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Unformatted text preview: Chapter 11 Equilibrium and Elasticity Note: In solving static equilibrium problems, torques may be calculated about any point.
The following solutions cannot possibly exhaust all of the valid ways to do any individual
problem. In many instances, an origin has been chosen to simplify the calculations. 11.1: Take the origin to be at the center of the Small ball; then, (1.00 kg)(0) + (lookaliﬂrﬁﬂ = 0 387 m 3““ 3.00 kg from the center of the small ball. 11.2: The calculation of Exercise 11.1 becomes 2 ——~—~W..W——~MW_.W.WW_.W ....... = .351
3““ 4.50 kg 0 m This result is smaller than the one obtained in Exercise 11.1. 11.3: In the notation of Example 11.1, take the origin to be the point S, and let the
child’s distance from this point be :0. Then, which is (L/ 2 — D / 2) / 2, halfway between the point 8 and the end of the plank. 11.4: a) The force is applied at the center of mass, so the applied force must have the
same magnitude as the weight of the door, or 300 N. In this case, the hinge exerts no force.
b) With respect to the hinge, the moment arm of the applied force is twice the distance
to the center of mass, so the force has half the magnitude of the weight, or 150 N. The
hinge supplies an upward force of 300 N — 150 N = 150 N. 11.5: F(8.0 In) sin40° = (2800 N)(10.0 m), so F = 5.45 kN, keeping an extra ﬁgure. 11.6: The other person lifts with a force of 160 N — 60 N = 100 N. Taking torques about
the point where the 60N force is applied, 160 N
100 N 11.7: If the board is taken to be massless, the weight of the motor is the sum of the
applied forces, 1000 N. The motor is a distance %§§§%ﬁl = 1.200 m from the end
where the 400N force is applied. 11.8: The weight of the motor is 400 N + 600 N — 200 N = 800 N. Of the myriad ways to
do this problem, a sneaky way is to say that the lifters each exert 100 N to lift the board,
leaving 500 N and 300 N to lift the motor. Then, the distance of the motor from the end
where the SOD—N force is applied is (20'1me = 0.75 m. The center of gravity is located at Mtlwﬁgbw m 0.80 m from the end where the 600 N force is applied. (100 N):c= (160 N)(1.50 m), or 3: = (1.50 m)( ) = 2.40 m. 281 282 Chapter 11 11.9: The torque due to T; is —Tmh = — £3” cot 9h, and the torque due to Ty is TyD = Lw.
The sum of these torques is L111 (1 — % cot 19). From Figure (11.9(b)), h. = D tan 6, so the
net torque due to the tension in the tendon is zero. 11.10: a) Since the wall is frictionless. the only vertical forces are the weights of the man
and the ladder, and the normal force. For the vertical forces to balance, n2 2 wt + mm =
160 N+740 N = 900 N, and the maximum frictional force is #3712 = (0.40)(900 N) = 360 N
(see Figure 11.7(b)). b) Note that the ladder makes contact with the wall at a height of
4.0 In above the ground. Balancing torques about the point of contact with the ground, (4.0 m)n1 = (1.5 m)(160 N) + (1.0 m) (3/5)(740 N) = 684 Nm, so 111 = 171.0 N, keeping extra ﬁgures. This horizontal force must be balanced by the fric
tional force, which must then be 170 N to two ﬁgures. 0) Setting the frictional force, and
hence 71,], equal to the maximum of 360 N and solving for the distance :1; along the ladder, (4.0 m)(360 N) = (1.50 m)(160 N) + $(3/5)(740 N), so :1: = 2.70 m, or 2.7 In to two ﬁgures. 11.11: Take torques about the left end of the board in Figure (11.21). a) The force F
at the support point is found from F(1.00 1n) = +(280 N)(1.50 m) + (500 N)(3.00 m), or
F = 1920 N. b) The net force must be zero, so the force at the left end is (1920 N) —
(500 N) — (280 N) = 1140 N, downward. 11.12: a) [
I000 . all}? i
 ' " Fact} 7m ..... 900 600
4m 3m .
2w ......... o ; : —1ooi ;  *2“) L : ' W. a = ‘ ‘ 5 :' 5 E
0 0.5 l 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7
I b) a: = 6.25 m when FA = 0, which is 1.25 [[1 beyond point B. c) Take torques about
the right end. When the beam is just balanced, FA = 0, so FB = 900 N. The distance that point B must be from the right end is then 991%?3 ”0 = 1.50 In. Equilibrium and Elasticity 283 11.13: In both cases, the tension in the vertical cable is the weight w. a) Denote the
length of the horizontal part of the cable by L. Taking torques about the pivot point,
TL tan 300° = wLf— w(L / 2), from which T = 2.6010. The pivot exerts an upward vertical
force of 210 and a horizontal force of 2.6011), so the magnitude of this force is 3.2811}, directed
37.6” from the horizontal. b) Denote the length of the strut by L, and note that the angle
between the diagonal part of the cable and the strut is 155.0". Taking torques about the
pivot point, TL sin 15.0° = wL sin 45.0° + (w/ 2)L sin 45°, so T = 4.101;). The horizontal
force exerted by the pivot on the strut is then Tcos 300° : 3.5511) and the vertical force
is (210) + Tsin 30° = 4.0511), for a magnitude of 5.3811), directed 48.8”. 11.14: a) Taking torques about the pivot, and using the 345 geometry,
(4.00 m)(3/5)T = (4.00 m)(300 N) + (2.00 m)(150 N), so T = 625 N. b) The horizontal force must balance the horizontal component of the force
exerted by the rope, or T(4/5) = 500 N. The vertical force is 300 N + 150 N — T(3/5) =
75 N, upwards. 11.15: To ﬁnd the horizontal force that one hinge exerts, take the torques about the
other hinge; then, the vertical forces that the hinges exert have no torque. The horizontal
force is found from £140.00 m) = (280 N )(0.50 m), from which FH = 140 N. The top
hinge exerts a force away from the door, and the bottom hinge exerts a force toward the
door. Note that the magnitudes of the forces must be the same, since they are the only
horizontal forces. 11.16: (a) Free body diagram of wheelbarrow: l
i i 0.70m j}?
450N son ”m“
k . .. J
_Y
2.0m
ZTwheel = 0
—(450 N)(2.0 1n) + (80 N)(0.70 m) + WL(0.70 m) = 0
WL = 1200 N (b) From the ground. 284 Chapter 11 11.17: Consider the forces on Clea. “r "1
6M 0.95 m % axis ew—xw—H} .
nr=89N,nf=157N “' m+m=wsow=246N
ET 2 0, axis at rear feet
Let a: be the distance from the rear feet to the center of gravity.
nf(0.95 m)  mw = 0
:1: = 0.606 In from rear feet so 0.34 m from front feet. 11.18: a) Denote the length of the boom by L, and take torques about the pivot point.
The tension in the guy wire is found from TL sin 60° = (5000 N)L00360.0° + (2600 N)(0.35L) cos 600°, so T = 3.41 kN. The vertical force exerted on the boom by the pivot is the sum of the
weights, 7.60 kN and the horizontal force is the tension= 3.41 kN. b) No; tan (ﬁ) 75 0. 11.19: To ﬁnd the tension TL in the left rope, take torques about the point Where the
rope at the right is connected to the bar. Then, TL(3.00 m) sin 150’ = (240 N)(1.50 m) +
(90 N)(0.50 In), so TL = 270 N. The vertical component of the force that the rope at the
end exerts must be (330 N) — (270 N) sin 150° 2 195 N, and the horizontal component of
the force is —(270 N) cos 150°, so the tension in the rope at the right is T3 = 304 N, and
9 = 399°. 11.20: The cable is given as perpendicular to the beam, so the tension is found by taking
torques about the pivot point; T(3.00 m) = (1.00 kN)(2.00 In) cos 25.0” + (5.00 kN)(4.50 m)
cos 25.0”, or T = 7.40 kN. The vertical component of the force exerted on the beam by the
pivot is the net weight minus the upward component of T, 6.00 kN—T cos 250" = 0.71 kN.
The horizontal force is Tsin 250" = 3.13 kN. 11.21: a) F1(3.00 In) — F2(3.00 111+!) = (8.00 N)(—l). This is given to have a magnitude
of 6.40 Nm, so I = 0.80 m. b) The net torque is clockwise, either by considering the
ﬁgure or noting the the torque found in part (a) was negative. c) About the point of
contact of ?2, the torque due to '3‘: is —F11, and setting the magnitude of this torque
to 6.40 Nm gives I = 0.80 m, and the direction is again clockwise. 11.22: From Eq. (11.10), to memcwmm _—_ F(1333 1114). y = _ =
FAJA (3.0 x 102 m)(50.0 x 104 m2) Equilibrium and Elasticity 285 Then, F : 25.0 N corresponds to a Young’s modulus of 3.3 x 10“ Pa, and F = 500 N
corresponds to a Young’s modulus of 6.7 X 105 Pa. 11.23: A = Fl“ z (400 NW 00 ml." = 1.60 x 10*6 1112 VA: (20 x 1016 Pa)(0. 25 ><10—2 m) and so at = «414/2 2 1.43 x10‘3 :11, or 1.4 mm to two ﬁgures. 11.24: a) The strain, from Eq. (11.12), is 130‘ _ —Y—I‘:q. For steel, using Y from Table (11.1)
andA=7r—=1._477x10 m3, At (4000 N)
13—0 :.(2 0 ><1011 Pa)(1. 77 x 10 4 1112) Similarly, the strain for copper (Y = 1.10 x 1011 Pa.) IS 2.1 x 104. b) Steel: (1.1 x 10—4) x
(0.750 m) = 8.3 X 10—3 m. Copper: (2.1 x10‘4)(0.750 m) = 1.6 x 10“1 m 11.25: From Eq. (11.10), = 1.1 x 10“. (6020 N)(4.00 In)
(0.50 x 10“1 m3)(0.20 x 10*2 m) Y = = 2.0 x 1011 Pa. 11.26: From Eq. (11.10), 2
Y = @WWPJ = 6.8 x 103 P3,. («(3.5 x 10‘3 m)2)(1.10 m) 11.27: a) The top wire is subject to a tension of (16. 0 kg)(9. 80 m/s2 ) = 3157 N and hence (157 N a tensile strain of (20 x 10Wi’W' “5' = 3.14 x 10—3 , or 3.1 x 10 3 to two ﬁgures. The bottom wire is subject to a tension of 98. 0 N, and a tensile strain of 1. 96 x 10”3
2.0 ><10_3 to two ﬁgures. b) (3.14 x 10 3)(0.500 m) = 1. 57 mm (1. 96 x 10 3)(0 500 m) = 0.98 mm 11.28: a) «(.12 ”WNW: 10 x 106 Pa. b) 123%?“ :08 x 106. c) (0.8 x 106)><
(2.50 m)— _ 2 x 106 m. 11.29: (2.3 — 1)(1.013 x 106 Pa)(50.0 m3) : 9.1 x 106 N. 11.30: a) The volume would increase slightly. b) The volume change would be twice
as great. 0) The volume change is inversely proportional to the bulk modulus for a given
pressure change, so the volume change of the lead ingot would be four times that of the gold. 11.31: a) 0.75 353035.? = 3.33 x 106 Pa. b) (3.33 x 106 Pa)(2)(200 x10“1 :03) = 133 kN. 11.32: 3) Solving Eq. (11.14) for the volume change, AV = 411/1316
= —(45.8 x 1011 Pa“1)(1.00 m3)(1.16 x 103 Pa — 1.0 x 105 Pa)
= —0.0531 1113. 286 Glimmer I b) The mass of this amount of water has not changed, but its volume has decreased t 1.000 m3 — 0.053 m3 = 0.947 m3, and the density is now Lﬁgﬁﬂ = 1.09 x 103 kg/nl3. 11.33: B = W06 Pa) WWWWWW :43 109p k: = .1 1010P ‘1.
(0.45 cm3) x a’ 2 X a l
B
11.34: a) Using Equation (11.17), F” (0 x 105 N) Shear strain 2 — — ~—  = 2.4 x 10—2. b) Using Equation (11.16), :1: = Shear strain  h = (.024)(.1m) = 2.4 x 10—3 m. 11.35: The area A in Eq. (11.17) has increased by a factor of 9, so the shear strain fo
the larger object would be 1/9 that of the smaller. 11.36: Each rivet bears onequarter of the force, so P
Shear stress = —” = .._........W.._ — 3
A 7r(.125 x 102m)2 — 511 X 10 Pa. 11.37: g = mﬁgﬁzﬁﬁi’gm = 3.41 X 107 Pa, or 3.4 x 107 Pa to two ﬁgures. 11.38: a) (1.6 x 10'3)(20 x 1010 Pa)(5 x 104i m2) = 1.60 x 103 N. b) If this were
the case, the wire would stretch 6.4 mm. c) (6.5 x 10—3)(20 x 1010 Pa)(5 x 10—6 m2)
6.5 x 103 N. F... (2.40 x 108 Pa)(3.00 x 104 m2)/3 2 2
11.39: = = m~#—~wm1www~w~e— w . = 1 .2 .
a m (1200 kg) 980 m/s 0 m/s 11.40: A = ﬁ§§% = 7.45 x 107 m2, so d = ./4A/« = 0.97 mm. 11.41: a) Take torques about the rear wheel, so that fwd = wzrcm, or 33m = fd.
b) (0.53) (2.46 m) = 1.30 In to three ﬁgures. 11.42: If Lancelot were at the end of the bridge, the tension in the cable would be (from
taking torques about the hinge of the bridge) obtained from ' T(12.0 N) = (600 kg)(9.80 m/s2)(12.0 m) + (200 kg)(9.80 m/s2)(6.0 m), so T = 6860 N. This exceeds the maximum tension that the cable can have, so Lance]
is going into the drink. To ﬁnd the distance 3: Lancelot can ride, replace the 12.0 multiplying Lancelot’s weight by :1: and the tension T by Tmax = 5.80 X 103 N and sol
for :5;
(5.80 x 103 N)(12.0 m) L (200 kg)(9.80 m/sz)(6.0 m) :1: = mw__WMWIE753~jm4~— vvvvvv = 9.84 m. Equilibrium and Elasticity 287 11.43: For the airplane to remain in level ﬂight, both 21F = 0 and 27' = 0. Taking the clockwise direction as positive, and taking torques about the center of mass,
Forces: —Ftai1 — W + Fwing = 0
Torques: w—(3.66 m)Em1 + (.3 1rn)JFL.,.img = 0 A shortcut method is to write a second torque equation for torques about the tail, and
solve for the Fwing: — (3.66 m)(6700 N) +(3.36 n1)1*",...ilclg : 0. This gives PM = 7300 N (up),
and Fm. = 6700 N — 7300 N = —600 N (down). Note that the rear stabilizer provides a downward force, does not hold up the tail of
the aircraft, but serves to counter the torque produced by the wing. Thus balance, along '
with weight, is a crucial factor in airplane loading. 11.44: The simplest way to do this is to consider the changes in the forces due to
the extra weight of the box. Taking torques about the rear axle, the force on the front wheels is decreased by 3600 Nég 3 = 1200 N, so the net force on the front wheels is 10, 780 N w— 1200 N = 9.58 X 103 N to three ﬁgures. The weight added to the rear wheels is
then 3600 N + 1200 N = 4800 N, so the net force on the rear wheels is 8820 N +4800 N = 1.36 x 10‘1 N, again to three ﬁgures.
b) Now we want a shift of 10,780 N away from the front axle. Therefore, Wégﬁ =
10,780 N and so 11! = 32,340 N. 11.45: Take torques about the pivot point, which is 2.20 m from Karen and 1.65 m from
Elwood. Then wglmd(1.65 m) = (420 N)(2.20 m) + (240 N)(0.2O In), so Elwood weighs
589 N. b) Equilibrium is neutral. 11.46: a) Denote the weight per unit length as a, so ml 2 oz(10.0 cm), 1112 = a(8.0 cm),
and 103 = at. The center of gravity is a distance arm, to the right of point 0, where _ ml (5.0 cm) + 102(95 cm) + w3(10.0 cm — 3/2) mlCH]. (10.0 cm) (5.0 cm) + (8.0 cm)(9.5 cm) + {(10.0 cm — 1/2) (10.0 cm) + (8.0 cm) +l Setting mm = 0 gives a quadratic in l, which has as its positive root 1 = 28.8 cm. b) Changing the material from steel to c0pper would have no effect on the length 1
since the weight of each piece would change by the same amount. 288 Chapter 1‘ 1 11.47: Let T’: = ‘?’, — 7%}, where T? is the vector from the point 0 to the point P.
The torque for each force with reSpect to point P is then 7’: = 73, X _F+,:, and so the net torque is In the last expression, the ﬁrst term is the sum of the torques about point 0, and the
second term is given to be zero, so the net torques are the same. 11.48: From the ﬁgure (and from common sense), the force TF"l is directed along the
length of the nail, and so has a moment arm of (0.0800 m) sin 60°. The moment arm of
TF3 is 0.300 In, so F2 = Frame—WW = (500 N)(0.231) = 116 N. 11.49: The horizontal component of the force exerted on the bar by the hinge must
balance the applied force ?, and so has magnitude 120.0 N and is to the left. Taking
torques about point A, (120.0 N)(4.00 m) + Fv(3.00 In), so the vertical component is
—160 N, with the minus sign indicating a downward component, exerting a torque in a
direction opposite that of the horizontal component. The force exerted by the bar on the
hinge is equal in magnitude and opposite in direction to the force exerted by the hinge on the bar. 11.50: a) The tension in the string is 1102 = 50 N, and the horizontal force on the bar
must balance the horizontal component of the force that the string exerts on the bar, and
is equal to (50 N) sin 37°: 30 N, to the left in the ﬁgure. The vertical force must be
(50 N) co337° + 10 N: 50 N up b) oraton(g— 3:) 59°. c) ,/(30 N)"2T‘(50 N)? = 58 N.
(1) Taking torques about (and measuring the distance from) the left end, (50 N)z =
(40 N)(5.0 In), so 3 = 4.0 m, where only the vertical components of the forces exert torques. 11.51: a) Take torques about her hind feet. Her fore feet are 0. 72 In from her hind feet, and so her fore feet together exert a force of ﬂfggéﬂﬂ— — 73 9 N, so each foot exerts a force of 36.9 N, keeping an extra ﬁgure. Each hind foot then exerts a force of
58.1 N b) Again taking torques about the hind feet, the force exerted by the fore feet is mo. 28 fwﬂﬂl— — 105.1 N, so each fore foot exerts a force of 52. 6 N and each hind foot exerts a force of 54.9 N. 11.52: a) Finding torques about the hinge, and using L as the length of the bridge and
“HIT and 1113 for the weights of the truck and the raised section of the bridge, TL sin 70° = mgr.) cos 30° + 103 GI.) cos 30°, so Equilibrium and Elasticity 289 (ng + émB) (9.80 111/52) cos 30° T = 1.... “We“ wwwwwwwwwwwww = 2.57 x105 N. b) Horizontal: Tcos(70°—30°) = 1.97 x 105 N. Vertical: w: + wB—T sin 40° = 2.46 x 105 N. 11.53: 3.) Take the torque exerted by $3 to be positive; the net torque is then —F1(:.r:)
sin ¢+F2(m+l) sinqb = F 3 sin qb, where F is the commou magnitude of the forces. b) 7'1 =
—(14.0 N)(3.0 m) sin 37° = #253 NH], keeping an extra ﬁgure, and 72 = (14.0 N)(4.5 n1)
sin 37° 2 37.9 Nrn, and the net torque is 12.6 Nm. About point P, 1'1 = (14.0 N)(3.0 m)
(sin 37°) = 25.3 Nrn and r2 = (—14.0 N)(1.5m)(sin 37°) = 12.6 Nm, and the net torque
is 12.6 NHL The result of part (a) predicts (14.0 N)(1.5 m) sin 37°, the same result. 11.54: a.) Take torques about the pivot. The force that the ground exerts on the ladder
is given to be vertical, and FV(6.0 In) sin9 = (250 N)(4.0 m) sin6+ (750 N)(1.50 n1) sin (9,
so R; = 354 N. b) There are no other horizontal forces on the ladder, so the horizontal
pivot force is zero. The vertical force that the pivot exerts on the ladder must be (750 N) +
(250 N) — (354 N) = 646 N, up, so the ladder exerts a downward force of 646 N on the
pivot. c) The results in parts (a) and (b) are independent of 6. 11.55: a) V = mg + w and H = T. To ﬁnd the tension, take torques about the pivot
point. Then, deonting the length of the strut by L, 2
T (0L) sinﬂ = 111 (EL) cost} + mg (36—!) c059, or T = (w + %) cot9. b) Solving the above for w, and using the maximum tension for T, w = Ttan — ¥ = (700 N) tan 550° — (5.0 kg)(9.80 m/sz) = 951 N. c) Solving the expression obtained in part (a) for taut? and letting “in H 0, tan!) = $T3 =
0.700, so 6 = 4.00°. 11.56: (a) and (b) Lower rod: 6.0 N A 213, = 0 : (6.0 N)(4.0 cm) = A(8.0 cm)
A=3.0N
EF=O:T3=6.0N+A=6.0N+3.0N=9.0N 290 Chapter 11 Middle md: T:
J 50cm 3.0m: .
P 1 T3=9.0N a: 273, = 0 : B(3.0 cm) = (9.0 N)(5.0 cm)
3 =15N
EF=0:T2 =B+T3=15N+9.0N=24N Upper nod: TI 2.0 an I 611cm
T= 24 N C 21", = 0 :(24 N)(2.0cm) = C(6.00m)
C = 8.0N
EF=0:T1=Tg+C=24N+8.0N=32N' (c) The center of gravity must lie somewhere directly below 5'1 or the mobile would
tilt. 11.57: 21' = 0, axis at hinge
T(6.0 m)(sin 40°) — w(3.75 m)(cos 30°) = 0
T 2 760 N Equilibrium and Elasticity 291 11.58: (a)
45.000N
Y
EllHinge =
T(3.5 m) sin 37° = (45, 000 N)(7.0 m) can 37°
T = 120, 000 N (b) 21F:c = U : H = T =120,000N
SE, = 0 : V = 45,000N
The resultant force exerted by the hinge has magnitude 1.28 x 105 N and direction 206° above the horizontal.
11.59: a) Z 'r = 0, axis at lower end of beam Let the length of the beam be L. T(sin 20°)L = —mg (g) cos 40" = 0 1)) Take +y upward.
EFy=0givaan—w+Tsin60°=Oson=73.6N
2F; 2 0 gives f;1 =Tcosf30° = 1372 N fs = pan, #3 =
The ﬂoor must be very rough for the beam not to slip. 292 Chapter 11 11.60: a.) The center of mass of the beam is 1.0 m from the suspension point. Taking
torques about the suspension point, w(4.00 m) + (140.0 N)(1.00 m) = (100 N)(2.00 m) (note that the common factor of sin 30" has been factored out), from which 10 2 15.0 N.
b) In this case, a common factor of sin 45° would be factored out, and the result would
be the same. 11.61: a) Taking torques about the hinged end of the pole (200 N)(2.50 m) + (600 N) X
(5.00 m) —Ty(5.00 m) = 0. Therefore the y—component of the tension is Ty = 700 N. The cccomponent of the tension is then T3 = «(1000 N)2 — (700 N)2 = 714 N. The height
above the pole that the wire must be attached is (5.00m)% = 4.90 m. b) The 3;
component of the tension remains 700 N and the xcomponent becomes (714 N) 1% E = 795 N, leading to a total tension of «(795 N)2 + (700 N)2 = 1059 N, an increase of 59 N. 11.62: A and B are straightforward, the tensions being the weights suspended; TA ='
(0.0360 kg)(9.80 m/s2) = 0.353 N, T3 = (0.0240 kg +0.0360 kg)(9.80 m/s2) = 0.588 N. To
ﬁnd To and TD, a trick making use of the right angle where the strings join is available; use
a coordinate system with axes parallel to the strings. Then, T C = TB cos 369° 2 0.470 N,
T9 = TB cos 53.1...
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