Ry + Fy = 0
3. Ry − Fy +W = 0
4. W −Ry + Fy = 0
5. Ry sin θ + Fy sin θ −W cos θ = 0
Explanation:
For the equilibrium in the y direction,
∑
Fy = 0 turns out to be
Ry + Fy −W = 0 .
018
Taking the origin of the torque equation atpoint O, what is the condition for rotationalequilibrium?
1. Fy ` cos θ − W`
2cos θ − Fx ` sin θ = 0
correct
2. 2Fy ` sin θ −W ` cos θ − 2Fx ` sin θ = 0
3. Fy ` cos θ −W ` sin θ + Fx ` sin θ = 0
4. Fy ` sin θ −W`
2sin θ − Fx ` sin θ = 0
5. W`
2− Fx ` cos θ − Fy ` cos θ = 0
Explanation:
For rotational equilibrium about point O,the net torque on the system about that pointmust vanish. The angle θ appears as follows
x
y
`Fx
Fy
WW cos θ
Fy cos θ
Fx sin θ
θ
θ
θ
θ
O
and we see that the forces Rx and Ry exertno torque on the point O. From the figure wehave, counterclockwise
` Fy cos θ −`
2W cos θ − ` Fx sin θ = 0 .
Tilting a Block
12:02, calculus, numeric, > 1 min.019
Given: g = 9.8 m/s2 .Consider the rectangular block of massm =
40 kg height h = 1 m, length l = 0.6 m. A
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 10
force F is applied horizontally at the upperedge.
l
mh
F
What is the minimum force required tostart to tip the block?Correct answer: 117.6 N.Explanation:
Basic Concepts: In equilibrium
∑
~F = 0
∑
~τ = 0
Part 1: Let the right lower corner of theblock be denoted as A. Then
∑
τA = hF −l
2mg = 0
Therefore
F =mg l
2h
= (40 kg) (9.8 m/s2)0.6 m
2 (1 m)
= 117.6 N .
020
What is the minimum coefficient of static fric-tion required for the block to tip with the ap-plication of a force of this magnitude?Correct answer: 0.3 .Explanation:
∑
Fy = NA −mg = 0
Therefore
NA = mg
and
∑
Fx = F − f
= F − µNA
= 0 .
Solving for µ,
µ =F
NA
=F
mg
=117.6 N
(40 kg) (9.8 m/s2)= 0.3
Bricks on the Brink
12:03, calculus, numeric, > 1 min.021
A uniform brick of length 20 cm is placedover the edge of a horizontal surface withthe maximum overhang x possible withoutfalling.
L
x
x
y
Find x for a single block.Correct answer: 10 cm.Explanation:
Basic Concepts: The definition of thecenter of mass (n bricks):
xcm ≡
n∑
i=1
ximi
n∑
i=1
mi
=1
n
n∑
i=1
xi ,
where xi is the center of mass position of theith brick and mi is the mass of the i
th brick.Solution: The center of mass of a single
brick is in its middle or1
2of a brick’s length
from its maximum overhang. Sincex1
L=1
2,
as measured from the maximum overhang,
xcm
L
∣
∣
∣
n=1=
12
1=1
2
022
Two identical uniform bricks of length 20 cm
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 11
are stacked over the edge of a horizontal sur-face with the maximum overhang x possiblewithout falling.
L
x
x
y
Find x for two blocks.Correct answer: 15 cm.Explanation:
Basic Concepts: Sincemi = m (all bricks
have the same mass),n∑
i=1
m = nm. Note:
The bricks will just balance when the centerof mass is over the fulcrum; i.e., the edge ofthe horizontal surface. Note: In the solutionbelow, measurement will be made from theleft edge of the top brick with the maximumoverhand. To calculate the center of massxcm
L=
1
nL
n∑
i=1
xi, when an additional brick
is positioned at the bottom of the stack (nbricks), the additional brick’s edge is placedat the edge of the horizontal surface of theprevious stack (n−1 bricks). The center of
mass of the additional brick is1
2of a brick’s
length plus the maximum overhang of theprevious stack (n−1 bricks).
Solution: The top brick can extend1
2of a
brick’s length from the maximum overhang.
When the top brick extends1
2its length past
the second brick, the center of mass of the top
two bricks is in their middle or3
4of a single
brick’s length from the maximum overhang.
Sincex2
L=
xcm
L
∣
∣
∣
n=1+1
2=1
2+1
2= 1, as
measured from the maximum overhang
xcm
L
∣
∣
∣
n=2=
12 + 1
2=3
4.
023
Three identical uniform bricks of length 20 cmare stacked over the edge of a horizontal sur-face with the maximum overhang x possiblewithout falling.
L
x
x
y
Find x for three blocks.Correct answer: 18.3333 cm.Explanation:
The top two bricks can extend3
4of a brick’s
length from the maximum overhang. When
the top two bricks extended3
4of their length
past the third brick, the center of mass of the
top three bricks is in their middle or11
12of a
brick’s length from the maximum overhang.
Sincex3
L=
xcm
L
∣
∣
∣
n=2+1
2=3
4+1
2=5
4, as
measured from the maximum overhang,
xcm
L
∣
∣
∣
n=3=
12 + 1 +
54
3=11
12.
024
n identical uniform bricks of length L arestacked over the edge of a horizontal surfacewith the maximum overhang x possible with-out falling.Find x for n blocks.
1.L
2
n∑
i=1
1
2 i− 1
2. L
n∑
i=1
1
i+ 1
3. L
n∑
i=1
1
2i
4.3L
2
n∑
i=1
1
i+ 2
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 12
5.L
2
n∑
i=1
1
icorrect
6. Ln∑
i=1
1
i! + 1
7.L
2
n∑
i=1
(
1
i+ 1+1
4
)
8.L
2
n∑
i=1
1
i!
9. L
n∑
i=1
i
2i
Explanation:
We can rewrite the first three solutions:
xcm
L
∣
∣
∣
∣
n=1
=1
2=1
2(1)
xcm
L
∣
∣
∣
∣
n=2
=3
4=1
2
(
3
2
)
=1
2
(
1 +1
2
)
xcm
L
∣
∣
∣
∣
n=3
=11
12=1
2
(
11
6
)
=1
2
(
6
6+3
6+2
6
)
=1
2
(
1 +1
2+1
3
)
.
For four bricks, the top three bricks can ex-
tend11
12of a brick’s length from the maximum
overhang. When the top three bricks extend11
12of their length past the fourth brick, the
center of mass of the top three bricks is in
their middle or25
24of a brick’s length from the
maximum overhang. Sincex4
L=
xcm
L
∣
∣
∣
n=3+1
2=11
12+1
2=17
12, as
measured from the maximum overhang,
xcm
L
∣
∣
∣
∣
n=4
=12 + 1 +
54 +
1712
4
=25
24=1
2
(
25
12
)
=1
2
(
12
12+6
12+4
12+3
12
)
=1
2
(
1 +1
2+1
3+1
4
)
.
For five bricks,x5
L=
xcm
L
∣
∣
∣
n=4+1
2=25
24+1
2=37
24, and
xcm
L
∣
∣
∣
∣
n=5
=12 + 1 +
54 +
1712 +
3724
5
=137
120=1
2
(
137
60
)
=1
2
(
60
60+30
60+20
60+15
60+12
60
)
=1
2
(
1 +1
2+1
3+1
4+1
5
)
.
For six bricks,x6
L=
xcm
L
∣
∣
∣
n=5+1
2=137
120+1
2=197
120, so
xcm
L
∣
∣
∣
∣
n=6
=12+ 1 + 5
4+ 17
12+ 37
24+ 197
120
6
=441
180=1
2
(
441
60
)
=1
2
(
180
180+90
180+60
180+45
180
+36
180+30
180
)
=1
2
(
1 +1
2+1
3+1
4+1
5+1
6
)
.
And so forth. . .Thus for n bricks, we have
xcm
L
∣
∣
∣
∣
n bricks
=1
2
(
1 +1
2+1
3+1
4
+1
5+1
6+ · · ·+
1
n
)
=1
2
n∑
i=1
1
i.
Alternate Perspective: Rewriting thesequence for the xcm’s we have
xcm
L=1
2
(
1,
1 +1
2,
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 13
1 +1
2+1
3,
1 +1
2+1
3+1
4,
1 +1
2+1
3+1
4+1
5,
1 +1
2+1
3+1
4+1
5+1
6, ...
)
=1
2
(
1,3
2,11
6,25
12,137
60,
441
180,664
210, ... ,
n∑
i=1
1
i
)
.
So n bricks
(
i.e. using1
2times the sum of
i elements in the series,1
i
)
, we have
xcm
L
∣
∣
∣
∣
n bricks
=1
2
(
1 +1
2+1
3+1
4
+1
5+1
6+ · · ·+
1
n
)
=1
2
n∑
i=1
1
i.
Note: The maximum overhang occurs asgiven, that is, when the top block is extendedas far out as possible on the second block, andthe top two blocks are extended as far outas possible on the third block, and so forth.Other mechanical arrangements do not give amaximum overhang. Furthermore, this seriesdoes not converge, so any overhang is possi-ble. A careful classroom demonstration usingonly four bricks will show the top brick com-pletely past the edge of the horizontal surface(x = L). How many bricks are required tohave the overhang x = 2L?
Dishonest Shopkeeper
12:03, calculus, numeric, > 1 min.025
Two pans of a balance are 50 cm apart. Thefulcrum of the balance has been shifted 1 cmaway from the center by a dishonest shop-keeper.By what percentage is the true weight of the
goods being marked up by the shopkeeper?
(Assume the balance has negligible mass.)Correct answer: 8.33334 percent.Explanation:
W− > standardweight
W′− > weight of goods sold
W (L/2− l) =W ′(L/2 + l)
Therefore
W =W ′(L/2 + l
L/2− l)
and(
W −W ′
W ′
)
× 100
=
(
L/2 + l
L/2− l− 1
)
× 100
=
(
(50 cm)/2 + (1 cm)
(50 cm)/2− (1 cm)− 1
)
× 100
= 8.33334 percent
Elongation of a Rod
12:04, trigonometry, numeric, > 1 min.026
In the figure the radius of the rod is 0.2 cm,the length of aluminum part is 1.3 m, thecopper part is 2.6 m. For aluminum, Young’smodulus is 7 × 1010 Pa, for copper, 1.1 ×1011 Pa.
Aluminum Copperr
La Lb
Determine the elongation of the rod if it isunder a tension of 5800 NCorrect answer: 1.9481 cm.Explanation:
The cross-sectional area of the wire A =π (0.2 cm)2 = 1.25664× 10−5 m2 m2, and thetension = 5800 N N throughout both pieces.Let us compute the elongation of each partseparately: For aluminum:
∆Lal =L0 F
Y A
=(1.3 m)(5800 N)
(7× 1010 Pa)(1.25664× 10−5 m2)= 0.00857163 m.
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 14
Similarly, for the copper part:
∆Lcu =L0 F
Y A
=(2.6 m)(5800 N)
(1.1× 1011 Pa)(1.25664× 10−5 m2)= 0.0109093 m.
So, the total elongation = 0.019481 m =1.9481 cm.
Serway CP 09 02
12:04, trigonometry, numeric, > 1 min.027
If the shear stress in steel exceeds about 4 ×108 N/m2, the steel ruptures.Find the shearing force necessary to shear
a steel bolt 1 cm in diameter.Correct answer: 31415.9 N.Explanation:
Given : r = 0.5 cm = 0.005 m and
Stress = 4× 108 N/m2 .
A
F
t
Force is
F = A (Stress)
= π r2 (Stress)
= π (0.005 m)2(
4× 108 N/m2)
= 31415.9 N .
028
Find the shearing force necessary to punch ahole 1 cm in diameter in a steel plate 0.5 cmthick.Correct answer: 62831.9 N.Explanation:
Given : t = 0.5 cm = 0.005 m .
The area over which the shear occurs is equalto the circumference of the hole times itsthickness:
A = (2π r) t .
Thus
F = A× (Stress)
= 2π r t (Stress)
= 2π (0.005 m)(0.005 m)
·(
4× 108 N/m2)
= 62831.9 N .
Force on a Table
15:01, trigonometry, numeric, > 1 min.029
Given: g = 9.8 m/s2 .A 0.75 kg physics book with dimensions of
24 cm by 20 cm is on a table. What force doesthe book apply to the table?Correct answer: 7.35 N.Explanation:
F = mg
030
What pressure does the book apply?Correct answer: 153.125 Pa.Explanation:
P =F
A=
F
l w
Dimensional analysis for P :
N
cm · cm·
(
100 cm
1m
)2
=N
m2= Pa
Holt SF 09Rev 19
15:01, highSchool, numeric, > 1 min.031
Given: g = 9.81 m/s2 .A submarine is at an ocean depth of 250 m.
Assume that the density of sea water is 1.025×103 kg/m3 and the atmospheric pressure is1.01× 105 Pa.a) Calculate the absolute pressure at this
depth.Correct answer: 2.61481× 106 Pa.
Create assignment, 22814, Homework 10, Nov 11 at 4:11 pm 15
Explanation:
Basic Concept:
P = P0 + ρgh
Given:
h = 250 m
ρsea water = 1.025× 103 kg/m3
g = 9.81 m/s2
P0 = 1.01× 105 Pa
Solution:
P = 101000 Pa
+ (1025 kg/m3)
· (9.81 m/s2)(250 m)
= 2.61481× 106 Pa
032
b) Calculate the magnitude of the force ex-erted by the water at this depth on a cir-cular submarine window with a diameter of30.0 cm.Correct answer: 184830 N.Explanation:
Basic Concepts:
P =F
A
A = πr2 = π
(
D
2
)2
Given:
D = 30.0 cm
Solution:
F = PA
= P
(
πD2
4
)
= (2.61481× 106 Pa)π(30 cm)2
4
·
(
1 m
100 cm
)2
= 184830 N
