Tài liệu Physics exercises_solution: Chapter 05 ppt

b If the mass of the light pulley may be neglected, the net force on the pulley is the vector sum of the tension in the chain and the tensions in the two parts of the rope; for the pulle

5.1: a) The tension in the rope must be equal to each suspended weight, 25.0 N b) If the

mass of the light pulley may be neglected, the net force on the pulley is the vector sum of the tension in the chain and the tensions in the two parts of the rope; for the pulley to be

in equilibrium, the tension in the chain is twice the tension in the rope, or 50.0 N

5.2: In all cases, each string is supporting a weight w against gravity, and the tension in each string is w Two forces act on each mass: w down and T ( w ) up

5.3: a) The two sides of the rope each exert a force with vertical component T sinθ, and

the sum of these components is the hero’s weight Solving for the tension T,

.N 1054.20.0

1sin 2

)sm(9.80kg)0.90(sin 2

1050.2(2

sm(9.80kg)(90.0arcsin2

5.4: The vertical component of the force due to the tension in each wire must be half of

the weight, and this in turn is the tension multiplied by the cosine of the angle each wire makes with the vertical, so if the weight is , cos and arccos32 48

4

3

5.5: With the positive y-direction up and the positive x-direction to the right, the

free-body diagram of Fig 5.4(b) will have the forces labeled n and T resolved into x- and

y-components, and setting the net force equal to zero,

.0sin

cos

0sincos

n F

n T

w

T T

sinsin

cossin

sin

and so nT cot wsincot wcos, as in Example 5.4

5.6: wsin αmgsin α(1390kg)(9.80m s2)sin 17.54.10103 N

5.7: a) cos ,or cos 5.23 104 N.

40 cos

) s m kg)(9.8 4090

cos

45

sin   adding the last two equations gives T A(cos30sin30)w, and so

.732

A   Then, T B T Acoscos4530  0.897w

b) Similar to part (a), T C w,T Acos60T Bsin45w, and

.045cos60

cos60

T

T B  B  

5.9: The resistive force is wsin(1600kg)(9.80m s2)(200m 6000m)523 N

5.10: The magnitude of the force must be equal to the component of the weight along the

incline, orWsinθ (180kg)(9.80m s2)sin11.0337 N

5.11: a) W 60N,TsinθW, soT (60N) sin45, or T 85 N

b) F1F2 T cosθ,F1F2 85Ncos4560 N

5.12: If the rope makes an angle  with the vertical, then 0 110

1 51sin  0 073

.0

)sm80.9kg)(

270.0())073(

The force of the pole on the ball is the tension times sinθ, or (0.073)T 0.193 N

5.13: a) In the absence of friction, the force that the rope between the blocks exerts on

block B will be the component of the weight along the direction of the incline,

α

w

T  sin b) The tension in the upper rope will be the sum of the tension in the lower

rope and the component of block A’s weight along the incline,

.sin2sin

5.14: a) In level flight, the thrust and drag are horizontal, and the lift and weight are

vertical At constant speed, the net force is zero, and so F  and f w b) When the L.plane attains the new constant speed, it is again in equilibrium and so the new values of

the thrust and drag, F and f , are related by F f; if F2F, f2f c) In order to increase the magnitude of the drag force by a factor of 2, the speed must increase by a factor of 2

5.15: a)

The tension is related to the masses and accelerations by

.2 2 2

1 1 1

a m g m T

a m g m T









b) For the bricks accelerating upward, let a1a2 a (the counterweight will

accelerate down) Then, subtracting the two equations to eliminate the tension gives

.sm96.2kg0.15kg0.28

kg0.15kg0.28sm80.9

or ,)(

)(

2 2

1 2

1 2

2 1 1

m m g a

a m m g m m

c) The result of part (b) may be substituted into either of the above expressions to find the tension T 191N As an alternative, the expressions may be manipulated to

eliminate a algebraically by multiplying the first by m and the second by 2 m and adding 1

(with a2 a1) to give

N

191kg)

0.28kg0.15(

)sm(9.80kg)(28.0kg)0.15(22

or ,02

)(

2

2 1

2 1

2 1 2

g m m T

g m m m

m T

In terms of the weights, the tension is

.22

2 1

1 2 2 1

2 1

m m

m w m m

m w T

5.16: Use Second Law and kinematics: agsin θ,2axv2, solve for θ

or ,2

so a2.50m s2, and so m(10 N) (9.80m s22.50m s2)1.37kg d) T mamg, so T mg, because a0

5.18: The maximum net force on the glider combination is

N, 7000N

25002

N 000,

so the maximum acceleration is 5.0m s2

kg

14007000Nmax  

a a) In terms of the runway length L and takoff speed , 22 max,

a a

m

160 )sm0.5(2

)sm40( 2

b) If the gliders are accelerating at amax, from

N

6000N

2500)

smkg)(5.0700

exactly half of the maximum tension in the towrope between the plane and the first glider

5.19: Denote the scale reading as F, and take positive directions to be upward Then,

.1

or

w ma w F

a) a (9.80m s2)((450 N) (550N)1)1.78m s2 , down

b) a(9.80m s2)((670N) (550 N)1)2.14m s2, up c) If F  ,0 ag and the student, scale, and elevator are in free fall The student should worry

5.20: Similar to Exercise 5.16, the angle is arcsin(2 2),

gt

L , but here the time is found in terms of velocity along the table, t v x,x

0

 being the length of the table and v the 0

velocity component along the table Then,

 

m)75.1(sm80.9

s)mm)(3.8010

50.2(2arcsin

2arcsin

2arcsin

2 2

2 2

2

2 0 2

x g L

5.21:

5.22:

5.23: a) For the net force to be zero, the applied force is

N .22)sm(9.80kg)2.11(20.0

k k

 f μ n μ mg F

b) The acceleration is kg,and 2axv2, so xv2 2kg, or x3.13m

5.24: a) If there is no applied horizontal force, no friction force is needed to keep the

box in equilibrium b) The maximum static friction force is, from Eq (5.6),

N, 16.0N) 40(40.0

(

s

balances the applied force of 6.0 N c) The maximum friction force found in part (b), 16.0 N d) From Eq (5.5), kn(0.20)(40.0N)8.0N e) The applied force is enough

to either start the box moving or to keep it moving The answer to part (d), from

Eq (5.5), is independent of speed (as long as the box is moving), so the friction force is 8.0 N The acceleration is ( ) 2.45m s2

F

5.25: a) At constant speed, the net force is zero, and the magnitude of the applied force

must equal the magnitude of the kinetic friction force,

N

)sm(9.80kg)00.6(12.0

k k

 f μ n μ mg

F b) F  f k ma,

so

N

)sm80.9)12.0(smkg)(0.18000

.6(

)(

2 2

k k

5.26: The coefficient of kinetic friction is the ratio f nk , and the normal force has

magnitude 85N25N110 N The friction force, from g a

f

N 28s

m80.9

sm9.0N85N

2 H

(note that the acceleration is negative), and so 11028 NN 0.25

k  



5.27: As in Example 5.17, the friction force is knkwcos and the component of the weight down the skids is wsin  In this case, the angle is arcsin(2.00 20.0)5.7.The ratio of the forces is 00..1025 1,

tan sin

5.7cos)25.0(5.7(sin)sm (9.80kg)(260)

5.28: a) The stopping distance is

m

53)sm80.9(80.0(2

)sm7.28(2

2

k

2 2

b) The stopping distance is inversely proportional to the coefficient of friction and proportional to the square of the speed, so to stop in the same distance the initial speed should not exceed

.sm 1680.0

25.0)sm7.28(dry k,

5.29: For a given initial speed, the distance traveled is inversely proportional to the

coefficient of kinetic friction From Table 5.1, the ratio of the distances is then 00..0444 11

5.30: (a) If the block descends at constant speed, the tension in the connecting string

must be equal to the hanging block’s weight, w B Therefore, the friction force kw A on

block A must be equal to w and B, w B kw A

(b) With the cat on board, ag(w Bk2w A) (w B 2w A)

B A

B A

f f

f T f

,8

32

2

2 0

2 0 4 1 2 0 2 2 0 r

Lg

v Lg

v v Lg

v v g





where L is the distance covered before the wheel’s speed is reduced to half its original

speed Low pressure, 18.1m; 0.0259

) s m m)(9.80 (18.1

s) m 50 3 ( 8

3

2 2





.00505.0m;

N160

2 k







g μ

F m

With the dolly: the total mass is 34.7kg5.3kg40.04kg and friction now is rolling friction, fr rmg

2 r

r

sm82.3

a

ma mg F





5.34: Since the speed is constant and we are neglecting air resistance, we can ignore the

2.4 m/s, and Fnetin the horizontal direction must be zero Therefore fr rn

N200

horiz 

 F before the weight and pressure changes are made After the changes,

,)

( n Fhoriz because the speed is still constant and Fnet 0 We can simply

divide the two equations:

N230N)

200(42.1(81.0(

)42.1)(

81.0(

horiz

N.

200 r



5.35: First, determine the acceleration from the freebody diagrams.

There are two equations and two unknowns, a and T:

a m T g m

a m T g m

B B

A A

:a g m  m m m a

(a) v  ax(2 ) 2 0.22m s

(b) Solving either equation for the tension gives T 11 N

5.36: a) The normal force will be w cos θ and the component of the gravitational force along the ramp is w sin The box begins to slip when θ wsinθ swcosθ, or

,35.0

tanθ s  so slipping occurs at θarctan(0.35)19.3, or 19 to two figures  b) When moving, the friction force along the ramp is kwcos θ, the component of the gravitational force along the ramp is wsin θ, so the acceleration is

.sm92.0)cos(sin

)cos)

5.37: a) The magnitude of the normal force is mg F sin θ

 The horizontal component

of F, F cosθ

must balance the frictional force, so

);

sin(

mg μ

5.38: a) There is no net force in the vertical direction, so nFsinw0, or

.sinsinθ mg F θ

mg F

N, 293)25sin)35.0(25(cos

)smkg)(9.8090

)(

35.0

of force along the ramp are the tension in the first rope (9 N, from part (a)), the

component of the weight along the ramp, the friction on block B and the tension in the second rope Thus, the weight of block C is

N,31.0)36.9(0.35)cos36.9

N)(sin (25.0

N9

)9.36cos9

.36(sinN

or 31 N to two figures The intermediate calculation of the first tension may be avoided to

obtain the answer in terms of the common weight w of blocks A and B,

)),cos(sin

w

giving the same result

(d) Applying Newton’s Second Law to the remaining masses (B and C) gives:

  1.54m s )

sincos

5.40: Differentiating Eq (5.10) with respect to time gives the acceleration

t

t m k t

m

e m

k v

Integrating Eq (5.10) with respect to time with y0 0 gives

]1

[

) ( t

t ) ( t

0

) ( t

t m k

t

t m k

e k

m t v

k

m v e

k

m t v

dt e

v y

5.41: a) Solving for D in terms of v , t

.mkg44.0)

sm42(

)sm(9.80kg)80(

m)kg25.0(

)smkg)(9.8045

D

mg v

5.42: At half the terminal speed, the magnitude of the frictional force is one-fourth the

weight a) If the ball is moving up, the frictional force is down, so the magnitude of the

net force is (5/4)w and the acceleration is (5/4)g, down b) While moving down, the frictional force is up, and the magnitude of the net force is (3/4)w and the acceleration is (3/4)g, down.

5.43: Setting F equal to the maximum tension in Eq (5.17) and solving for the speed vnet

gives

,sm0.26kg)

(0.80

m)N)(0.90600

or 26 m/s to two figures

5.44: This is the same situation as Example 5.23 Solving for  yieldss

.290.0)smm)(9.80(220

s)m0.25(

2

2 2

Rg v



5.45: a) The magnitude of the force F is given to be equal to 3.8w “Level flight” means

that the net vertical force is zero, so Fcosβ(3.8)wcos w,, and





arccos(13.8) 75

(b) The angle does not depend on speed

5.46: a) The analysis of Example 5.22 may be used to obtain tan(v2 gR), but the

subsequent algebra expressing R in terms of L is not valid Denoting the length of the horizontal arm as r and the length of the cable as l,Rrlsinβ The relation v 2TR is

2 2

2 4 ( sin ) 4

gT l r gT

tan )sm(9.80

)30m)sin (5.00m

00.3(4tan

)sin(

4

2

2 2

Note that in the analysis of Example 5.22, β is the angle that the support (string or cable)

makes with the vertical (see Figure 5.30(b)) b) To the extent that the cable can be considered massless, the angle will be independent of the rider’s weight The tension in the cable will depend on the rider’s mass

5.47: This is the same situation as Example 5.22, with the lift force replacing the tension

in the string As in that example, the angle β is related to the speed and the turning radius by tan 2

gR v



 Solving for β ,

h)))km6.3(s)m(1(hkm240(arctan

2 2

5.48: a) This situation is equivalent to that of Example 5.23 and Problem 5.44, so

S Tg R

  A platform speed

of 40.0 rev/min corresponds to a period of 1.50 s, so

.269.0)sm80.9(s)(1.50

m)150.0(4

2 2

60 0  0 067 m

5.49: a) Setting arad  in Eq (5.16) and solving for the period T gives g

s,1.40sm9.80

m4002

g

R π T

so the number of revolutions per minute is (60s min) (40.1s)1.5rev min

b) The lower acceleration corresponds to a longer period, and hence a lower rotation rate, by a factor of the square root of the ratio of the accelerations,

min

rev92.09.83.70min)

5.50: a) 2 T R 2(50.0m) (60.0s)5.24m s b) The magnitude of the radial force

is mv2 Rm42R T2 w(42R gT2)49N (to the nearest Newton), so the apparent weight at the top is 882N49N833N, and at the bottom is 882N49N931N

c) For apparent weightlessness, the radial acceleration at the top is equal to g in

magnitude Using this in Eq (5.16) and solving for T gives

.s14sm9.80

m0.502



g

R T

d) At the bottom, the apparent weight is twice the weight, or 1760 N

5.51: a) If the pilot feels weightless, he is in free fall, and a gv2 R, so

sm3.38)smm)(9.80150



 Rg

the sum of the net inward (upward) force and the pilot’s weight, or

N,3581

m)150)(

sm80.9())sm(h)km((3.6

h)km280(1

N700

1

2 2

or 3580 N to three places

5.52: a) Solving Eq (5.14) for R,

m

230)sm80.94(s)m0.95(

sm(9.80kg)0.50(5

to three figures

5.53: For no water to spill, the magnitude of the downward (radial) acceleration must be

at least that of gravity; from Eq (5.14), v  gR  (9.80m s2)(0.600m) 2.42m s

5.54: a) The inward (upward, radial) acceleration will be 4.64m s2

m) (3.80 s) m 2 4 ( 2 2





R

bottom of the circle, the inward direction is upward

b) The forces on the ball are tension and gravity, so T mgma,

N

1051sm80.9

sm4.64N) 71(1)

5.55: a)

T is more vertical so supports more 1

of the weight and is larger

You can also see this from F xma x:

2 2

1

1 2

532.160

cos

40

cos

060cos40

cos

T T

T

T T

40sin60

T

ma

The tension in the lower chain balances the weight and so is equal to w The lower pulley must have no net force on it, so twice the tension in the rope must be equal to w,

and so the tension in the rope is w 2 Then, the downward force on the upper pulley due

to the rope is also w, and so the upper chain exerts a force w on the upper pulley, and the tension in the upper chain is also w.

5.57: In the absence of friction, the only forces along the ramp are the component of the

weight along the ramp, wsin , and the component of F

along the ramp,

balance the weight, ncos w Eliminating n gives the same result.

5.58: The hooks exert forces on the ends of the rope At each hook, the force that the

hook exerts and the force due to the tension in the rope are an action-reaction pair

The vertical forces that the hooks exert must balance the weight of the rope, so each hook exerts an upward vertical force of w on the rope Therefore, the downward force that 2the rope exerts at each end is Tendsinθw 2, so Tend w (2sinθ)Mg (2sinθ)

b) Each half of the rope is itself in equilibrium, so the tension in the middle must balance the horizontal force that each hook exerts, which is the same as the horizontal component

of the force due to the tension at the end; TendcosθTmiddle, so

.)tan 2()

sin2(cos

(c) Mathematically speaking,   because this would cause a division by zero in the 0equation for Tend or Tmiddle Physically speaking, we would need an infinite tension to keep a non-massless rope perfectly straight

5.59: Consider a point a distance x from the top of the rope The forces acting in this

point are T up and M m(L Lx)g downwards Newton’s Second Law becomes

M g m M

5.60: a) The tension in the cord must be m g in order that the hanging block move at 2

constant speed This tension must overcome friction and the component of the

gravitational force along the incline, so m2g m1gsinμ k m1gcos and

)cos(sin

1

2 m  μ k 

b) In this case, the friction force acts in the same direction as the tension on the block

of mass m , so 1 m2g (m1gsinαμkm1gcosα), or m2 m1(sinαμk cos)

c) Similar to the analysis of parts (a) and (b), the largest m could be is 2

1(sin scos )

m    and the smallest m could be is 2 m1(sin  scos )

5.61: For an angle of 45.0, the tensions in the horizontal and vertical wires will be the same a) The tension in the vertical wire will be equal to the weight w12.0N; this must

be the tension in the horizontal wire, and hence the friction force on block A is also 12.0

N b) The maximum frictional force is μsw A (0.25)(60.0N)15 N; this will be the tension in both the horizontal and vertical parts of the wire, so the maximum weight is 15 N

5.62: a) The most direct way to do part (a) is to consider the blocks as a unit, with total

weight 4.80 N Then the normal force between block B and the lower surface is 4.80 N, and the friction force that must be overcome by the force F is

N, 1.44

or N, 1.440N)

80.4

between block B and the lower surface is still 4.80 N, but since block A is moving

relative to block B, there is a friction force between the blocks, of magnitude

N,0.360N)

5.63: (Denote F

by F.) a) The force normal to the surface is nFcosθ; the vertical component of the applied force must be equal to the weight of the brush plus the friction force, so that FsinθwμkFcosθ, and

N, 161.53cos)51.0(53.1sin

N 00.12cos

w F

keeping an extra figure b) Fcosθ(16.91 N)cos53.110 N

5.64: a)

N 101.3dynes13

)scm980)(

g10210)(

5.62(

5.62)5.62(

4

2 6

140)140(

4 max

g)ms)(14005

.0(2

1

g)ms)(62.5(1.2

g)ms)(77.52

.1(21

)3()2()1(

5.65: a) The instrument has mass mw g 1.531kg Forces on the instrument:

2sm07.13

ma mg T

ma

s25.3gives

?,sm 07.13,

sm 330,

0

0

2 0

a v

v

t a

v

v

y y

y

y y

y

Consider forces on the rocket; rocket has the same a y Let F be the thrust of the

rocket engines

N1072.5)sm07.13sm (9.80kg)000,25()

dt

t dv

a ( ) 3.0m s22(0.20m s3) 3.0m s2(0.40m s3)

At t 4.0s,a3.0m s2 (0.40m s3)(4.0s)4.6 m s2 From Newton’s Second Law, the net force on you is

N 1040

or N 1036

)smkg)(4.672

()sm 9)(

kg72(weight

scale

scale net

ma w F

F

5.67: Consider the forces on the person:

2sm88.560.0so6

mg n

ma mg n

ma

sm0.5 gives)(

2

?,0 ,sm88.5 m,0.3

0

2 0 2

0

2 0

y y

y y

y

v y

y a v

v

v v

a y

y

5.68: (a) Choosing upslope as the positive direction:

ma mg

mg f

m8.9



v at the top

sm11

or sm5.11sm132

sm132m)0.8)(

sm25.8(22

2 2 0

2 2 2

2 0

(b) For the trip back down the slope, gravity and the friction force operate in opposite

directions:

2 2

k net

sm55.3))799.0)(

30.0()602.0)((

sm8.9()37cos30.037sin

(

37cos37

μ mg

F

Now

sm7.5

or sm54.7sm8.56

sm8.56

m)0.8)(

sm55.3(20)(2

and,0m,0.8,

0

2 2

2 2

2 0

2 0 2 0 0

x x

v

5.69: Forces on the hammer:

ma T

ma

F

mg T

mg T

74sinso074

singives

Divide the second equation by the first:

2sm8.2and

It’s interesting to look at the string’s angle measured from the perpendicular to the top of the crate This angle is of course 90—angle measured from the top of the crate The free-body diagram for the washer then leads to the following equations, using Newton’s Second Law and taking the upslope direction as positive:

slope w

string

slope w

string

string slope

w

w string slope

w

coscos

)sin(

sin

0cos

cos

sinsin

θ g m T

θ g a m θ

T

θ T θ

g m

a m θ

T θ

g m

slope

slope string

0cos

sintan

g

θ g a

 For the crate, the component of the weight along the slope is m gc sinslope and the normal force is mcgcosθslope Using Newton’s Second Law again:

slope

slope k

c slope c

k slope c

cossin

cossin

θ g

θ g a

a m θ

g m θ

g m

which leads to the interesting observation that the string will hang at an angle whose

tangent is equal to the coefficient of kinetic friction:

40.022tan)6890tan(

tan string



5.71: a) Forces on you:

2 k

k

sm094.3)cos(sin

sin

cos gives

ma f α mg

ma F

mg n ma

F

x x

y y

Find your stopping distance

v x 0,a x 3.094m s2,v0x 20m s,xx0 ?

2 2 ( 0)gives 0 64.6m,

0

2 v  a xx xx 

stop before you reach the hole, so you fall into it

5.72: The key idea in solving this problem is to recognize that if the system is

accelerating, the tension that block A exerts on the rope is different from the tension that block B exerts on the rope (Otherwise the net force on the rope would be zero, and the

rope couldn’t accelerate.) Also, treat the rope as if it is just another object Taking the

“clockwise” direction to be positive, the Second Law equations for the three different

parts of the system are:

Block A (The only horizontal forces on A are tension to the right, and friction to the left):

To solve for a and eliminate the tensions, add the left hand sides and right hand sides of

R B A

A L R B

m m m

m m m R

B A L

d R B

R B

m m m m m

g

a  



 As the system moves, d will increase, approaching

L as a limit, and thus the acceleration will approach a maximum value of

.) ( A B R

5.73: For a rope of length L, and weight w, assume that a length rL is on the table, so that

a length (1r) L is hanging The tension in the rope at the edge of the table is then

,

)

1

( r w and the friction force on the part of the rope on the table is fs srw This must

be the same as the tension in the rope at the edge of the table, so

)

1(1and)

1

 rw r w r  Note that this result is independent of L and w for a

uniform rope The fraction that hangs over the edge is 1 r s (1s); note that if

.01

5.74: a) The normal force will be mgcosFsin , and the net force along (up) the ramp is

).cos(sin

)sin(cos

)sin cos

(sin

cos mg s mg αF  F  s  mg  s 

F

In order to move the box, this net force must be greater than zero Solving for F,

.sin cos

cossin

expression must be positive, or cos ssin, and  s cot b) Replacing s with kwith in the above expression, and making the inequality an equality,

.sincos

cossin

2axv is useful, but we also need a The acceleration

must be large enough to cause the box to begin sliding, and so we must use the force of static friction in Newton’s Second Law: smgma,or asg Then,

,)