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Unformatted text preview: 5.58. Model: We will model the sculpture as a particle of mass m. The ropes that support the sculpture Win assumed to have zero mass. Visualize: Pictorial representation Physical representation
y
a 3 Tan 30° T2
Tension T1 Tension T2 600 Weight {43 Solve: Newton‘s ﬁrst law in component form is (FM). 2 EPA. = T“. + T2. + w" : — T1 sin 30° + T2 sin 60° + 0 N = 0 N
(Fug). = 2R, = T”. + T2). + w}. = Tl cos 30° + T2 cos 60° —— w = 0 N
Using thexcomponent equation to obtain an expression for T1 and substituting into the y—component equation yields:
I 0 lb
T3 : (sin60°)(cos)30°) : 50 2 S = 2501135 + cos60°
sm 30° Substituting this value of T2 back into the xcompoent equation, T, = Tzlsfnm = 250 lbs an60 = 433 lbs sm 30° sm 30° We will now ﬁnd a rope size for a tension force of 433 lbs, that is, the diameter of a rope with a safety rating of 433
lbs. Since the cross~sectional area of the rope is ﬂail, we have 1/2
33 .
d: =0.37linch
7r(4,0001bs/mch ) Any diameter larger than 0.371 inch will ensure a safety rating of at least 433 lbs. The rope size corresponding to a
diameter of 3/8 of an inch will therefore be appropriate. Assess: If only a single rope was used to hang the sculpture, the rope would have to support a weight of 500 lbs.
The diameter of the rope for a safety rating of 500 lbs is 0.399 inches, and the rope size jumps from a diameter of
3/8 to 4/8 of an inch. Also note that the weight of the sculpture is distributed in the two ropes. It is the sum of the y—
components of the tensions in the ropes that will equal the weight of the sculpture. 5.60. Model: The ball hanging from the ceiling of the truck by a string is represented as a particle. Visualize: y
g g i 0=10°
'e' ’9‘ o '9 e o x
Truck at rest or moving Truck accelerating T>
with constant velocity forward ‘1», F“a Freebody diagram
while accelerating
Solve: (a) You cannot tell from within the truck. Newton’s first law says that there is no distinction between “at
rest” and “constant velocity.” In both cases, the net force acting on the ball is zero and the ball hangs straight down.
(1)) Now you can tell. If the truck is accelerating, then the ball is tilted back at an angle.
(c) The ball moves with the truck, so its acceleration is 5 m/s2 in the forward direction.
((1) The free—body diagram shows that the horizontal component of T provides a net force in the forward direction.
This is the net force that causes the ball to accelerate in the forward direction along with the truck. (e) Newton‘s second law for the ball is av : 0 mzlsz = (Fuel. ___ T, +Wy = Tcole°mg
m m m ‘ m m m (F...)X 1;. = Tsin10° a: : .r We can solve the second equation for the magnitude of the tension: 1k 9.8 /2
T: mg =(__g_)(_tl_s_)=9_95N
cosiO° cole" Then the ﬁrst equation gives the acceleration of the ball and truck:
a _ Tsin10° _ (9.95 N)sin10° J: = 1.73 m/s2
m m The truck’s velocity cannot be determined. 6.29. Model: The particle model for the ball and the constantacceleration equations of motion in a plane assumed. Visualize: Pictorial representation
Known x0: [0:0 y0=2.0m 0=5°
v0=20.0 m/s
x1=7.0m v1x=vox ay=—g Solve: The initial velocity is
vol, = v0 cos5° = (20 m/s)cos 5° = 19.92 m/s voy = v0 sin 5° = (20 m/s)sin5° = 1.743 m/s The time it takes for the ball to reach the net is
x, = x0 + vain—10):) 7.0 m = 0 m+(19.92 m/s)(rl — 0 5) => 2 = 0.351 s
The vertical position at I, = 0.35] s is 7 Y1 = Yo + Vat(’1 —t0)+%ay(tl _to)
= (2.0 m) + (1.743 m/s)(0.351 s — 0 s) + %(—9.8 m/s2)(0.3518 — 0 s)2 = 2.01 m Thus the ball clears the net by 1.01 m. .
Assess: The vertical free fall of the ball, with zero initial velocity, in 0.351 s is 0.6 m. The ball will clear approximately 0.4 m if the ball is thrown horizontally. The initial launch angle of 5° provides some initial V6
velomty and the ball clears by a larger distance. The aboveresult is reasonable. ___________________—————— 6.42. Model: We will assume a particle model for the sand, and use the constantacceleration kin equations.
Visualize:  Fictorial representation \r In )‘D‘ f I] 2
Solve: Using the equation .‘cl = .ra + l’m (ll — to) +511 (:1 — to) ,
I. = 0 m + (co cosl 5°)(t. — 0 s) + 0 m = (60 rm!s](t:o515"')ti /—~~a. We can ﬁnd I, from the y—equation. but note that PRr = —vcl sin15° beeause the sand is launched’lat an 011313
horizontal. .;I\ yl = ya + who] — to) + 1la_,(r, — to)2 :> 0 in = 3.0 m — (an sin]5“)iI aggrf
= 30 m — (6.0 mls)(sin15°}r. — {49.8 misty? :5 4.91‘,1 + 1551'I — 30 = 0 :> fl = 0.6399 s and — 0.956 s (unphysical) ____________.._.————__ Substituting this value of :1 in the x—equation gives the distance
d = x, = (6.0 mfs)cosl§“(0.6399 s) = 3.71 m 6.58. Model: Let Alberta’s frame be S and Fred's framn: be 3'. Frame 8' is moving relative to S with veloci .
Assume that ‘ﬁjpﬁeld" is in the .rdirection. Visualize: Pictorial representation
Fred : ‘
r»»«——....L_M.._ 2?. Alberto 130cm n a
if” = — 4.0 mrsj
.1" ‘t' = 20 mos
V I” Find
.1 xi?! 0
Solve: The Galilean transformation of velocity is i? = f” + 17, where l7 = —(4.0 ails); and ﬁ’ = (20 mis]ms_
(20 mls)sinﬂ} . Thus,
l; = (20 mls)cosﬂi +[(20 nu's)sin9 — (4.0 mils”; According to Alberto, Fred released the ball as he was directly upfield from Alberto. He observes the ball I _
toward him along the .rdirection. That is v! = 0 mls. This means ' (20 m!s)sin9 — (4.0 mls) = 0 this => 9 = sin—(é) = 11.5° Alberto must throw the ball 1 LS“ toward the side opposite the one he is running toward. 6.63. Model: Use the particle model for the motorcycle daredevil and apply the kinematic equations of motion.
Visualize: Pictorial representation Known
‘a=.“o=’a=0 rat: ~10 m’s rat: 0
a = 20“ '
nx=0 or: —g Find
310‘! Solve: We need to ﬁnd the coordinates of the landing ramp [.l'l, y.). We have x. = x9 + volt] —ID) = 0 m + (40 mishl y, 2 yo + who, — r0)+%a_v(r, — ta.)1 2 0 m + 0 m +:1(—9.3 mislhf = —(4.9 rm’sz)“1 This means we must ﬁnd n. Since Fl makes an nngle of 20° below the horizontal we can find 1'”. as follows: lvlrl 1. = tan 20° :5 vb. = —v,, 131120” = —[40 mishan 20° = “14.56 miss
I. We now use this: value of 1.1).. or = —9.8 ml’s2 . and rm. = 0 rnls in the following equation to obtain II: vb. :1,“ + air. 40) => —l4.56 mls = 0 min —(9.8 misﬁt! => 1.: 1.436 5
Now, we are able to obtain x, and y, using the above .1' and yequalions:
x, = (40 mls)(l .486 s) = 59.4 In _,.I = _(4_9 misz](l.436 s)2 = 40.32 In That is. the landing ramp should be placed 10.32 to lower and 59.4 m away from the edge of the horizontal
nintfnrm ' odel: Use the particle model for the car and the model of kinetic friction
. Physical representation  Known
r=200m
1' =25mfs
Pk: 1“ Find w I A. Flat B. Bottom C. Top
e WI" apply Newton’s second law to all three cars. 2% ="I+(J'L)(+w1 =0 N~ft+0N=mn, =11!'i'LIﬁELlgt'.I =n+0N——mg=0 N l"1PGI1enl e nation — '
q means n #1113. Since ft = gtkn . we have ft = ukmg. From the .rcomponent equation, _I —#m ,
“— ‘= ng—Hgg=9.8ml'5' .r
1" m Car B: Car B is in circular motion with the center of the circle above the car. 2;“; =nr+(fk)r+wr=n+DN—mg=mar= r 2F; =ni+(fk)r+w,=0N—j;+0N=+nrml From the r—equalion v 25f: =Pkﬂ=ﬂrm[8+ ) r 1
ml" n : mg+
r
Substituting back into the Fequation, . ‘2 2
a =_£L=_ﬂ[g+‘——]=—pk[9.8mfsz+(25mls) ]=r12.9mt's1
200m 1
H1 III 1" Car C: Car C is in circular motion with the center of the circle below the car. 2F; 2n,+(fk)’+wr=—n+0N+mg=mar = mvz r
2E=ﬂl+(ﬁ)l+ll'l=0Nrfk+0N=lﬂﬂl From the requation n = m(g — I'ZIJ'). Substituting this into the tequation yields 7.55. Model: Model the block as a particle and use the model of kinetic friction.
VisualiZE: Pictorial repmentaﬁon Physical representation Top View Edge View Known
m=5ﬂﬂg r: 1.2 In. F, = 4.0 N T
pt :06 “'11 Air Find
8' at Tm Suite: The only radial force is tension, so we can use Newton’s second law to ﬁnd the angular velociw mm at
which the tube breaks: 2F, =T=mrozr:>mm = F“ = —3—0N— =9.12 radls
' mr (0.50 kg](l.2 ml The compressed air and friction exert tangential forces. and the second law along the tangential direction is Zr: = 1: —ft = E —,ttln : F; ﬂitting = trimI
F. , 4.0M
a: —{0.60){_9.80 mils! ')=2.12 ms2
at 0.50 kg The time needed to accelerate to 9.] 2 radls is given by new # (1.2 )(9.12 radls)
a 2.12 rat's1 I m,=mm=0+[§L]r,=rl= =5.I65 r y During this interval, the block turns through angle 1 .
‘C‘ = 3.75 rev r l.2 m 1 2. . 1 1
A9 = 9, — 90 = (1105+ Lli] :; = 0+ % ﬂ“— (5.16 s)‘ = 23.52 rad x
. 2amd 7.62. Model: Use the particle model for a ball in motion in a vertical circle and then as a projectile.
Visualize: Pictorial representation Physicai repraentaﬁon In Eu “it. At top of circle ‘tdl
_ . _. . I .t
0
Solve: For the circular motion. Newton’s second law along the rdirection is
ttttr2
2E = T+ w = '
r Since the string goes slack as the panicle makes it over the top, T: 0 N. That is, w = mg: "W'— 3 tr' 2 gr =1i(9.8 mlsz)(0.5 m) = 2.2111115
r The ball begins projectile motion as the string is released. The time it takes for the ball to hit the ﬂoor can be f r
as follows: y, = yo + 15:0, — to) + %a_‘.(t,  I“): => 0 m = 2.0 at + D m + 5(—9.8 m I 53X:l — D 5)2 => 1‘. = 0.639 s The place where the ball hits the ground is
x, = x0 + Path 40) = 0 nt+(+2.21 m I s)(0.639 s —0 s) = +1.41 m
The ball hits the ground 1.41 m to the right of the point beneath the center of the circle. 7.63. Model: Use the particle model for the ball.
Vimalize: Pictorial representation Physical representation ..
ﬂ Solve: (a) Newton's second law along the r and z—directions is 2P; =rtcose=mratz ZE=HSin3—w=ON
Using It‘ 2mg and dividing these equations yields:
R_
tantiI = g, = y
m)’ r
where you can see from the ﬁgure that lane 2 (R—y)/r. Thus a) = —)’ (h) n) will be minimum when (R — y) is maximum or when 3‘ =0 In. Then mm = g I R . (c) Substituting into the above expression. 0): g =’ 9.8mt's‘ =9_9£dx ﬁﬂsxlrev =945rpm
R—y 0.2m—0.lnt 5 1mm 23rrad ...
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 Winter '08
 Susskind,L
 Physics, mechanics

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