Tài liệu Physics exercises_solution: Chapter 11 docx

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 300N.. b With respect to the hinge, the moment arm of

11.1: Take the origin to be at the center of the small ball; then,

m387.0kg

00.3

)m580.0)(

kg00.2(kg)(0)00.1(

x

from the center of the small ball

11.2: The calculation of Exercise 11.1 becomes

m351.0kg

50.4

)m580.0)(

kg00.2()m280.0)(

kg50.1()0)(

kg00.1(

x

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 pointS and let the ,child’s distance from this point be x Then,

m,125.12,0)

2(

M

mx D

M s

which is (L 2D 2) 2,halfway between the point S 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 300N 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

N

150 The hinge supplies an upward force of 300N150N150N

11.5: F(8.0m)sin40(2800N)(10.0m),soF 5.45kN,keeping an extra figure

11.6: The other person lifts with a force of 160N60N100N Taking torques about the point where the 60-Nforce is applied,

m

40.2N100

N160)m50.1(or),m50.1)(

N160

11.7: If the board is taken to be massless, the weight of the motor is the sum of the

applied forces, 1000N The motor is a distance( 2 00(1000m )( 600N)N )1.200mfrom the end where the 400-N force is applied

11.8: The weight of the motor is 400N600N200N800N Of the myriad ways

to do this problem, a sneaky way is to say that the lifters each exert 100Nto the lift the board, leaving 500Nand 300Nto the lift the motor Then, the distance of the motor from the end where the 600-N force is applied is ( 2 00(800m )(N300) N ) 0.75m.The center of gravity is located at ( 200 N )( 1 0(m1000)  ( 800N)N )( 0 75 m )0.80m

from the end where the 600N force

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,

N,900N740N160

with the wall at a height of 4.0 m above the ground Balancing torques about the point of contact with the ground,

m,N684)N740))(

53)(

m0.1()N160)(

m5.1()m0.4

so n1 171.0N, keeping extra figures This horizontal force about must be balanced by the frictional force, which must then be 170 N to two figures c) Setting the frictional force, and hence n , equal to the maximum of 360 N and solving for the distance x along 1

the ladder,

),N740)(

53()N160)(

m50.1()N360)(

m0.4

so x = 2.70 m, or 2.7 m 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

N

1920

or ),m00.3)(

N500()m50.1)(

N280()

zero, so the force at the left end is (1920N)(500N)(280N)1140N, downward

11.12: a)

b) x6.25m when F A 0,which is 1.25 m beyond point B c) Take torques about the right end When the beam is just balanced, F A 0,soF B 900N.The distance that

point B must be from the right end is then ( 300(900N )( 4N 50) m ) 1.50m

11.13: In both cases, the tension in the vertical cable is the weight ω a) Denote the .length of the horizontal part of the cable by L Taking torques about the pivot point,

),2(0

30

force of 2w and a horizontal force of 2.60w, so the magnitude of this force is 3.28w, 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 15.0 Taking torques about the pivot point, TLsin 15.0wLsin 45.0(w 2)Lsin 45,soT 4.10w The horizontal force exerted by the pivot on the strut is then T cos30.03.55ω and the vertical force is (2w)Tsin 304.05w, for a magnitude of 5.38w directed , 48.8

11.14: a) Taking torques about the pivot, and using the 3-4-5 geometry,

),N150)(

m00.2()N300)(

m00.4()53(m00.4

so T 625N b) The horizontal force must balance the horizontal component of the force exerted by the rope, or T(4 5)500N The vertical force is

N,75)53(N

150

N

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 FH(1.00m)(280N)(0.50m),from which FH 140N 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:

N1200

0)m70.0()m70.0)(

N80()m0.2)(

N450(

(b) From the ground

11.17: Consider the forces on Clea.

N246so

N157

n

n

n n

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

,0.60cos)35.0N)(

2600(0.60cosN)5000(60

TL

so T 3.14kN The vertical force exerted on the boom by the pivot is the sum of the

F

Ftan H

11.19: To find the tension T in the left rope, take torques about the point where the L

rope at the right is connected to the bar Then,

N

270so

m),N)(0.5090

(m)N)(1.50240

(150sin

sin N)270

N)

270

 so the tension is the rope at the right is TR 304N.andθ 39.9

11.20: The cable is given as perpendicular to the beam, so the tension is found by

taking torques about the pivot point;

kN.40.7

or ,0.25cos m)50.4kN)(

00.5(0.25cosm)00.2kN)(

00.1

magnitude 6.40N.m,sol 0.80m b) The net torque is clockwise, either by

considering the figure or noting the torque found in part (a) was negative c) About the point of contact ofF2,the torque due to F1isF1l, and setting the magnitude of this torque to 6.40Nmgivesl0.80m, and the direction is again clockwise

11.22: From Eq (11.10),

)

m1333()m100.50m)(

100.3(

m)200.0

2 4 2

Then, F 25.0N corresponds to a Young’s modulus of 3.3104 Pa,andF 500Ncorresponds to a Young’s modulus of 6.7105 Pa

11.23: 1.60 10 m ,

)m1025.0)(

Pa1020(

)m00.2)(

N400

2 10

Fl A

and so d  4A 1.4310 3 m,or 1.4 mm to two figures

11.24: a) The strain, from Eq (11.12), is

Pa100.2(

)N4000

2 4 11

Similarly, the strain for copper (Y 1.101011Pa) is2.110 4.b) Steel:

m103.8)m750.0

m1050.0(

)m00.4)(

N5000

2 2

)m105.3((

)m0.45)(

sm80.9)(

kg0.65

2 3

11.27: a) The top wire is subject to a tension of (16.0kg)(9.80m s2)157Nand

) m 10 5 2 )(

Pa 10 20 (

) N 157 ( 7 2 3.14 10 ,or 3.1 1010







bottom wire is subject to a tension of 98.0 N, and a tensile strain of 1.96103, or

10

11.28: a) 1.6 106Pa

) m 10 5 12 (

) 80 9 )(

kg 8000

(

2 2 2 s m

11.30: a) The volume would increase slightly b) The volume change would be twice

as great c) The volume 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) 3.33 106 Pa

m 10 0.75

N 250

2 4

)Pa101.0Pa1016.1)(

m00.1)(

Pa108.45(Δ

3

5 8

3 1

b) The mass of this amount of water not changed, but its volume has decreased to

,m947.0m053.0

3 3

)Pa106.3)(

cm600

3

6 3

11.34: a) Using Equation (11.17),

]Pa105.7)][

m005)(

m10[(

)N109

b) Using Equation (11.16),x Shear stain  h(.024)(.1m)2.410 3 m

11.35: The area A in Eq.(11.17) has increased by a factor of 9, so the shear strain for the larger object would be 19 that of the smaller

11.36: Each rivet bears one-quarter of the force, so

)m10125(

)N1020.1

2 2

4 4

11.37: 3.41 107Pa,or 3.4 107Pa

) m 10 92 0 ( ) N 8 90 (

11.38: a) (1.610 3)(201010Pa)(510 6m2)1.60103N b) If this were the case,

the wire would stretch 6.4 mm

c) (6.510 3)(201010Pa)(510 6m2)6.5103N

)kg1200(

3)m1000.3(Pa1040.2

11.40: 7.45 10 7 m2,so 4 0.97mm

Pa 10 4.7 N 350

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

so T 6860N This exceeds the maximum tension that the cable can have, so Lancelot is

going into the drink To find the distance x Lancelot can ride, replace the 12.0 m

multiplying Lancelot’s weight by x and the tension by 5.80 103 N

kg600(

)m0.6)(

sm80.9)(

kg200()m0.12)(

N1080.5(

2

2 3

sm80.9)(

kg200()m0.12)(

sm80.9)(

kg600()

11.43: For the airplane to remain in level flight, both F 0and0.

Taking the clockwise direction as positive, and taking torques about the center of mass,

Forces:Ftail W Fwing 0

Torques: (3.66m)Ftail(.3m)Fwing 0

A shortcut method is to write a second torque equation for torques about the tail, and solve for theFwing:(3.66m)(6700N)(3.36m)Fwing 0 This gives

.N(down)600

N7300N

6700and

),up(N

F

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 3600N1.003.00mm 1200N, so the net force on the front wheels

is10,780N1200N9.58103 Nto three figures The weight added to the rear wheels

is then 3600N1200N4800N, so the net force on the rear wheels is

N,1036.1N4800

N

b) Now we want a shift of 10,780 away from the front axle Therefore, N

N780

11.45: Take torques about the pivot point, which is 2.20 m from Karen and 1.65 m

from Elwood Then wElwood(1.65m)(420N)(2.20m)(240N)(0.20m),so Elwood weighs 589 N b) Equilibrium is neutral

11.46: a) Denote the weight per unit length

asα,sow1α(10.0cm), w2 α(8.0cm),andw3 αl

The center of gravity is a distance xcmto the right of point O where

3 2 1

3 2

1 cm

)2cm0.10()cm5.9()cm0.5(

w w w

l w

w w

cm0.8()cm0.10(

)2cm0.10()cm5.9)(

cm0.8()cm0.5)(

cm0.10(

l

l l

Setting xcm 0 gives a quadratic in l which has as its positive root , l 28.8cm

b) Changing the material from steel to copper would have no effect on the length l

since the weight of each piece would change by the same amount

is the vector from the point O to the point P.

The torque for each force with respect to point P is then i ri Fi

i

i i

F R F r

F R F

In the last expression, the first term is the sum of the torques about point O, 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 F1

is directed along the length of the nail, and so has a moment arm of (0.0800 m) sin60 The moment arm of

500(m)

300.0(

60sin m)0800.0(

1

F

11.49: The horizontal component of the force exerted on the bar by the hinge must

balance the applied force F

, and so has magnitude 120.0 N and is to the left Taking torques about point A,(120.0N)(4.00m)FV(3.00m), so the vertical component is N

160

 , 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 w2 50N,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 (50N)sin 3730N, to the left in the figure The vertical force must be

N.58N)50(N)30(c).59N30

N50arctan b)

up

N,50N1037

N40

(

)

N

50

forces exert torques

11.51: a) Take torques about her hind feet Her fore feet are 0.72 m from her hind feet,

and so her fore feet together exert a force of ( 190(0N).72( 0m) 28 m) 73.9N, 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

N,1.105m

72

.

0

m) 09 0 ( N) 25

30cos)sm80.9

B 2

1 T 4

11.53: a) Take the torque exerted by F2

to be positive; the net torque is then ,

sinsin

)(sin

)

b) 1 (14.0N)(3.0m)sin3725.3Nm, keeping an extra figure, and

2 (14.0N)(4.5m)sin3737.9Nm, and the net torque is 12.6Nm About point

P,1 (14.0N)(3.0m)(sin37)25.3Nm, and

2 (14.0N)(1.5m)(sin37)12.6Nm, and the net torque is 12.6Nm.The result of part (a) predicts (14.0N)(1.5m)sin37 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, andFV(6.0m)sinθ (250N)(4.0m)sinθ

N

354so

,sin)m50

ladder, so the horizontal pivot force is zero The vertical force that the pivot exerts on the ladder must be(750N)(250N)(354N)646N,up, so the ladder exerts a downward force of 646N on the pivot c) The results in parts (a) and (b) are independent of θ.

11.55: a) V mgw andH T.To find the tension, take torques about the pivot point Then, denoting the length of the strut by L,

.cot4

or,cos6

cos3

2sin

32

θ

mg w T

θ

L mg θ L w θ L T

,

B15N F 0:T2 BT3 15N9.0N24N

Upper rod:

)cm0.6()cm0.2)(

N24(:0





N0.8



C

F 0:T1 T2C 24N8.0N32 N

37cos)m0.7)(

N45,000(

37m)sin 5

11.59:

a)  0, axis at lower end of beam

Let the length of the beam be L.

N270020

sin

40cos

040cos2)

T

b) Take +y upward.

N137260

cosgives

0

N6.73so060sin gives

n T

w n

F y

N73.6

N1372

s s

n

f μ

n

μ

f

The floor must be very rough for the beam not to slip

11.60: a) The center of mass of the beam is 1.0 m from the suspension point Taking

torques about the suspension point,

)m00.2)(

N100()m00.1)(

N0.140()m00.4

w

(note that the common factor of sin 30 has been factored out), from which w15.0N

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

0)m00.5()m00.5()N600()m50

the tension is T y 700N The x-component of the tension is then

N714)

N700()N

1000



x

attached is (5.00m)714700 4.90m b) The y-component of the tension remains 700 N and the x-component becomes (714N)44..4090mm 795N, leading to a total tension of

N,1059N)

700(

kg0360.0kg0240.0(N,353.0)sm80.9)(

470.036.9

the point where string F is attached;

m,N0.833

m)5000)(

sm809kg)(

1200(

)m200.0(53.1sin )

m800.0(36.9sin )

m000.1(

.

T T

so T E 0.833N.T F may be found similarly, or from the fact that T E T F must be the total weight of the ornament (0.180kg)(9.80m s2)1.76N,from which T F 0.931N

11.63: a) The force will be vertical, and must support the weight of the sign, and is 300

N Similarly, the torque must be that which balances the torque due to the sign’s weight about the pivot, (300N)(0.75m)225Nm b) The torque due to the wire must balance the torque due to the weight, again taking torques about the pivot The minimum tension occurs when the wire is perpendicular to the lever arm, from one corner of the sign to the other Thus, T (1.50m)2(0.80m)2 225Nm,or T 132N.The angle that the wire makes with the horizontal is 90arctan (1.500.80)62.0.Thus, the vertical component of the force that the pivot exerts is (300 N) –(132 N) sin 62.0183Nand the horizontal force is (132N)cos62.062 N , for a magnitude of 193 N and an angle of 71 above the horizontal