CH 12 P1-75.pdf

CHAPTER 12

PROBLEM 12.1

The acceleration due to gravity on Mars is 3. 3.75 75 m/ m/ss 2 . Know Knowin ing g that that the the mass of a silver bar has been officially designated as 20 kg, determine, on Mars, its weight in newtons.

SOLUTION

m W

=

=

20 kg, g mg

=

=

3.75 m/s 2

(20) (3.75)

PROPRIETARY PROPRIET ARY MATERIAL. ©

W

=

75 N �

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PROBLEM 12.2

The value of the acceleration of gravity at any latitude φ is given by

(

)

2

2

g = 9.7087 1 + 0.0053 sin φ m/s , where the effect of the rotation of the

earth as well as the fact that the earth is not spherical have been taken into account. Knowing that the mass of a gold bar has been officially designated as 2 kg, determine to four significant figures its mass in kilograms and its weight in newtons at a latitude of ( a) 0° , (b) 45° , (c) 60°.

SOLUTION

m

At all latitudes, (a) φ

=

0°,

(

) = 9.7807 m/s

g = 9.7807 1 + 0.0053 sin 2 φ W

=

(

W (c)

=

(

W

=

) = 9.8066 m/s

) = 9.8196 m/s

W = 19.61 N �

2

mg = ( 2.000 )(9.8196)

PROPRIETARY MATERIAL. ©

W = 19.56 N �

2

mg = ( 2.000 )(9.8066)

φ = 60°, g = 9.7807 1 + 0.0053 sin 2 60°

2.000 kg �

2

mg = ( 2.000 )(9.7807 )

(b) φ = 45°, g = 9.7807 1 + 0.0053 sin 2 45°

=

W = 19.64 N �


PROBLEM 12.3 A and a lever scale B having equal lever arms are fastened A spring scale scale A and scale B having to the roof on an elevator, and identical packages are attached to the scales as shown. Knowing that when the elevator moves downward with an acceleration of of 0.6 0.6 m/s m/s22 the the spring springscale scaleindicates indicatesa aload loadofof3.2 3.2kg, kg b)) the load indicated by the determine (a) (a) the weight of the packages, ( b spring scale and the mass needed to balance the lever scale when the elevator moves upward with an acceleration of .6 m/s m/s22. .

SOLUTION

Assume g = 32.2 9.81 ft/s m/s22

=

m

Σ F =  

W 1 −

a 

 = g 

F s

W =

or

W g

F s a 1− g

W

−F =

ma : W

s

=

g

3.2 × 9.81

1−

.6 9.81

W = m

W

=

33.4

=

g

Σ F =

9.81

33.4 N

�

= 35.4 N

�

= 3.4 kg

ma : F s

−W =

W g

a

 a  = W 1 +   g  .6  = 33.4  1 +   9.81 

F s

a

F s

For the balance system B,

ΣM = 0

=

Fw

0: bFw

− bF = 0

F p

But, Fw

 a   a  = W 1 +  and F = W 1 +   g   g 

so that

Ww

w

=W

p

p

and mw


=

p

p

W p g

m

=

3.4 kg 3.

�


PROBLEM 12.4

A Global Positioning System (GPS) satellite is in a circular orbit 20157 km above the surface of the earth and completes one orbit every 12 h. Knowing that the magnitude of the linear momentum of the satellite is 340194 kg. m/s and the radius of the earth is 6370 km determine (a) the mass of the satellite, (b) the weight of the satellite before it was launched from earth.

SOLUTION

Periodic time: Radius of Earth: Radius of orbit:

Velocity of satellite:

τ =

R

=

12 h

=

43200 s

6370 km

r = 6370

v

=

2π r

+

=

20157

(a)

mv m

43200

=

3858.2 m/s

=

340194 kg.s

=

mv v

26527 km

(2π ) (26527 × 103)

τ

It is given that

=

340194 3858.2

=

88.17 kg

m (b)

W

=

mg = (88.17) ( 9.81)

=

=

88.17 kg

�

865 N W = 865 N �



PROBLEM 12.5 80 N block starts from rest and moves upward when constant forces The 80-N of 20 N and 40 N are applied to supporting ropes. Neglecting the masses of the pulleys and the effect of friction, determine the speed of the block after it has moved .457 m.

SOLUTION

+

∑ F = ma y

y

:

20

+ 20 + 20 + 40 − 80 =

a y = a y =

dv dt

=

dy dv dt dy

( 9.81)(20) 80

=v

80 9.81

ay

= 2.45 m/s2

dv dy

v dv = a y d y

∫

v

0

v dv =

∫

v

0

a y d y ,

v == 2a y y


=

1 2

v2 = a y y

(2)( 2.45)(.457)

v =

1.5 m/s �


PROBLEM 12.6

A motorist traveling at a speed of 108 km/h suddenly applies the brakes and comes to a stop after skidding 75 m. Determine ( a) the time required for the car to stop, (b) the coefficient of friction between the tires and the pavement.

SOLUTION

Data: v0 = 108 km/h = 30 m/s, x f = 75 m acceleration: a = v (a) Assume constant acceleration.

dv dx

=

dv

= constant

dt

x

0

∫ v0 v dv = ∫ f a dx 0

1

− v02 = a x f 2

v02

a=−

2 x f

t

0

=−

∫ v0 dv = ∫ f a dt 0

(30) = − 6 m/s ( 2)(75)

2

− v0 = a t f t f = − (b)

+

v0

=

a

∑ F y = 0:

− 30 −6

t f = 5.00 s �

N − W = 0 N = W

∑ F x = ma : µ =

µ =


−

−

ma N

(− 6) 9.81

=−

− µ N = ma ma W

=−

a g

µ

= 0.612 �


PROBLEM 12.7

A 1400-kg automobile is driven down a 4 ° incline at a speed of 88 km/h when the brakes are applied, causing a total braking force of 7500 N to be applied to the automobile. Determine the distance traveled by the automobile before it comes to a stop.

SOLUTION

(a)

+

∑ F = ma : a=−

F f

=−

m

− F f + W sin α = ma +

W sin α m

7500 N 1400 kg

=−

F f m

+ g sin α

+ (9.81 m/s 2 ) sin 4° = − 4.6728 m/s 2

a = 4.6728 m/s 2

4°

v0 = 88 km/h = 24.444 m/s a=v

From kinematics,

∫

x f 0

a dx =

dv dx 0

∫ v0 v dv

1 a x f = − v02 2 x f = −


v02 2a

2

( 24.444) =− ( 2)( − 4.6728)

x f = 63.9 m �


PROBLEM 12.8

ts car, its velocity is reduced from 112.16 In the braking test of aaspor sports car its to 112.16 km/h km/h to zero in a distance of 51.8 m with slipping impending. Knowing that the coefficient of kinetic friction is 80 percent of the coefficient of static friction, determine (a) the coefficient of static friction, ( b) the stopping distance for the same initial velocity if the car skids. Ignore air resistance and rolling resistance.

SOLUTION

friction: (a) Coefficient of static friction. Σ F y =

N

0:

v0

− W =

v02 2

2

=

2 vv22 v20 −v −

2 ( s

−

=

s0 )

0

µ s = −

mat W

W

mat : = −

at ( s − s0 ) 2

( 31.16 ) (2) (51.8)

−

Forbraking brakingwithout withoutskidding: skidding For Σ Ft =

=

= 112.16 km/h = 31.16 m/s m/s.

v2

at

N

0, 0

µ = µ s,

− µ s N =

at g

= − 9.37 m/s

so that

mat

µ s N =

m| at |

9.37

=

2

µ s = .955

9.81

�

(b) Stopping distance with skidding. skidding: Use

µ = µ k =

(0.80)( .955 ) =.764 Σ F =

ag t =

−

mat :

µ k N

m

µ k N = −mat

= −µ k

= − 7.49 m/s

2

Since acceleration is constant,

( s − s0 ) =

v2

−

v02

2at

=

2

( 31.16 ) ( 2) ( − 7.49 )

0

−

s


−

s0

= 64.8 m �


PROBLEM 12.9

= 0 A .09-kg .09 kg model model rocket rocket is is launched launched vertically verticallyfrom fromrest restatattime timet t 0 with with aa = t >> 1 constant thrust of 4 N N for for one second and no thrust for t 1 s. s. Neglecting a) the air resistance and the decrease in mass of the rocket, determine ( a) h reached by the rocket, (b) maximum height h reached ( b) the time required to reach this maximum height.

SOLUTION

Σ F =

For the thrust phase,

ma : Ft

−W =

ma

=

W g

a

a

4 F = g  − 1 = ( 9.81)  − 1 = 34.6 m/s   W   .09 × 9.81  

v

=

at =

y

=

1

t

2

At t = 1 s,

2

(34.6) (1) = 34.6 m/s

at 2 =

1

2

( 34.6 ) (1) = 17.3 m

2

For the free flight phase, t > 1 s. a

=

v At v

= 0,

t

v2

(a) ymax

v1

= − g = − 9.81 m/s2

+ a (t − 1) = 34.6 + ( − 9.81)( t − 1) 34.6

−1 =

− v12 =

= 3.527 s,

9.81

2a ( y

=−

v2

4.527 s

− y1 ) = −2 g ( y − y1 )

−v

2 1

y

−

y1

=

h

= 61 + 17.3

2 g

(b) As already determined,


t =

=−

0

2

− ( 34.6 ) = 61 m ( 2)( 9.81) h = 78.3 m � t = 4.53 ss � t 4.527 =


PROBLEM 12.10

A 40-kg package is at rest on an incline when a force P is applied to it. Determine the magnitude of P if 4 s is required for the package to travel 10 m up the incline. The static and kinetic coefficients of f riction between the package and the incline are 0.30 and 0.25, respectively.

SOLUTION

Kinematics: Uniformly accelerated motion. ( x0 x

=

x0

+

v0t

Σ F y =

N

=

Σ F x =

+

1 2 at , 2 N

0:

−

P sin 50°

−

=

2 x 2

t

=

mg cos 20°

=

( 2 )(10 ) 2 (4 )

=

0) =

1.25 m/s2

0

P sin 50° + mg cos 20° ma : P cos 50° − mg sin 20° −

or P cos 50° − mg sin 20°

P =

a

or

0, v0

=

ma

+

For motion impending, set a

ma

( P sin 50° + mg cos 20° ) =

− µ

mg ( sin 20° cos50°

µ N =

+ µ cos 20°

)

− µ sin 50° =

0 and

µ = µ s =

0.30.

( 40 )(0 ) + (40 )(9.81)(sin 20° + 0.30cos 20° )

P =

cos 50° − 0.30 sin 50°

For motion with a

=

P =

ma

1.25 m/s 2 , use µ

= µ k =

=

593 N

0.25.

( 40 )(1.25 ) + (40 )(9.81)(sin 20° + 0.25cos 20° ) cos 50°

−

0.25sin 50° P = 612 N �



PROBLEM 12.11

If an automobile’s braking distance from 100 km/h is 60 m on level pavement, determine the automobile’s braking distance from 100 km/h when it is (a) going up a 6° incline, (b) going down a 2-percent incline.

SOLUTION

Calculation of braking force/mass ( Fb / m ) from data for level pavement. v0 == 100 m/s = 27.778 v 100 km/h km/hr= 27.778 m/s v2 2 a

=

F br

−

v02

2 ( x

−

x0 )

ma :

(a) Going up a 6° incline. (θ Σ F =

a

ma :

= −

m

−

x

−

x0

=

−

=

v02

2a

x0 ) 2

(27.778) ( 2)( 60)

0

−

ma

6.43 m/s 2 6°) mg sin θ

=

ma

g sin θ

= −6.43 −

v2

=

−

m/s 2

− F br −

F br

a (x

=

− F br =

= −a =

m

2

v2

= −6.43 Σ F x =

v02

−

9.81sin 6° =

= −7.455

m/s 2

2

(27.778) ( 2)( −7.455)

0

−

x (b) Going down a 2% incline. ( tan θ Σ F =

a

ma :

= −

F br m

= − 6.43 −

x

−

x0

=

v2

−

2a


v02

=

mg sin θ

−

=

=

51.7 m �

)

ma

g sin θ 9.81sin ( −1.145° )

= − 6.234

m/s 2

2

(27.778) (2)( −6.234)

0

x0

= −0.02, θ = − 1.145 °

− F br − −

−

x

−

x0

=

61.9 m �


PROBLEM 12.12

The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between the blocks and the incline, determine ( a) the acceleration of each block, (b) the tension in the cable.

SOLUTION

Let the positive directions of x A and x B be down the incline.

a A

+ 3 xB = constant

x A

Constraint of the cable:

+ 3a = B

0

or

aB

−T =

For block A:

Σ F =

ma : m A g sin 30°

For block B:

Σ F =

ma : m B g sin 30° − 3T a A

Eliminating T and solving for

(3m

A

a A g (a) a A a B

=

g

g

=−

1 3

aA

m Aa A

=

mB aB

(1)

= − mB aA

(2)

,

− m g ) sin 30° =  3m +  B

A

m B 3

 a  

A

(3m − m ) sin 30° (30 − 8) sin 30° = = 0.33673 3m + m / 3 30 + 2.667 A

B

A

B

= (0.33673)(9.81) = 3.30 m/s 2 1

= − (3.30 ) = −1.101 m/s 3

2

a A

= 3.30 m/s

2

30° �

a B

= 1.101 m/s

30° �

2

(b) Using equation (1), T

=

 

m A g  sin 30°

−

a A 

 = (10)(9.81)(sin 30° − 0.33673)

g 

T = 16.02 N �



PROBLEM 12.13

The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and assuming that the coefficients of friction between both blocks and the incline are µ s = 0.25 and µ k = 0.20, determine (a) the acceleration of each block, ( b) the tension in the cable.

SOLUTION

Let the positive directions of x A and x B be down the incline. Constraint of the cable: a A Block A:

Σ F y =

x A +

3aB

0: N A

Σ F x =

+

−

3 xB =

=

constant a B

0

mA g cos 30°

ma : mA g sin 30° −

= − =

1 3

aA

0

µ N A −

T

=

m Aa A

Eliminate N A. m A g (sin 30°

Σ F y = 0:

Block B:

Σ F =

NB

− µ cos 30°) − T =

mA aA

− mB g cos 30° = 0

ma : m B g sin 30°

+ µ N B − 3T = mB aB = −

m B aA 3

Eliminate N B. m B g (sin 30° +

°) − 3T = −

µ cos 30

m B a A 3

Eliminate T . m   (3m A g − mB g ) sin 30° − µ (3mA g + mB g ) cos 30° =  3mA + B  aA 3   Check the value of µ s required for static equilibrium. Set a A = 0 and solve for

µ .

µ

Since

µ s

Calculate

=

(3m A − mB ) sin 30° (75 − 20) = tan 30° = (3m A + mB ) cos 30° (75 + 20)

0.334.

= 0.25 < 0.334, sliding occurs. aA g

for sliding. Use

µ

=

µ k

= 0.20. continued



PROBLEM 12.13 CONTINUED

a A g

=

a B (b) T

A

B

A

A

= (a) a A

(3m − m ) sin 30° − µ ( 3m + m ) cos 30° 3m + m / 3 B

B

(30 − 8) sin 30° − ( 0.20)( 30 + 8) cos 30° = 0.13525 30 + 2.667

= (0.13525)(9.81) = 1.327 m/s 2

a A

= 1.327 m/s

2

30° �

1 = −    (1.327 ) = − 0.442 m/s  3 

a B

= 0.442 m/s

2

30° �

=

m A g (sin 30°

2

− µ cos 30°) − m

A

aA

= (10 )(9.81)(sin 30° − 0.20 cos 30°) − (10)(1.327) T = 18.79 N �



PROBLEM 12.14

A light train made up of two cars is traveling at 88 km/h when the brakes are applied to both cars. Knowing that car A has a weight of 24947.56 kg and car B has a weight of 19958 kg and that the braking force is 31137.5 N on each car, determine (a) the distance traveled by the train before it comes to a stop, (b) the force in the coupling between the cars while the train is slowing down.

SOLUTION

m A = 24947.56 kg

Data :

m B = 19958 kg v0 = − 88 kg/h = 24.4 m/s (a) Use both cars together as a free body. Consider horizontal force components only. Both cars have same acceleration.

∑ F x = ∑ max : − Fb − Fb = mAax + mB ax a x =

Fb + F b m A + mB

a x = v

∫

x f 0

x f = −

a x dx = v02 2a x

31137.5 + 31137.5

=

24947.56

+ 19958

= 1.38 m/s2

dv dx 0

∫ v0

v dv

ax x f =

v02 2

2

=−

( − 24.4 ) = − 215.7 m (2)( 1.38 ) 215.7 m to

the left �

(b) Use car A as free body. F c = coupling force.

∑ F x = ∑ max :

Fc − Fb = mAax

Fc = mAax − F b = ( 24947.56 )(1.38 ) + 31137.5 = 65.6 kN F c =


65.6 kN

tension �


PROBLEM 12.15

Solve Prob. 12.14, assuming that the brakes of car B fail car B fail to operate. Problem 12.14: A

traveling at at 88 55 km/h mi/h light train made up of two cars is traveling when the brakes are applied to both cars. Knowing that car A has A has a weight of 55,000 lb and car B has a weight of 44,000 lb and force 24947.56 kg and car B has a weight of 199.58 kgthat andthe thatbraking the braking a) is 7000 lb on each car, determine ( the distance traveled by the force is 31137.5 kg on each car, determine (a) the distance traveled bytrain the before it comes to a stop, ( b) the in theincoupling between the cars train before it comes to a stop, ( b)force the force the coupling between the while the train is slowing down. cars while the train is slowing down.

SOLUTION

Data:

m A = 24947.56 kg m B = 19958 kg v0 = 24.4 m/s

(a) Use both cars together as a free body. Consider horizontal force components only. Both cars have same acceleration.

∑ F x = ∑ max : − Fb − Fb = mAax + mB ax a x =

F b m A + mB

a x = v

∫

x f 0

x f = −

a x dx = v02 2a x

31137.5 + 31137.5

=

19958 24947.56 +

= .69 m/s 2

dv dx 0

∫ v0

v dv

ax x f =

v02 2

2

=−

( − 24.4 ) = − 431.4 m ( 2)(.69 ) 431.4 m

to the left �

(b) Use car B as a free body. F c = coupling force.

∑ F x = ∑ max : − Fc = mB ax − F c = ( 19958 )(.69) = 13.8 kN F c =


13.8 kN

compression �


PROBLEM 12.16

Aweighs Block Block A weighs356 356NN, and block B B weighs weighs 71 71 N N.. The coefficients coefﬁcients of of friction between all surfaces of contact are µ s = 0.20 and µ k = 0.15.

Knowing that P = 0, determine (a) the acceleration of block B, (b) the tension in the cord.

SOLUTION

2 x A

Constraint of cable: a A

+

aB

+

( xB

=

−

xA )

0 0,

xA

=

xB

+

aB

or

= constant.

= −a A

Assume that block A moves down and block B moves up. Block B:

Σ F y =

0:

N AB

Σ F x =

ma :

− W B cos θ =

−T +

µ N AB

0 W B

+ WB sin θ =

g

aB

Eliminate N AB and a B . −T +

Block A:

W B (sin θ

Σ F y =

0:

N A

N AB

=

Σ F x =

−W B

(sin θ

NA

+

µ cosθ )

−

N AB

mA aA :

+

−

µ cosθ )

WB ) sin θ

−

=

(WB

− WB

(W B

µ (WA

aB g

+ W A

− T + WA sin θ −

−µ

(W A

WB

− W A cos θ =

WA cosθ

+

=

g

aA

g

0

) cosθ

FAB

FA

−

+ WA sinθ −

=

W A g

aA

µWB cosθ

aA g

WA ) cosθ

=

W A

3WB ) cosθ

=

(WA

+

+

a A

= −WB

+ WB

)

aA

g

Check the condition of impending motion. µ

=

(W A

0.20,

aA

WB ) sin θ s

− 0.20

µ s −

=

=

aB

(WA

+

=

0,

θ

=

θs

3WB ) cosθ s = 0 continued




tan θ s

=

θ s

=

Calculate

aA g

g

=

a A

(b)

T

=

W B

=

(0.20)(128) 64

=

0.40

25°. The blocks move.

<

θ

=

µk

=

0.15 and θ

285 sin 25° −

−

=

µ k (WA +

+

25°. 3WB ) cosθ

W B

(0.15)(569) cos 25° 427

= .10093

(.10093 )( 9.81) = .99 m/s2

W B (sin θ

= 71

=

3W B )

W A

= − .99 m/s

=

21.8°

−

+

(W A − WB ) sin θ

=

a B

W A

using µ

a A

(a)

0.20 (W A

2

+

a B = .99 m/s

µ cosθ ) + W B

2

25° �

aA g

(sin 25° + 0.15 cos 25° ) + (71)( .10093 ) T =


46.8 N

�


PROBLEM 12.17

Block A weighs 350 N, and block B weighs 71 N. The coefficients of friction between all surfaces of contact are µ s = 0.20 and µ k = 0.15. Knowing that P = 44.5 N →, determine (a) the acceleration of block B, (b) the tension in the cord.

SOLUTION +

( xB

aB

=

2 x A

Constraint of cable:

a A

+

−

xA )

0,

xA

=

+

xB

aB

or

constant.

=

= −a A

Assume that block A moves down and block B moves up. Block B:

Σ F y =

N AB

0:

Σ Fx =

ma x :

−

W B cosθ

−T +

0

=

µ N AB

+

WB sin θ

µ cosθ )

=

WB

aB g +

=

W B g

aB

Eliminate N AB and a B . W B (sin θ

−T +

Block A:

Σ F y =

0:

NA

N A

=

N AB

=

(W B

Σ F x =

mA a A :

−W B

−µ

(W A

− WB

) sin θ

(W B −

N AB

−

WA cos θ

+

WA cosθ

−

P sin θ

+

WA ) cosθ

−T +

(sin θ

+

+

−

P sinθ

−

FAB

WA sin θ

µ cosθ ) − WB

WA ) cosθ

+

−

µ ( WA

+

+

a A

+

g

µ P sin θ

3WB ) cosθ

+

−

= −WB

P sin θ

FA

+

WA sin θ

+

+

=

P cosθ

−

P cosθ

P ( µ sinθ

aA g

0

=

W A g

aA

µWB cosθ

=

W A

aA g

cosθ ) = (WA

+

WB )

aA g

Check the condition of impending motion. µ

(W A

−

=

WB ) sin θ

µ s −

=

0.20, a A

µ s (WA

+

aB

=

0, θ

3WB ) cosθ

+

Ps ( µ s sinθ

=

=

25° +

cosθ ) = 0 continued




P s

=

=

µ s (WA

+

3WB ) cosθ µ s sin θ

a A g

=

0.20 sin 25°

=

aA using µ g

(W A

(WA

−

WB ) sinθ

cosθ

+

(0.20 )(569) cos 25° − 285 sin 25°

Blocks will move with P = Calculate

−

−

=

WB ) sin θ

cos 25°

=

0.15, θ

µ k (WA

+

=

25°, and P =

3WB ) cosθ

W A 285 sin 25° −

= − 17.46 N < 44.5 N

44.5 N.

µk −

+

+

+

44.5 N .

P ( µ k sinθ + cosθ )

W B

(0.15)(569) cos 25° + ( 44.5)( 0.15sin 25° + cos 25°) 427

= .26016

a A

=

(a)

a B

(b)

T

(.26016 )( 9.81) 2

= − 2.55 m/s

=

W B (sin θ

= 71

+

2

= 2.55 m/s

2

, µ cosθ )

a B = 2.55 m/s

+

W B

25° �

aA g

(sin 25° + 0.15 cos 25° ) + (71)( .26016) T =


58.13 N �


PROBLEM 12.18

Boxes A and B are at rest on a conveyor belt that is initially at rest. The belt is suddenly started in an upward direction so that slipping occurs between the belt and the boxes. Knowing that the coefficients of kinetic friction between the belt and the boxes are ( µ k ) A = 0.30 and

( µ k ) B = 0.32, determine the initial acceleration of each box.

SOLUTION

Assume

a B > a A so

that the boxes separate. Boxes are slipping.

µ = µ k

Σ F y = 0:

N

N

Σ F x =

ma :

− mg cos15° = 0 =

mg cos15°

µ k N − mg sin15° =

ma

µ k mg cos15° − mg sin15° =

ma

a = g ( µ k cos15° − sin15° ) , For box A, a A

a B

µ k = 0.30

(0.30 cos15° − sin 15° )

= 9.81

For box B,

independent of m.

or

a A =

0.304 m/s 2

15° �

µ k = 0.32

(0.32 cos15° − sin 15° )

= 9.81


or

a B =

0.493 m/s 2

15° �


PROBLEM 12.19

The system shown is initially at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys, determine ( a) the acceleration of each block, (b) the tension in each cable.

SOLUTION

Let y be downward position for for all blocks. y be positive Let the positive downward position all blocks. Constraint of cable attached to mass A:

y A

a A

yC

Constraint of cable attached to mass C :

For each block

Σ F =

ma :

TA

3aB

+

3 yB

+

aC

+

aB

=

=

constant

0

yB

+

=

or =

a A

= −3a B

aC

= − aB

constant

0

or

Block A:

W A

−

=

mA a A ,

or

TA

=

WA

−

m Aa A

=

WA

− 3mA aB

Block C :

WC

− TC =

mC aC ,

or

TC

=

WC

−

mC aC

=

WC

−

(WC

−

mC aB )

WB

−

3WA

−

WC

W B

+

9WA

+

W C

Block B: W B

−

3TA

W B

−

3 (WA

−

TC −

mB aB

=

3mA aB )

(b) Tensions. Tensions:

−

a B

or (a) Accelerations. Accelerations:

=

g

=

a A

= −

aC

= − − .752

= 88 −

T C

= 88 −

=

mB aB =

265 − 265 −

(3)( − .752)

88

265 + 795 + 88

( − . 07 665 )( 9 .8 1) = − .752 m/s

a B

T A

mC aB

= 2.256 m/s

2

2

2

= − .076 65

22 a B = .75 .752m/s m/s

�

22 2.26 m/s a A = 2.256 m/s

�

22 .75 m/s m/s aC = .752

�

(

)

9.81

(2.256)

T A

= 67.76 N

�

(.752 )

T C

= 81.25 N

�

88

88 9.81

= .752 m/s



PROBLEM 12.20

Each of the systems shown is initially at rest. Neglecting axle friction and the masses of the pulleys, determine for each system ( a) the acceleration of block A, (b) the velocity of block A after it has moved through 1.5 m, (c) the time required for block A to reach a velocity of 3 m/s.

SOLUTION

Let y be positive downward for both blocks. Constraint of cable: y A + yB = constant a A + aB = 0

For blocks A and B,

Σ F = ma : W A aA g

Block A: W A − T =

Block B: P + W B − T = P + W B − WA +

Solving for a A ,

aA =

a B = − a A

or

W A − WB − P W A + W B

WB g

T = W A −

or aB = −

W B g

W A aA g

aA

W A W a A = − B aA g g

g

(1) 2

v A2 − (v A )0 = 2a A  y A − ( y A )0 

with

v A = 2aA  y A − ( yA )0    v A − (v A )0 = a At t =

with

( vA )0 = 0 (2)

( vA )0 = 0

v A a A

(3) continued




(a) Acceleration of block A. System (1): W A

= 441 N

By formula (1),

(a )

By formula (1),

(a )

System (3): W A

=

(b) v A at y A

−

A

WB

4895 =

1.5 m.

=

=

0, P =

=

−

4895

P = 0

N,

( 9.81)

223

441

4895 N,

(a )3

( y A )0

WB

+

441 − 223

=

A 2

By formula (1),

441

441 N,

=

= 223

441 − 223

=

A 1

System (2): W A

WB

,

3.22 m/s 2

�

(a )

=

4.85 m/s

2

�

(a )3

=

.23 m/s

A 1

223 N

(9.81 )

A 2

4669.5 N,

4 669.5 N

+

=

(a )

4669.5

P = 0

(9.81)

A

2

�

Use formula (2).

System (1):

(v )1

=

( 2)( 3.22)(1.5)

( v A )1

=

3.1 m/s

�

System (2):

(v )2

=

( 2)(4.85) ( 1.5)

(v A )2

=

3.8 m/s

�

System (3):

(v )3

=

( 2)( .23) ( 1.5)

(v A )3

(c) Time at v A

A

A

A

=

3 m/s.

=

System (2): t 2

=

System (3): t 3

=

.83 m/s

t 1

=

0.932 s �

t 2

=

.618 s

t 3

=

13.04 s �

Use formula (3).

3

System (1): t 1

�

=

3.22 3 4.85 3 .23


�


PROBLEM 12.21

The flat-bed trailer carries two 1360.8 kg beams with the upper beam secured by a cable. The coefficients of static friction between the two beams and between the lower beam and the bed of the trailer are 0.25 and 0.30, respectively. Knowing that the load does not shift, determine ( a) the maximum acceleration of the trailer and the corresponding tension in the cable, (b) the maximum deceleration of the trailer.

SOLUTION

(a) Maximum acceleration. The cable secures the upper beam; only the lower beam can move.

For the upper beam,

Σ F = 0: y

N1

− W = 0 N1

For the lower beam,

Σ F = 0: y

Σ F = x

a

= 0.85

W m

Σ F x =

ma : T

mg

− N − W = 0

or N 2

1

ma : 0.25 N1

= (0.85) (9.81 ) = 8.3 m/s

For the upper beam, T

N2

=W =

=

2W

+ 0.30 N = (0.25 + 0.60 )W = ma 2

2

a

= 8.3 m/s

2

�

− 0.25 N1 = ma

13349.45 = 0.25W + ma = (0.25)( 13349.45) +   ( 8.3 ) = 14.6 kN  9.81 

T =

14.6 kN �

(b) Maximum deceleration of trailer. Case 1: Assume that only the top beam slips. As in Part ( a) N1 = mg .

Σ F = a

ma : 0.25W

=

ma

= 0.25g = 2.45 m/s

2

continued PROPRIETARY MATERIAL. ©



Case 2: Assume that both beams slip. As before N 2

Σ F =

a

=

=

2 W .

(2m ) a : ( 0.30)( 2W ) 0.30g =

2.9 m/s

=

( 2m) a

2

a

The smaller deceleration value governs.


2

= 2.45 m/s

�


PROBLEM 12.22

The 10-kg block B is supported by the 40-kg block A which is pulled up an incline by a constant 500 N force. Neglecting friction between the block and the incline and knowing that block B does not slip on block A, determine the smallest allowable value of the coefficient of static friction between the blocks.

SOLUTION

Since both blocks move together, they have a common acceleration. Use blocks A and B together as a free body. Σ F = Σ

ma : P

−

a

=

=

m A g sin 30° P m A

+

mB

−

mB g sin 30° = ( mA

− g sin 30° =

500 50

−

+

mB ) a

9.81 sin 30°

5.095 m/s 2

Use block B as a free body. Σ F =

m B a cos 30° :

F f

(10)(5.095) cos 30° = 44.124 N

=

Σ F =

N B

=

m B a sin 30° : mB ( g

+

= 123.575

Ff

NB

=

−

mB a cos 30°

mB g

=

mB a sin 30°

a sin 30°) = 10 (9.81 + 5.095 sin 30°) N

Minimum coefficient of static friction: µ min =


F f N B

=

44.124 123.575

µ min =

0.357 �


PROBLEM 12.23

A package is at rest on a conveyor belt which is initially at rest. The belt is started and moves to the right for 1.5 s with a constant acceleration of 3.2 m/s2 . The belt then moves with a constant deceleration a 2 and comes to a stop after a total displacement of 4.6 m. Knowing that the coefficients of friction between the package and the belt are µ s = 0.35 and µ k = 0.25, determine (a) the deceleration a 2 of the belt, (b) the displacement of the package relative to the belt as the belt comes to a stop.

SOLUTION

(a) Kinematics of the belt. vo

0

=

a1

1. Acceleration phase with v1

=

vo

+

a1t1 = 0

x1

=

xo

+

vot1

+

1 2

+

a2

=

v22

2 − v1 =

−

2 ( x2

a1t 12

2a2 ( x2

0

=

0

+

=

+

4.8 m/s 1 2

2

(3.2 )(1.5)

=

0 since the belt stops.

−

x1)

=

3.6 m

2

v12

−

3.2 m/s 2

(3.2 )(1.5)

2. Deceleration phase. v2 v22

=

x1 ) t2

0 − ( 4.8)

=

−

2 ( 4.6 − 3.6 ) t 1

=

v2

−

v1

a2

= −11.52

0

4.8 −11.52

=

−

a 2 = 11.52

=

m/s 2

�

0.41667 s

(b) Motion of the package. 1. Acceleration phase. Assume no slip.

Σ F y =

0: N

−W =

Σ F x =

ma : F f

=

( a p )1 = 3.2 m/s 2

0 or N

W

=

mg

0.35 W

=

0.35 mg

=

m (ap )

1

The required friction force is F f . The available friction force is F f m

=

( a p )

1


<

µ s N

m

µ s N =

= µ s g =

( 0.35)(9.81)

=

3.43 m/s 2



Since 3.2 m/s 2

3.43 m/s 2, the package does not slip.

<

(v p )1 = v1 = 4.8 m/s

( x p )1 = 3.6 m.

and

(a p )2

2. Deceleration phase. Assume no slip. Σ F x =

F f

=

m µ s N

m

µ s mg

=

m

= µ s g =

ma :

− Ff =

( a p )

2

<

m/s 2

m (a p )

2

= −11.52 m/s

3.43 m/s 2

Since the available friction force

= −11.52

2

11.52 m/s 2

µ s N is

less than the required

friction force F f for no slip, the package does slip.

(a p )2

<

11.52 m/s2, m (a p ) :

Σ F x =

(a p )

= −

2

µ k N

m

2

= −µ k g = −

Ff

= µ k N

− µ k N =

m (ap )

2

( 0.25)(9.81)

= −2.4525

m/s 2

+ ( −2.4525)( 0.41667) = 3.78 m/s 2 (v p )2 = ( v p )1 + ( a p )2 (t2 − t1 ) = 4.8

( x p )

2

=

( x p ) + ( v p ) ( t

=

3.6

1

+

1

2

−

2

t1 )

+

1

( a p ) (t 2 2

2

−

2

t 1 )

1

(4.8)(0.41667) + ( −2.4525)( 0.41667)

2

2

=

5.387 m

Position of package relative to the belt

( x p )2 − x2

=

5.387


−

4.6

=

0.787

x p/belt

=

0.787 m

�


PROBLEM 12.24

To transport a series of bundles of shingles A to a roof, a contractor uses a motor-driven lift consisting of a horizontal platform BC which rides on rails attached to the sides of a ladder. The lift starts from rest and initially moves with a constant acceleration a1 as shown. The lift then decelerates at a constant rate a2 and comes to rest at D, near the top of the ladder. Knowing that the coefficient of static friction between the bundle of shingles and the horizontal platform is 0.30, determine the largest allowable acceleration a1 and the largest allowable deceleration a2 if the bundle is not to slide on the platform.

SOLUTION

Acceleration

F1

a1 : Impending slip. Σ F y =

mA ay : N1 N1

W A

=

m A ( g

Σ F x =

F1

= µ s N

a1

=

WA

−

=

+

= µ s N 1 =

a1 sin65° )

+

or m Aa1 cos 65°

cos 65° − 0.30 sin 65°

mA a1 sin 65°

mAa1 sin65°

mAax : F1

0.30 g

=

0.30 N1

=

=

=

mAa1 cos 65°

0.30mA ( g

+

(1.990 )(9.81) a1 =

Deceleration

a2 : Impending slip. Σ F y =

may : N1

Σ F x =

F2 a2

= µ S N 2

=

−

max : F2

or

cos 65° + 0.30 sin 65°

=

= µ S N 2 =

WA

= −mA a2 sin 65°

N1

=

=

W A

−

19.53 m/s 2

19.53 m/s 2

65° �

0.30 N 2

mAa2 sin65°

mAa2 cos 65° =

0.30mA ( g

(0.432 )(9.81) =

−

a2 cos 65° )

4.24 m/s 2

a2 =


=

F2

m Aa2 cos 65°

0.30 g

a1 sin 65° )

4.24 m/s 2

65° �


PROBLEM 12.25

To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Knowing that the coefficients of friction between the bottom sheet of plywood and the bed are µ s = 0.40 and µ k = 0.30, determine (a) the smallest acceleration of the truck which will cause the stack of plywood to slide, ( b) the acceleration of the truck which causes corner A of the stack of plywood to reach the end of the bed in 0.4 s.

SOLUTION

Let

a P be the acceleration of the

truck, and

a P / T be the acceleration of the plywood relative to the truck.

(a) Find the value of

aT so

that the relative motion of the plywood with

respect to the truck is impending. aP F1

aT be the acceleration of the

plywood,

= µ s N 1 =

=

aT and

0.40 N1

ΣF y =

mP a y :

N1 ΣF x =

N1

−

WP cos 20°

mP ( g cos 20°

=

−

= − mP aT sin 20°

aT sin 20°)

max : F1 − WP sin 20° = mP aT cos 20°

F1 = mP ( g sin 20° + aT cos 20° ) mP ( g sin 20° aT

=

+

aT cos 20° )

(0.40 cos 20° − cos 20°

+

0.40 mP ( g cos 20°

=

sin 20°)

0.40 sin 20°

g

=

−

( 0.03145)( 9.81)

aP / T

=

=

( xP / T )o

2 xP / T 2

t

=

+

( vP / T ) t

( 2 )(1) 2 (0.4)

=

a P = aT + a P / T =

F y N 2

N 2

− =

=

+

1 2

aP / T t 2

12.5 m/s2 ,

( aT

)

→ +

=

0

+

0

+

a P / T =

(12.5 m/s2

=

0.309

0.309 m/s2

aT =

(b) xP / T

aT sin 20° )

1 2

s�

aP / T t 2

12.5 m/s2

20°

20° )

mP a y :

WP cos 20°

= −mP aT sin 20 °

mP ( g cos 20°

−

aT sin 20° ) continued




Σ F x = Σmax :

F2

−

F2

W P sin 20°

=

m P ( g sin 20° F2

For sliding with friction m P ( g sin 20° aT

+

=

=

mP aT cos 20°

=

aT cos 20°

−

+

aT cos 20°

= µ k N 2 =

aP / T )

=

+

mP aP / T

−

aP / T )

0.30 N 2

0.30m P ( g cos 20°

(0.30 cos 20° − sin 20°) g cos 20°

−

+

aT sin 20° )

a P / T

0.30 sin 20°

(−0.05767 )(9.81) + ( 0.9594)(12.5)

=

aT =


+

11.43 11.43 m/s 2

�


PROBLEM 12.26

The propellers of a ship of mass m can produce a propulsive force F0 ; they produce a force of the same magnitude but opposite direction when the engines are reversed. Knowing that the ship was proceeding forward at its maximum speed v0 when the engines were put into reverse, determine the distance the ship travels before coming to a stop. Assume that the frictional resistance of the water varies directly with the square of the velocity.

SOLUTION

At maximum speed a = 0.

F0 = kv02 = 0

k =

F 0 v02

When the propellers are reversed, F 0 is reversed.

Σ F = ma : − F0 − kv2 = ma x

− F0 − F0

v2 2 0

v

= ma

dx =

∫ x = −

mv02 F0

1 2

(

x

0

vdv a

dx = −

ln v02 + v 2

)

0

v

0

a−

=

mv02 F 0

= −

F 0 mv02

(v

2 0

+ v2 )

mv02 vdv

(

F0 v02 + v 2 0

vdv

∫ 0 v + v v

)

2 0

2

mv02  mv02 ln v02 − ln 2v02  = ln 2  2 F0  2F 0

( )

x = 0.347


m0v02 F 0

�


PROBLEM 12.27

A constant force P is applied to a piston and rod of total mass m to make them move in a cylinder filled with oil. As the piston moves, the oil is forced through orifices in the piston and exerts on the piston a force of magnitude kv in a direction opposite to the motion of the piston. Knowing that the piston starts from rest at t = 0 and x = 0, show that the equation relating x, v, and t , where x is the distance traveled by the piston and v is the speed of the piston, is linear in each of the variables.

SOLUTION

Σ F = ma : dv dt

∫

t 0

dt

=

=

a

P P

=

− kv = ma

− kv m

m dv m m ∫ P − kv = − k ln ( P − kv) = − k ln ( P − kv ) − ln P  v

v

0

0

t = P

−

m k

ln

− kv =

x

=

t

∫

v dt 0

=

Pt

=

Pt

k

k

− kv P

= P

Pt k

+

(e − m

−

kv


or

e−kt / m

m

x

P

m

or t

− 0

kt / m

P

ln

v

=

P

− kv = − kt m

m

P 1 − e −kt / m k

(

)

t

− e−kt / m      0 k m

− 1) =

k

Pt k

, which is linear.

−

P m

(1 − e−

kt / m

) �


PROBLEM 12.28

A 4-kg projectile is fired vertically with an initial velocity of 90 m/s, reaches a maximum height and falls to the ground. The aerodynamic drag 2 D has a magnitude D = 0.0024 v where D and v are expressed in newtons and m/s, respectively. Knowing that the direction of the drag is always opposite to the direction of the velocity, determine (a) the maximum height of the trajectory, (b) the speed of the projectile when it reaches the ground.

SOLUTION

Let y be the position coordinate of the projectile measured upward from the ground. The velocity and acceleration a taken to positive upward. D = kv 2.

ΣF y = ma :

(a) Upward motion.

− D − mg =

ma

 D kv  = −  g +  = −  g +   m  m       k  = −  g + kv  = −  v + mg    m  m k   2

a dv

v

v

∫

0

v0

v

dy

v dv mg 2

+

k

=−

2

+

dy

m

=−

h

k m

∫ dy 0

k

 ln  v +  2 2

mg  k

0

kh

=−   v m 0

mg 1 2

h

k

v dv mg 2

1

2

k mg

ln

v02

+

k

=−

1 2

ln

 kv  kh + 1 = −   m  mg  2 0

 (0.0024)(90)   kv  4 ln  1 = ln  + 1 = +  2k  mg  ( 2)( 0.0024)  (4 )(9.81)  = 335.36 m h = 335 m � 2 0

m

2

continued

2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.


261


ΣF = ma :

(b) Downward motion.

v

a

=

dv

=

dy

v dv mg

∫

v f

D

−g =

m kv 2 m

m

v dv mg

0

− mg =

= −k

− v2

k

D

=−

−

∫

 mg − v 2  f   − ln  k  = mg 2      k 

  

ln  1 −

=±

v f

= =

g

dy

kh m

kh m

 2  = − kh  mg m 

kv2f

1−

v f

−

h

v

1

m

dy

 mg − v 2  f = ln    0 2  k

1

kv 2

= − k  mg − v2   m k 

g

k 0 m

ma

kv 2f mg

mg

= e− 2 kh/m

1 − e − ( k

2

kh/m

)

( 4 )(9.81)  − 2 0.0024)(335.36 )/4  1 − e ( )(  0.0024  73.6 m/s

v f

=

73.6 m/s �



262

PROBLEM 12.29

A spring AB AB of constant k is attached to a support A A and to a collar of mass m. The unstretched length of the spring is �. Knowing that the collar is released from rest at x = x0 and neglecting friction between the collar and the horizontal rod, determine the magnitude of the velocity of the collar as it passes through point C .

SOLUTION

Choose the origin at point C , and let x let x be be positive to the right. Then Then x is a x is position coordinate of the slider B slider B,, and x0 is its initial value. Let L Let L be be the stretched length of the spring. Then, from the right triangle L

The elongation of the spring is e exerted by the spring is F

s

By geometry,

cosθ

ke ke

=

(

�

ΣF =

ma :

+

−�

�

2

v

2

v

1 2

�

v2

k = −  0 − �2 − m

v2

=

(2 m

=

k 

m 

v dv

+

x2

�

−

)

2

1 2

+ x x02

2

x �

+ x 2

2

ma

=

ma

�x � 0

= ∫

x

0

2

+ x

2

   

a dx

 k  1  dx = − m  2 x − � 

+�

�

+

x02

+

x02

+

− 2�

�

2

+ x02 ) − 2�

�

x02

)

=

cosθ

s

2

2

�

(�

v

0

�x

0

k

2

−F

x

x2

0

x

0

�

 = − k  x − m

 = − ∫  x − m  k

�, and the magnitude of the force

−

+ x 2

2

∫ 2

(

x2

+

x

=

a

1

k

2

L

=

=

x

−k

�

=

2

2

2

+

x02

+

�

�

2

+

2

x

  

0

x

0

  

) 2

 

answer: v

=

k

( m

�

2

+

x02

−�

)�

2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission permission..

PROPRIETARY MATERIAL. MATERIAL. ©

263

PROBLEM 12.30

The system of three identical 10-kg blocks is supported in a vertical plane and is initially at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys, determine ( a) the change in position of block A after 0.5 s, (b ( b) the tension in the cable.

SOLUTION

B,, and C respectively Let y A, y B, and y Let y and yC be the position coordinates of blocks A blocks A,, B respectively measured downward from the upper support. Then the corresponding velocities and accelerations are positive downward. Constraint of cable: Differentiating twice,

− yB + y A + 2 yB + yC = constant a A − aB + a A + 2 aB + aC = 0 2a A + aB + aC = 0

y A

(1)

Draw free body diagrams of each of the blocks.

Block A. A.

Σ F = ma :

m A g

− 2T = mAaA a A

B. Block B.

Σ F = ma :

m B g

Σ F = ma :

mC g

2T

(2)

m A

− T = mB aB T

= g −

a B Block C Block C .

= g −

m B

(3)

(4)

− T = mC aC aC

= g −

T mC

Substitute (2), (3) and (4) into (1).

  T   T  +  g −  +  g −  = 0  m   m   m   4 1 1  + + 4 g −   T = 0 m m  m  4 1 1  (4)(9.81) −  + +  T = 0 T = 65.4 N  10 10 10   − 

2  g

2T

A

B

A

C

B


C



Substitute into (2),

a A

=

9.81 −

(a) Change in position.

y A

−

( y )0

y A

−

( y )0 A

=

1 2

A

( − 3.27 ) ( 0.5)

2

(2)( 65.4) 10 =

1 2

= − 3.27

m/s 2

aAt 2

= − 0.409

m ∆ y =

0.40 0. 409 9m �

T = 65. 5.4 4 N �

(b) Te Tens nsio ion n in in the the ca cabl blee.



PROBLEM 12.31

The coefficients of friction between block B and block A are and µ k

=

µ s =

0.12

0.10 and the coefficients of friction between block A and the

incline are µ s = 0.24 and µ k = 0.20. The masses of block A and block B are 10 kg and 5 kg, respectively. Knowing that the system is released from rest in the position shown, determine ( a) the acceleration of A, (b) the velocity of B relative to A at t = 0.5 s.

SOLUTION

W A = mA g = 98.1 N

W B = mB g = 49.05 N

Assume that block B slides downward relative to block A. Then the friction force F 1 is directed as shown. Its magnitude is F1 ΣF y =

ΣF x = =

0.10 N1.

° = 49.05cos30° = 42.48 N. 0: N1 − WB cos30° = 0, N1 = WB cos 30 F 1

a B

= µ k N 1 =

1 m B

=

( 0.10)( 42.48)

m B aB : WB sin 30°

=

−

4.248 N.

F1

1

(WB sin 30° − F 1 ) = ( 49.05sin 30° − 5

=

mB aB

4.248) = 4.055 m/s2

Assume that block A slides downward relative to the fixed plane. The friction force F 2 is directed as shown. Its magnitude is F2

F y

=

0: N 2

−

= µ k N 2 =

0.20 N2 .

N1 − WA cos 30° = 0, N 2

F 2

=

ΣF x =

a A

=

=

a B / A

=

( 0.20 )(127.44)

=

25.49 N

m AaA : WA sin 30°

1 m A 1 10 aB

42.48 + 98.1cos 30° = 127.44 N.

=

−

F2

+

F1

(WA sin30° − F2 + F1 )

=

mA aA

(98.1sin 30° − 25.49 + 4.248 ) = 2.781 m/s 2 −

aA

=

4.055

−

2.781

=

1.274 m/s2

Since both a B / A and a A are positive, the directions of relative motion are as assumed above. A =

2.78 m/s 2

30° �

B / A =

0.637 m/s

30° �

(a) Acceleration of block A.

a

(b) Velocity of B relative to A at t = 0.5 s. v B / A

=

aB / A t = (1.274)( 0.5)


v


PROBLEM 12.32

The weights of blocks A, B, and C are W A = W C = 89 N, and W B = 44.5 N. Knowing that P = 222.4 N and neglecting the masses of the pulleys and the effect of friction, determine ( a) the acceleration of each block, ( b) the tension in the cable.

SOLUTION

Let the positive direction for position coordinates, velocities, and accelerations be to the right. Let the origin lie at the fixed anchor. 3( xC


ΣF x =

−

xA )

+

( xC

−

xB )

4aC

−

2aB

( −xB )

=

constant

3a A

=

0

−

(1)

max :

Block A:

3T

=

m Aa A

or

aA

=

Block B:

2T

=

m B aB

or

aB

=

− 4T =

mC aC

or

aC

=

Block C :

+

P

3T m A

2T m B P

3T

=

=

89

g

2T 44.5

− 4T

(2)

g

(3)

− 4T

P

=

20

89

g

(4)

Substituting (2), (3), and (4) into (1),

 P − 4T  − 2  2T  − 3  3T  = 0  89   44.5   89 

4

 16 + 4 +  89 44.5  T

 T =  89  9

4P 89

= (0.12121) P = ( 0.12121)(222.4) =

(3)(27)( 9.81)

(a) From (2),

a A

=

From (3),

a B

=

From (4),

aC

 222 .4 − ( 4 )( 27) ( 9.81) =  =

89

( 2)( 27) ( 9.81) 44.5

89

27 N

=

8.9 m/s 2

aA

= 8.9 m/s

=

11.9 m/s2

a B

= 11.9 m/s

2

12.6 m/s

aC

=

2

�

2

�

12.6 m/s2

�

T = 27 N

(b) As determined above, PROPRIETARY MATERIAL. ©

�


PROBLEM 12.33

The coefficients of friction between the three blocks and the horizontal surfaces are µ s s = 0.25 and µ k = 0.20. The weights of the blocks are W W 20N,lb,and andW W 10 lb. 89 44.5 N. Knowing that the blocks are initially W C C = B B = A =W A to the the right right through through0.73 2.4 ft at rest and that C moves to m in in 0.4 0.4 s, s, determine ((a) ( b) the tension in the cable, ( c) a) the acceleration of each block, (b) c) the P . Neglect axle friction and the masses of the pulleys. force P =

=

=

=

=

SOLUTION

m A = mC =

89 9.81

=

44.5

9.07 kg , mB =

9.81

4.54 kg

=

Let the positive direction for position coordinates, velocities, and accelerations be to the right. Let the origin lie at the fixed anchor. 3( xC


−

4aC

Block A: Σ F y = 0:

xA ) −

+

( xC

2aB

xB )

−

+

( − xB )

3a A

=

0

µk N A

=

µ κ W A

−

constant

=

(1)

N A − W A = 0 N A = WA, FA =

Σ F x = mAax : 3T − FA = mAa A 3T − µ kWA

a A = Block B: N B = WB ,

m A

FB =

=

3T

mA

− 0.20 g

(2)

µ kW B

Σ F x = mB aB : 2T − FB = mB aB a B =

2T − µ kWB

m B

Block C : NC = WC , FC = µ kW C

=

2T

mB

− 0.20 g

(3)

Σ F x = mC a A : P − 4T − FC = mC aC Kinematics: xC = ( xC ) + ( vC ) + 0 0

aC =

1 2

a Ct 2 = 0 +

2  xC − ( xC )0 



2

t


 = ( 2)( .72) = 2

(0.4 )

1 2

aC t 2

9 m/s

2

(4)

(5)



Substitute (2), (3) and (5) into (1).

 (4)( 9 ) − (2) 

2T 4.54

− 0.20 g  − (3)   

3T 9.07

2T = (4)( 9 ) − ( 2)  − (0.2 )(9.81) − (3)   4.54   = 36 − 1.873 T + 9.81 = 0

− (0.2 )(9.81)  T = 24.5 N

3T 9.07

= 4T + FC + mC aC = ( 4)( 24.5) + (0.20)( 89) + ( 9.07)( 9) = 197.4 N

From (4),

P

From (2),

a A

=

From (3),

a B

=

(a)

− 0.20 g  

(3)( 24.5 )

− (0.20)(9.81) =

6.14 m/s

− (0.20)(9.81) =

8.8 m/s

9.07

(2) ( 24.5 ) 4.54

Acceleration vectors.

2

2

a A

=

6.14 m/s

a B

=

8.8 m/s

aC

=

9 m/s

2

2

2

t t t

Since a A, aB , and aC are to the right, the friction forces F A , FB , and F C are to the left as assumed. (b)

Tension in the cable.

(c)

Force P.


T P

=

=

24.5 N t

197.4 N

t


PROBLEM 12.34

A 25-kg block A rests on an inclined surface, and a 15-kg counterweight B is attached to a cable as shown. Neglecting friction, determine the acceleration of A and the tension in the cable immediately after the system is released from rest.

SOLUTION

Let the positive direction of x and y be those shown in the sketch, and let the origin lie at the cable anchor. Constraint of cable: x A

+

=

yB / A

+

constant or aA

aB / A

=

0, where

the positive directions of a A and aB / A are respectively the x and the y directions. Then First note that Block B:

a B / A

(1)

a B = a A + a B / A =

( aA

20° )

mB ( aB ) x : mB g sin 20°

ΣF x =

( a B / A

+

+

N AB

=

mB g sin20°

15 a A

+

N AB

=

50.328 −

T

=

+

T

=

mB g cos20°

15 a B / A

+

T

=

138.276 N AB

20° )

mB aA

(2)

m B aB / A

+

=

N AB

m B a A

mA aA : mA g sin 20°

ΣF x =

−

mB ( aB ) y : mB g cos 20°

ΣF y =

Block A:

= −a A

mB aB / A

(3) +

T

m Aa A

−

N AB

+

T

=

m A g sin20°

25 a B / A

−

N AB

+

T

=

83.880

=

mA aA

(4)

Eliminate a B / A using Eq. (1), then add Eq. (4) to Eq. (2) and subtract Eq. (3). 55 a A

= −4.068

aA

or

From Eq. (1),

a B / A

From Eq. (3),

T

=

=

= −0.0740

aA =

0.0740 m/s2

20° �

0.0740 m/s 2

137.2 N


m/s2 ,

T = 137.2 N �


PROBLEM 12.35

A 250-kg crate B is suspended from a cable attached to a 20-kg trolley A which rides on an inclined I-beam as shown. Knowing that at the instant shown the trolley has an acceleration of 0.4 m/s 2 up to the right, determine (a) the acceleration of B relative to A, (b) the tension in cable CD.

SOLUTION

Motion of B relative to A. Particle B is constrained to move on a circular path with its center at point A. ( a B / A )t is the component of a B / A lying along the circle, say to the left in the diagram and toward point A. Initially,

=

(a)

a B

Crate B:

aA

+

aB / A

a B / A

= (aA

(a B / A )n is directed

0, since the system starts from r est.

=

25° ) + ( a B / A

)

Σ F x = Σmax : 0 = mB aB / A − mB aA cos 25° a B / A = a A cos 25° = 0.4cos 25° = 0.363 a B / A

= 0.363 m/s 2

t

F y = mB ( aB ) y : TAB − mB g = mB aA sin 25° T AB = mB ( g + aA sin 25° )

= 250 (9.81 + 0.4sin 25° ) = 2495 N (b) Trolley A:

Σ F = m Aa A : TCD − (T AB + mA g ) sin 25° = mAaA

TCD = (TAB + m A g ) sin 25° + mAaA

=  2495 + ( 20)(9.81)  sin 25° + ( 20)( 0.4) T CD = 1145 N t



PROBLEM 12.36

A 2-kg ball revolves in a horizontal circle as shown at a constant speed of 1.5 m/s. Knowing that L = 600 mm, determine ( a) the angle θ that the cord forms with the vertical, ( b) the tension in the cord.

SOLUTION

The ball moves at constant speed on a circle of radius = L sin θ

ρ

a

Acceleration (toward center of circle).

+

Σ F y =

ma y :

T cosθ − W = 0

Σ F y =

max :

T sin θ = ma W cosθ

(a)

tan θ sin θ = θ

(b)

mv2

T =

=

WL

=

v2 gL

sin θ

=

mv2

=

ρ

cosθ

=

mg cosθ

v2 ρ

T =

W cosθ

mv2 L sin θ

2

=

(1.5) = 0.38226 (9.81)( 0.6 )

34.21°

W

=

θ =

( 2)(9.81) cos34.21°


T

=

=

34.2° t

23.7 N t


PROBLEM 12.37

A single wire ACB of length 2 m passes through a ring at C that is attached to a sphere which revolves at a constant speed v in the horizontal circle shown. Knowing that θ 1 = 60° and θ 2 = 30° and that the tension is the same in both portions of the wire, determine the speed v.

SOLUTION

Let ρ be the radius of the horizontal circle. The length of the wire is L

=

ρ sin θ1

ρ . sin θ 2

+

Σ F y =

0: T cosθ1

Σ F x =

max : T sin θ1

cosθ1 Lg

T cosθ 2

mg cosθ1 + cosθ 2

mg (sin θ1

=

+

=

T

v2

Solving for ρ ,

+

sin θ1 sin θ 2 cosθ1 + cosθ 2


+

sin θ 2 )

cosθ 2 =

+

=

( 2)(9.81)

−

ρ

=

L sinθ1 sinθ 2 sin θ1

+

sin θ 2

mg = 0

T sin θ 2

=

mv2 (sin θ1

man +

mv2

=

ρ

sin θ 2 )

L sinθ1 sinθ 2 sin 60° sin 30° cos 60° + cos30°

=

6.2193 m 2/s 2

v

=

2.49 m/s t


PROBLEM 12.38

Two wires AC and BC are tied to a 6.8 kg sphere which revolves at a constant speed v in the horizontal circle shown. Knowing that θ 1 = 50° and θ 2 = 25° and that d = 1.2 m , determine the range of values of v for which both wires are taut.

SOLUTION

θ3 �1

sin θ 2

ρ

θ1

=

=

θ 2

=

50°

d

=

�1 sinθ 1

=

=

ΣF x =

max : TAC sin θ 2

T AC cos θ 2

=

g ρ tan θ 2

v

=

2.05 m/s

Case 2: T AC

=

0. T BC cosθ 1

v

=

=

3.28 m/s

=

sin25° 0: TAC cosθ 2

g ρ tan θ 1

�1

(1.2)(sin 25°)( sin 50°)

v2

=

=

−

=

25° d sin θ 2

sin θ 3

sinθ 3

ΣF y =

0.

25°

d sin θ 2 sinθ 1

W

=

=

+

−

W

TBC sin θ 1

=

or T AC

0

W tan θ 2

( 9.81)( 0.92 ) tan 25°

W

=

0

T BC sin θ1

=

W tan θ 1

−

0.92 m

TBC cosθ 1

+

=

T AC sin θ 2

v2

−

or

sin θ 3

=

Case 1: T BC

−

(9.81)( 0.92 ) tan 50°

=

=

Wv 2 g ρ

(1) (2)

cosθ 2

Wv 2 g ρ

4.2 m2/s2

=

W

cosθ 1

Wv 2 g ρ

10.75 m2/s 2 2.05 m/s ≤ v


0

W

=

T BC

or

=

=

=

≤

3.28 m/s �


PROBLEM 12.39

During a hammer thrower’s practice swings, the 7.3 kg head A of the hammer revolves at a constant speed v in a horizontal circle as shown. If ρ = . 9 m and θ = 60° , determine (a) the tension in wire BC , (b) the speed of the hammer’s head.

SOLUTION

(a)

Σ F y =

0: T sin θ T

(b)

Σ F x =

v2

=

=

=

W

=

0

W sin θ

=

71.6 N sin 60°

−

man : T cos θ ρT cosθ m ρ g tan θ

=

=

=

m

or T

82.7 N

=

2.26 m/s t

v2 ρ

ρW cosθ

m sin θ

(0.9)(9.81) tan 60°

=

5.097 m2/s2 v


t

=


PROBLEM 12.40

A 1-kg sphere is at rest relative to a parabolic dish which rotates at a constant rate about a vertical axis. Neglecting friction and knowing that r = 1 m, determine (a) the speed v of the sphere, (b) the magnitude of the normal force exerted by the sphere on the inclined surface of the dish.

SOLUTION

y

=

r2 2

dy

,

dx

=

Σ F y =

N

=

Σ F x =

r = 1

(a) v 2

=

(b) N

=

=

tan θ

0: N cosθ

−

cos 45°

=


θ =

45°

mg = 0

cosθ max : N sin θ

=

man

v2 r

gr tan θ

(9.81)(1)(1.0000) = 9.81 m 2/s 2 mg

or

mg

mg tan θ = m v2

=

(1)( 9.81) cos 45°

v

=

3.13 m/s t

N

=

13.87 N t


PROBLEM 12.41

A 1-kg collar C slides without friction along the rod OA and is attached to rod BC by a frictionless pin . The rods rotate in the horizontal plane. At the instant shown BC is rotating counterclockwise and the speed of C is 1 m/s, increasing at a rate of 1.3 m/s Determine at this instant, ( a) the tension in rod BC , (b) the force exerted by the collar on rod OA.

SOLUTION

Geometry � OC

2

= � OB

=

2

+ � BC

2

( 0.3)

+

2

−

2

(0.6)

° 2� OB � OC cos30

−

( 2)( 0.3)( 0.6) cos 30°

=

0.13823 m2

� OC = 0.37179 m � OC

sin 30° sin β

=

β

=

� OB sin30°

N

=

ΣFn = =

0.37179

=

0.40345

1.3 m/s 2

=

v2

=

=

ρ

2

2 vC

� BC

=

(1)

0.6

=

1.667 m/s2

1 kg ΣFt =

(a) T

( 0.3) sin30°

23.79°

an =

sin β

=

� OC

Acceleration components: at

Mass m

� OB

=

1.667

+

mat :

N cos β

1.3 cos23.79° man : T

=

−

=

(1)(1.3)

N sin β

=

(b) Force exerted by rod on collar is 1.421 N


1.3

1.421 N

1.421 sin 23.79°

Force exerted by collar on rod:

=

(1)(1.667 )

=

1.667

T

=

2.24 N �

(30° + β )

=

53.8°

1.421 N

53.8° �


PROBLEM 12.42

The 0.5-kg flyballs of a centrifugal governor revolve at a constant speed v in the horizontal circle of 150-mm radius shown. Neglecting the mass of links AB, BC , AD, and DE and requiring that the links support only tensile forces, determine the range of the allowable values of v so that the magnitudes of the forces in the links do not exceed 75 N.

SOLUTION

W

=

mg = (0.5 )(9.81) ρ =

Σ F y =

150 mm

0 : TDA cos 20°

0.93969 T DA

−

−

=

4.905 N

=

0.150 m

TDE cos 30°

0.86603 T DE

=

−

W

=

0

4.905

T DA

=

0.92160 T DE + 5.2198

(1a)

T DE

=

1.08506 T DA

(1b)

Σ F x =

v2

=

man

=

mv2 ρ

0.10261 T DA

−

5.6638

: TDA sin 20° +

+

TDE sin 30°

=

0.5 0.150

v2

0.15 T DE

(2)

Try T DA

=

75 N. By Eq. (1b), T DE

=

75.72 N

>

75 N (unacceptable)

Try T DE

=

75 N. By Eq. (1a), T DA

=

74.34 N

<

75 N (acceptable)

By Eq. (2), v 2

=

(0.10261)( 74.34) + ( 0.15)( 75)

v

=

4.34 m

Try T DA

=

0. By Eq. (1 b ), T DE

= −5.6638

Try T DE

=

0. By Eq. (1 a ), T DA

=

5.2198

(acceptable )

=

(0.10261)(5.2198) + ( 0.15)( 0)

v

=

0.732 m/s

≤

T BA, TBC , TDA, TDE ≤ 75 N,


18.877 m2 / s 2

(unacceptable )

By Eq. (2), v 2

For 0

=

=

0.5356 m 2 / s 2

0.732 m/s

≤

v

≤

4.34 m/s t


PROBLEM 12.43

As part of an outdoor display, a 5-kg model C of the earth is attached to wires AC and BC and revolves at a constant speed v in the horizontal circle shown. Determine the range of the allowable values of v if both wires are to remain taut and if the tension in either of the wires is not to exceed 116 N.

SOLUTION

W

=

ρ = Σ F y =

mg = (5)( 9.81)

49.05 N

=

0.9 m 0 : TCA cos 40°

0.76604 TCA

−

−

0.96593 TCB

TCB cos15° =

W

=

−

W

=

0

49.05

TCB

=

0.79307 T CA

−

50.780

(1a)

TCA

=

1.26093 T CB

+

64.030

(1b)

Σ F x =

v2

=

max : TCA sin 40° ρ

m

TCB sin15°

=

0.9 (TCA sin 40° 5

+

=

0.115702 TCA

0.046587 T CB

=

116 N. By (1b), T CA

Try TCA

=

116 N. By (1a), T CB

+ =

=

41.216 N (acceptable)

=

(0.115702)(116) + ( 0.046587)( 41.216)

v

=

3.92 m/s

0. By (1a ), T CB

= −50.78 N

Try TCB

=

0. By (1b ), T CA

=

By (2),

v 2

64.03 N

( 0.115702)(64.030 ) + 0

v = 2.72 m/s TCA , T CB ≤ 116 N,


ρ

(2)

v2

=

=

210.3 N (unacceptable)

=

≤

man

T CB sin15° )

Try TCA

For 0

=

(TCA sin 40° + T CB sin15°)

Try TCB

By (2),

+

mv2

=

15.34 m2 / s2

(unacceptable ) (acceptable )

=

7.408 m 2 / s 2 2.72 m/s

≤

v

≤

3.92 m/s t


PROBLEM 12.44

A small sphere of weight W is held as shown by two wires AB and CD. If wire AB is cut, determine the tension in the other wire ( a) before AB is cut, (b) immediately after AB has been cut.

SOLUTION

a

(a) Before wire AB is cut. Σ F x =

0 :

− T AB

Σ F = y

0 :

T AB sin 50°

=

0

cos50° + T CD cos 70° = 0 +

(1)

TCD sin 70° − W = 0

(2)

Solving (1) and (2) simultaneously, T AB

=

0.395 W

(b) Immediately after wire AB is cut. an

=

v2

=

T AB

=

0, v

=

0.742 W t

TCD

=

0.940 W t

0

0

ρ

+ Fn

=

TCD

=

man

=

TCD

− W cos 20° =

TCD

= W cos20°


0: 0


PROBLEM 12.45

A series of small packages being moved by a thin conveyor belt that passes over a 300-mm-radius idler pulley. The belt starts from rest at time 2 t = 0 and its speed increases at a constant rate of 150 mm/s . Knowing that the coefficient of static friction between the packages and the belt is 0.75, determine the time at which the first package slips.

SOLUTION

ΣF y = ma y : N

− mg = − man = − N

= mg −

mv 2 ρ

mv 2 ρ

ΣF x = max :

At onset of slipping,

Ft

= mat

Ft

= µ s N

mat

vs2

 = ρ  g − 

vs

= 1.6979 m/s

Time at slipping.

at 

= (0.300)  9.81 −   µ s 

t s

=

vs at

0.150 

 = µ s  mg − 

 = 2.883 m 0.75 

=

1.6979 0.150


mvs2  ρ

  

2

/s 2

t s

= 11.32 s �


PROBLEM 12.46

An airline pilot climbs to a new flight level along the path shown. Knowing that the speed of the airplane decreases at a constant rate from 165 m/s at point A to 146 m/s at point C , determine the magnitude of the abrupt change in the force exerted on a 90.7 kg passenger as the airplane passes point B.

SOLUTION

8°

Angle change over arc AB. θ =

180°

π = 0.13963 rad

Length of arc: s AB = ρθ = ( 6.4 ) (1000) ( 0.13963) = 893.6 m s BC = (0.5)(1000) = 500 m, s AC = 893.6 + 500 = 1393.6 146

∫ 165

v dv =

1393.6

∫ 0

at ds

1462

or

−

2

1652 2

= at (1393.6 )

at = −2.12 m/s2 v B

∫ 165 v dv =

893.6

∫ 0

at ds

v B2

or

2

v B2 = 23436.136 m2/s2

−

1652 2

= (− 2.12 ) ( 893.6)

vB = 153 m/s

Mass of passenger: m = 90.7 kg v = 153 m/s,

Just before point B.

an =

v2

ρ

ρ = (6.4)(1000) = 6400 m 2

=

(153 ) 6400

= 3.65 m/s2

ΣF y = N1 − W = −m ( an )1 : N1 = 889.7 − (90.7 ) (3.65) = 558.6 N ΣF x = Ft = mat : ( Ft )1 = (90.7 ) ( −2.12 ) = −192.3 N Just after point B.

v = 153 m/s,

ρ = ∞,

ΣF y = 0 : N2 − W = 0

an = 0

N 2 = W = 889.7 N

ΣF x = mat : Ft = mat = ( 90.7 ) (− 2.12 ) = − 192.3 N F t does not change.

N increases by 331.1 N

magnitude of change of force = 331 N �



PROBLEM 12.47

An airline pilot climbs to a new flight level along the path shown. The motion of the airplane between A and B is defined by the relation s 3t (180 t ), where s is the arc length in meter , t is the time in seconds, and t 0 when the airplane is at point A. Determine the force exerted by his seat on a 165- N passenger (a) just after the airplane passes point A, (b) just before the airplane reaches point B. =

−

=

SOLUTION

s

=

at

(

3 180t

−

dv =

dt

=

−

)

t 2 m,

v

=

6 m/s 2

ρ

=

ds

v

s

v

0

8°π 180°

∫ A B v dv = ∫ AB a 2

−

vA2 2

−

2t ) m/s,

6400 m

Length of arc AB. s AB = ρθ AB = ( 6400)

v B2

3 (180

=

dt

t

=

893.6 m

ds

= 2at s AB or vB2 = v A2 + 2at sAB = 5402 + (2 )( −6)( 893.6) = 280876.8,

v B = 530 m/s

For passenger, W = 165 N, m =

W g

=

165

= 16.8 kg

9.81

mv 2

Σ F = man : W cosθ − N = N = W cosθ −

mv2

ρ

ρ

(1)

Σ F = mat : P − W sin θ = mat P = W sin θ + mat (a) Just after point A, t = 0,

v = 540 m/s,

(2)

θ = 8 ° 2

(16.8 )(540)

= − 602.05 N

From Eq. (1),

N = 165cos8° −

From Eq. (2),

P = 165sin 8° + (16.8)( −6) = −77.8 N

F =

N 2 + P 2 = 607 N , tan β =

β − 8° = 74.6°

6400

602.05 77.8

= 7.73843, β = 82.6 ° F

= 607 N

74.6° t

continued




(b) At point B. θ

=

0, v

= 530 m/s

From Eq. (1),

N = 165 −

From Eq. (2),

P = (16.8

F =

2

(572.36)

(

+ 100.8

2

)

(16.8)(530 )

2

= − 572.36 N

6400

)(−6 ) = −100.8 N

= 581.2 N,

tan β

=

572.36 100.8

= 5.678,

F = 581.2 N


β = 80° 80°

t


PROBLEM 12.48

During a high-speed chase, an 1100-kg sports car traveling at a speed of 160 km/h just loses contact with the road as it reaches the crest A of a hill. (a) Determine the radius of curvature ρ of the vertical profile of the road at A. (b) Using the value of ρ found in part a, determine the force exerted on a 70-kg driver by the seat of his 1400-kg car as the car, traveling at a constant speed of 80 km/h, passes through A.

SOLUTION

(a) v

= 160 km/h =

44.44 m/s

Wheels do not touch the road.

Σ F = −ma y

ρ

(b) v

=

80 km/h

=

n

=

− mg = −mv

:

/ ρ

2

v2 g

=

(44.44) 9.81

=

201.4

ρ

=

201 m �

N

=

515 N t

22.22 m/s m

=

70 kg for passenger

Σ F = −ma y

N

2

n

− mg = −

N

=

:

mv 2 ρ

  

m  g −

v 2  ρ

  

 = (70)  9.81 − 


22.22 201.4

2

   


PROBLEM 12.49

0.2-kg A small 0.2 kg sphere B is given a downward velocity v 0 and swings freely in the vertical plane, first about O and then about the peg A after the cord comes in contact with the peg. Determine the largest allowable velocity v 0 if the tension in the cord is not to exceed 10 N.

SOLUTION

m = 0.2 kg,

W = mg = (0.2)(9.81) = 1.962 N

Σ Ft = mat : W cosθ = mat at =

W cosθ = g cosθ m

θ v s ∫ v0 v dv = ∫ at ds = ∫ at rdθ 0

1 2

v2 −

1 2

v02 =

θ

∫

0

0

g cosθ rdθ = gr sin θ

v 2 = v02 + 2 gr sin θ , where r = 0.6 m for 0 ≤ θ ≤ 90 ° v = vmax when the cord touches the peg or θ = 90°. 2 vmax = v02 + 2 gr

(1)

When the cord touches the peg, the radius of curvature of the path becomes ρ = 0.3 m.

Σ F y = man : Tmax − W = 2

vmax =

mvmax2 ρ

ρ (Tmax − W ) m

(2)

2

Eliminating vmax from from equations Equations(1) (1)and and(2), (2), v02 =

ρ (Tmax − W ) m

− 2 gr =

= 0.285 m2 /s2

(0.3)(10 − 1.962) 0.2

− ( 2)(9.81)( 6.0) v0


= 0.534 m/s t


PROBLEM 12.50

A 1 N block B fits inside a small cavity cut in arm OA, which rotates in the vertical plane at a constant rate such that v = 2.7 m/s. Knowing that the spring exerts on block B a force of magnitude P = 0.6 N and neglecting the effect of friction, determine the range of values of θ for which block B is in contact with the face of the cavity closest to the axis of r otation.

SOLUTION

Mass of block B. Acceleration of block B.

m B =

1N 9.81 m / s an =

v2 ρ

at = 0

= .102 kg.

2

=

2.7 m / s .6 m

= 1215 . m / s2

v = constant.

since

∑ F = ma :

Q − P − W sin � = − man Q = P + W sin � − man

But,

Q ≥ 0.

P + W sin θ − man > 0 sin θ ≥

sin θ ≥

man W

−

P W

=

. ) (.102)(1215 1

−

.6 1

.6393 39.7°


≤ θ ≤ 140.3° �


PROBLEM 12.51

A 240-N 240 N pilot flies a jet trainer in a half vertical loop of 1097-m 1097 m radius so that the speed of the trainer decreases at a constant rate. Knowing that the pilot’s apparent weights at points A and C are 760 N and 160 N, respectively, determine the force exerted on her by the seat of the trainer when the trainer is at point B.

SOLUTION

At position A, the vertical component of apparent weight is shown as N A. Σ F =

an

=

man : N A

−W =

W g

an

760 − 240 N A − W g = (9.81) = 21.255 m / s2 W 240

22 va r ann = . = (1097 m)(21255 v A== ρ ) = 23316.735 m 2 / s 2 A

At position C , the vertical component of apparent weight is shown as N C . Σ F =

an v C2

=

man : NC

+

W

=

W g

an

160 + 240 NC + W g = (9.81) = 16.35 m / s 2 W 240 . ) = 1793595 . m 2 / s2 (1097)(1635

= ρ an =

Length of arc ABC : s AC Calculate at , using vC2

(

= πρ = π 1097

−

v A2

at

=

=

vC2

2at sAC −

vA2

=

2 s AC

) = 3446.3 m

17935.95 − 23316.735

(2)(3446.3)

2

= − .78 m/s

At position B, s AB v B2

=

v 2A

+

2at s AB

=

=

π

2

ρ =

π

2

(1097)

= 1723.2 m

23316.735 +(2 ) (.78 )(17232 . ) = 26004 .9 m / s


2

2



Effective forces at B: man

=

mat

=

W g

Σ F =

P

−

W

=

at

240 26004.9

=

9.81 1097 240

=

9.81

= 580 N

(.78) = 19 N

mat :

mat or P Σ F =

W v B2 g ρ

=

W

+

mat

man : N B

=

=

240 − 19 = 221 N

man

= 580 N

Force exerted by seat: F

=

N B

2

P 2

=

tan β

=

+

5 80 2 + 221 = 620.6 N 221 580

β

= 20.8°

F = 620.6 N


20.8°

t


PROBLEM 12.52

A car is traveling on a banked road at a constant speed v. Determine the range of values of v for which the car does not skid. Express your answer in terms of the radius r of the curve, the banking angle θ , and the angle of static friction φ s between the tires and the pavement.

SOLUTION

The road reaction consists of normal component N and friction component F. The resultant R makes angle φ s with the normal. Case 1: v

=

vmax 0: R cos (θ

Σ F y =

R

n

vmax 2 r

cos (θ

vmax Case 2: v

=

−

mg

φ ) s

=

−

mg

φ )

=

ma

)

v

≤

0

=

φ )

+

s

ma : R sin (θ

g tan (θ

+

gr tan (θ

=

+

n

mg tan (θ

=

=

φ s )

mg

=

Σ F x =

ma

+

ma

n

φ )

+

s

φ ) s

φ )

+

s

vmin 0: R cos (θ

Σ F y =

R

n

vmin 2 r vmin

=

=

=

cos (θ

−

0

=

φ ) s

ma : R sin (θ

−

n

mg tan (θ g tan (θ

−

gr tan (θ

s

n

φ )

−

s

φ ) s

−

gr tan (θ


φ s )

mg

=

Σ F x =

ma

−

φ ) s

−

φ

s

≤

gr tan (θ

+

φ

s

) t


PROBLEM 12.53

A curve in a speed track has a radius of 200 m and a rated speed of 180 km/h. (See Sample Prob. 12.6 for the definition of rated speed.) Knowing that a racing car starts skidding on the curve when traveling at a speed of 320 km/h, determine (a) the banking angle θ , (b) the coefficient of static friction between the tires and the track under the prevailing conditions, (c) the minimum speed at which the same car could negotiate that curve.

SOLUTION

Weight. Acceleration.

W = mg v2

a=

ρ

∑ F x = max :

F + W sin θ = ma cosθ mv2

F = ∑ F y = ma y :

ρ

cosθ − mg sin θ

(1)

N − W cosθ = ma sin θ N =

mv 2 ρ

sin θ + mg cosθ

(2)

(a) Banking angle. Rated speed v = 180 km/h = 50 m/s. F = 0 at rated speed. 0= tan θ =

mv2

cosθ − mg sin θ

ρ v2 ρ g

=

(50)2 (200)(9.81)

= 1.2742

θ = 51.9° �

θ = 51.875°

(b) Slipping outward. F = µ N

v = 320 km/h = 88.889 m/s µ =

µ =

F N

=

v 2 cosθ − ρ g sin θ v 2 sin θ + ρ g cosθ

(88.889)2 cos51.875° − (200)(9.81)sin51.875°

(88.889) 2 sin51.875° + (200)(9.81)cos51.875° = 0.44899

µ = 0.449 t

continued PROPRIETARY MATERIAL. ©



F

(c) Minimum speed. −

µ =

v2

v 2 cosθ 2

v sin θ

−

ρ g sin θ

+

ρ g cosθ

ρ g (sin θ

=

=

cosθ

+

v

=

µ cosθ )

µ sin θ

(200)(9.81)(sin51.875°

−

0.44899cos51.875°)

cos51.875° + 0.44899sin51.875°

= 1029.87 m

−

= − µ N

2

/s2 v

32.09m/s


= 115.5

km/h �


PROBLEM 12.54

Tilting trains such as the Acela, which runs from Washington to New York to Boston, are designed to travel safely at high speeds on curved sections of track which were built for slower, conventional trains. As it enters a curve, each car is tilted by hydraulic actuators mounted on its trucks. The tilting feature of the cars also increases passenger comfort by eliminating or greatly reducing the side force Fs (parallel to the floor of the car) to which passengers feel subjected. For a train traveling at 200 km/h on a curved section of track banked at an angle θ = 8° and with a rated speed of 120 km/h, determine (a) the magnitude of the side force felt by a passenger of weight W in a standard car with no tilt ( φ = 0 ), (b) the required angle of tilt φ if the passenger is to feel no side force. (See Sample Problem 12.6 for the definition of rated speed.)

SOLUTION

Rated speed: v R = 120 km / h = 33.3 m / s, 200 km / h 55.56 m /s From Sample Problem 12.6, v R2 = g ρ tanθ

ρ =

or

v R2 g tan θ

(33.3)

=

2

9.81 tan 8º

= 804.3 m

Let the x-axis be parallel to the floor of the car.

ΣF x = max : Fs + W sin (θ + φ ) = man cos (θ + φ ) =

mv 2

ρ

cos (θ + φ )

(a) φ = 0.

 F s = W 

2  . ) (5556 cos 8º − sin 8º   (9.81)(804.3) 

 (183.33)2  = W  cos 8° − sin 8°  (32.2)( 2674)  = 0.247 W

Fs = 0.247 W �

(b) For F s = 0,

v2 g ρ

cos (θ + φ ) − sin ( θ + φ) = 0

tan (θ + φ ) =

v2

2

. ) (5556 = 0.39035 = . )(8043 . ) g ρ (981

θ + φ = 21.3° φ = 21.3° − 8°

φ = 13.3° �



293

PROBLEM 12.55

Tests carried out with the tilting trains described in Prob. 12.54 revealed that passengers feel queasy when they see through the car windows that the train is rounding a curve at high speed, yet do not feel any side force. Designers, therefore, prefer to reduce, but not eliminate that force. For the train of Prob. 12.54, determine the required angle of tilt φ if passengers are to feel side forces equal to 12 percent of their weights. Tilting trains such as the Acela, which runs from Washington to New York to Boston, are designed to travel safely at high speeds on curved sections of track which were built for slower, conventional trains. As it enters a curve, each car is tilted by hydraulic actuators mounted on its trucks. The tilting feature of the cars also increases passenger comfort by eliminating or greatly reducing the side force F s (parallel to the floor of the car) to which passengers feel subjected. For a train traveling at 125 200 km/h mi/h on a curved section of track banked at an angle angle q θ == 8°8 °and andwith witha arated ratedspeed speedofof 120 75km/h, mi/h, determine (a) the magnitude of the side force felt by a passenger of weight W in a standard car with no tilt ( φ = 0 ), (b) the required angle of tilt φ if the passenger is to feel no side force. (See Sample Problem 12.6 for the definition of rated speed.) Problem 12.54:

SOLUTION

Rated speed: v R = 120 km / h = 33.3 m / s, 200 km / h = 55.56 m /s From Sample Problem 12.6, v R2 = g ρ tan θ

or

ρ =

v R2 g tan θ

=

(33.3) 9.81 tan 8º

= 804.3 m

Let the x-axis be parallel to the floor of the car.

Σ F x = max : Fs + W sin (θ + φ ) = man cos (θ + φ ) =

Solving for F s ,

mv 2 ρ

cos (θ + φ )

 v2  F s = W  cos (θ + φ ) − sin (θ + φ )  g ρ  2

v2

. ) (5556 = = 0.39035 and F s = 0.12 W so that Now . )(8043 . ) g ρ (981

0.12 W = W 0.39035 cos (θ + φ ) − sin ( θ + φ )  Let u = sin (θ + φ ). Then, 0.12 = 0.39035 1 − u 2 − u


cos (θ + φ ) = 1 − u 2 . or

0.39035 1 − u 2 = 0.12 + u



Squaring both sides, 0.15237 1 − u 2

(

) = 0.0144 + 0.24u + u 2

or 1.15237u 2

0

+

0.24u

+

0.13797

=

The positive root of the quadratic equation is u Then, θ

+

φ = sin −1 u φ

=

=

0.2572.

14.90°

14.90° − θ


=

= 14.90° −

8°

φ = 6.90° �


PROBLEM 12.56

A small 250-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at a constant rate of 7.5 rad/s. Determine the three values of θ for which the collar will not slide on the rod, assuming no friction between the collar and the rod.

SOLUTION

If the collar is not sliding, it moves at constant speed on a circle of radius ρ = r sin θ . v = ρω Normal acceleration: acceleration. an =

v2 ρ

=

ρ2ω 2 ρ

= (r sin θ ) ω 2 ∑ F y = ma y :

N cosθ − mg = 0 N =

mg cosθ

Σ F x = max:

N sin θ = ma mg sin θ cosθ Either

sin θ = 0

or

cosθ =

2

= (m r sin θ )ω

g r ω 2

θ = 0° or 180°

or

cosθ =

9.81 (0.5)(7.5) 2

= 0.3488

θ = 0°, 1 80°, and 69.6° �



PROBLEM 12.57

For the collar and rod of Prob. 12.56, and assuming that the coefficients of friction are µ s = 0.25 and µ k = 0.20, indicate whether the collar will slide on the rod if it is released in the position corresponding to (a) θ = 75°, (b) θ = 40°. Also, determine the magnitude and direction of the friction force exerted on the collar immediately after release.

SOLUTION

If the collar is not sliding, it moves at constant speed on a circle of radius ρ = r sin θ . v = ρω From Prob. 12.56

r = 500 mm = 0.500 m, ω = 7.5 rad/s,

m = 250 g = 0.250 kg.

Normal acceleration:

a = n

v2 ρ

=

ρ2ω 2

= (r sin θ )ω 2

ρ

Σ F =

ma:

F − mg sin θ = − m (r sin θ) ω 2 cosθ

F = m ( g − r ω 2 cosθ ) sin θ

Σ F =

ma :

N − mg cos θ = m(r sin θ )ω 2 sin θ

N = m( g cos θ ) + r ω 2 sin 2 θ ) F = (0.25) 9.81 − (0.500 cos 75°)(7.5)2  sin 7 5°

(a) θ = 75°.





= 0.61112 N N = (0.25) 9.81cos75° + (0.500sin 2 75°)(7.5)2 

 = 7.1950 N



µ s N = (0.25)(7.1950) = 1.7987 N Since F < µ N , the collar does not slide. s

F = 0.611 N

75° t continued




F = (0.25) 9.81 − (0.500 cos 4 0°) (7.5)2  sin 4 0°

(b) θ = 40°.





= − 1.8858 N N = (0.25) 9.81cos 4 0° + (0.500 sin 2 40°)(7.5)2 





= 4.7839 N µ s N = (0.25)(4.7839) = 1.1960 N Since F > µ s N , the collar slides. Since the collar is sliding, F = µ k N .

∑ F n = +

ma:

N − mg cosθ = man sin θ N = mg cos θ + m (r sin θ )ω 2 sin θ

= m  g cos θ + (r sin 2 θ )ω 2  = (0.25) 9.81cos40°+ (0.500sin 2 40°)(7.5)2  = 4.7839 N F = µ k N = (0.20)(4.7839) = 0.957 N F = 0.957 N


40° t


PROBLEM 12.58

A small block B fits inside a slot cut in arm OA which rotates in a vertical plane at a constant rate. The block remains in contact with the end of the slot closest to A and its speed is 1.28 m/s for 0 ≤ θ ≤ 150°. Knowing that the block begins to slide when θ = 150°, determine the coefficient of static friction between the block and the slot.

SOLUTION

Draw the free body diagrams of the block B when the arm is at θ v�

=

Σ Ft =

N

=

Σ Fn =

F

Form the ratio

µ s

=

F N

F N

=

at

=

mat :

g =

0,

−W

=

150°.

2

9.81 m/s

sin 30°

+

N = 0

W sin30° man : W cos 30° − F

W cos30° −

=

m

v2 ρ

=

Wv2 ρ g

Wv2 ρ g

, and set it equal to µ s for impending slip.

=

cos 30° − v 2 / ρ g sin 30°

2

=

cos 30° − ( 1.28) /(.3)(9.81) sin 30° µ s


= .618

t


PROBLEM 12.59

A 2.7 kg block is at rest relative to a parabolic dish which rotates at a constant rate about a vertical axis. Knowing that the coefficient of static friction is 0.5 and that r = 1.8 m, determine the maximum allowable speed v of the block.

SOLUTION

Let β be the slope angle of the dish. tan β At r = 1.8 m , tan β

=

0.3

or

β

=

=

dy dr

=

1 6

r

16.7°

Draw free body sketches of the sphere. Σ F y =

N =

Σ Fn =

v2

=

−

µs N sin β − W = 0

W cos β

−

µ s sin β

man : N sin β

+

µ s N cos β =

W (sin β

+

µ s N cos β )

cos β

−

µ s sin β

=

mv 2

ρ

=

Wv 2

ρ g

Wv2

ρ g

µ s cos β sin 16.7º + 1.5 cos 16.7º = 16.6 m 2 / s 2 = (1.8)(9 .8 1) cos 16.7º − 0.5 sin 16.7º cos β − µ s sin β

sin β

ρ g

0: N cos β

+

v


=

4.07 m/st


PROBLEM 12.60

Four seconds after a polisher is started from rest, small tufts of fleece from along the circumference of the 25.4-cm-diameter 25.4 cm-diameter polishing pad are observed to fly free of the pad. If the polisher is started so that the fleece along the circumference undergoes a constant tangential acceleration of 3.7 m/s,2 determine (a) the speed v of a tuft as it leaves the pad, ( b) the magnitude of the force required to free the tuft if the average weight of a tuft is 1.5 × 10−6 kg.

SOLUTION

Uniformly accelerated motion on a circular path: path.

ρ =

D 2

=

cm 12.7 cm.

W = 14.715 × 10−6 N g = 9.81 m/s2 m

W

=

g

=

×

1.5 ×10−6 kg

(a) For uniformly accelerated motion, v

v0

=

+

at t

= 0 +

(3.7)( 4) v

(b)

Σ Ft =

Σ Fn =

mat : F t

man : Fn

=

=

=

14.8 m/s t

(1.5 × 10 −6 )(3.7) = 5.55 × 10 −6 N man

=

mv 2

1.5 × 10 −6 )(14.8) ( = .127

ρ =

2.587 × 10 −3 N

Magnitude of force: F

=

Ft

2

+

F n

2

=

2

10−6 (2587) + (5.55)

2

6 F = 2587 × 10− N t



PROBLEM 12.61

A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts to slide on the turntable 12 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.23 m/s2, determine the coefficient of static friction between the trunk and the turntable.

SOLUTION

Uniformly accelerated motion on a circular path. v

Ft

=

mat

=

Fn

=

man

=

F

=

Ft

2

+

=

v0

=

0

W g

at t

g ρ 2

+ (.23)(12) = 2.76 m / s

at : Ft

Wv2

Fn

+

ρ = 2.4 m

: F n

= W

=

at g

=

.23

W

9.81

W (2.76 )

=

0.0233 W

2

(9.8 1)(2.4)

=

.3235 W

= .324 W

This is the friction force available to cause the trunk to slide. The normal force N is calculated from equilibrium of forces in the vertical direction. Σ F y =

0:

Since sliding is impending,


N

−

W

µ s =

=

F W

0

= .324

N

=

W µ s = .324

t


PROBLEM 12.62

The parallel-link mechanism ABCD is used to transport a component I between manufacturing processes at stations E , F , and G by picking it up at a station when θ = 0 and depositing it at the next station when θ = 180°. Knowing that member BC remains horizontal throughout its motion and that links AB and CD rotate at a constant rate in a vertical plane in such a way that v B = 0.7 m/s, determine (a) the minimum value of the coefficient of static friction between the component and BC if the component is not to slide on BC while being transferred, (b) the values of θ for which sliding is impending.

SOLUTION

For constant speed, at an

=

0

=

v B2 ρ

with vB

Σ F x = Σ F y = Ratio

With

F

g ρ v B

2

N

mg

=

0.7 m/s, ρ = 0.2 m

max: F

=

man cosθ

− W = −man sin θ

man cosθ

=

N

may :

=

− man sinθ

=

cosθ g an

= mg − man sin θ

N

=

− sinθ

cosθ g ρ − sinθ 2 vB

(9.81)(0.2 ) F cosθ = 4.0041, the ratio becomes = N 4.0041 − sin θ (0.7) 2

For no impending slide, µ s

≥

F

=

N

cos θ 4.0041 − sin θ

To find the value of θ for which the ratio is maximum set the derivative with respect to θ equal to zero.

 ± cosθ  = ± 1 − 4.0041 sin θ =   d θ  4.0041 − sin θ  (4.0041 − sinθ ) d

2

sin θ F

=

4.0041

=

0.24974

cos14.446°

θ

= 14.446°,

θ

= 180° − 14.446° = 165.554°,

N

=

1

=

4.0041 − 0.24974 F N

=


0.258

0.258

(a) Minimum value of µ s for no slip. (b) Corresponding values of θ .

0

θ

( µ s )min = 0.258 = 14.5° and 165.5°

t t


PROBLEM 12.63

Knowing that the coefficients of friction between the component I and member BC of the mechanism of Prob. 12.62 are µ s = 0.35 and µ k = 0.25, determine (a) the maximum allowable speed v B if the component is not to slide on BC while being transferred, (b) the values of θ for which sliding is impending.

SOLUTION

For constant speed, at an

=

=

v B

0

2

with vB

ρ

Σ F x = Σ F y =

Ratio

Let u

may : N

F N

=

=

g ρ v B

−

W

so that

F N

=

0.7 m/s, ρ

max : F

=

=

N

cosθ g an

− sin θ

=

0.2 m

man cosθ

= −man sin θ

man cosθ mg − man sin θ

2

=

=

=

mg

man sin θ

−

cosθ g ρ − sin θ 2 vB

cosθ

− sin θ

u

Determine the value of θ at which F / N is maximum.

 cosθ  =   dθ  u − sin θ  d

cos2 θ

The corresponding ratio F N

=

±

− (u − sinθ )( sin θ ) 1 − u sin θ = = (u − sin θ ) (u − sin θ ) 2

v B

2

=

1 − u −2 u − u −1

=

arctan ( 0.35)

± u− sin θ =± = ± tanθ − cosθ 1−u 1

2

F N

= 19.29°,

= (9.81)(0.2) sin19.29° =

= tan θ = u −1

=

v B 2 g ρ

= 0.35

µ s

=

sin θ ,

0.648 m 2/s 2 v B


0

F . N

(a) For impending sliding to the left: θ

2

=

0.805 m/s t



For impending motion to the right: θ u −1

=

v2 2 g ρ

=

sin θ ,

=

F N

arctan (−0.35)

v B

2

=

=

= −

tan θ

=

µ

s

160.71°

(9.81)(0.2) sin160.71°

=

0.648 m 2/s 2 =

(b)

0.35

=

0.805 m/s t

For impending sliding to the left, θ For impending sliding to the right, θ


=

=

19.3 ° t

160.7 ° t


PROBLEM 12.64

In the cathode-ray tube shown, electrons emitted by the cathode and attracted by the anode pass through a small hole in the anode and then travel in a straight line with a speed v0 until they strike the screen at A. However, if a difference of potential V is established between the two parallel plates, the electrons will be subjected to a force F perpendicular to the plates while they travel between the plates and will strike the screen at point B, which is at a distance δ from A. The magnitude of the force F is F = eV / d , where – e is the charge of an electron and d is the distance between the plates. Neglecting the effects of gravity, derive an expression for the deflection δ in terms of V , v0 , the charge – e and the mass m of an electron, and the dimensions d , �, and L.

SOLUTION

Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x = � at the right edge of the plate. At the screen, x

=

�

2

L

+

Horizontal motion: There are no horizontal forces acting on the electron so that a x = 0. Let t 1

=

0 when the electron passes the left edge of the plate, t

when it passes the right edge, and t For uniform horizontal motion, x

=

v0t ,

t1

so that

t 2 when it impacts on the screen.

=

�

=

= t 1

t 2

and

v0

=

�

2v0

+

L v0

.

Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection produced by the electric force. While the electron is between plates (0 ≤ t ≤ t 1 ) , the vertical force on the electron is F y

=

eV / d . After it passes the plates ( t1

≤

t

≤

t 2 ) , it is

zero. For 0

≤

t

≤ t 1,

Σ F y =


may : a y

v y

=

(v )

y

=

y0

y

F y m

=

eV md eVt

a yt

=

0

(v y ) t

+

1 a yt 2 2

0

+

=

+

0

+

md =

0

+

0

+

eVt 2 2md



At t

=

For t1

At t

=

t1,

(v y )

=

eVt1

t2 ,

a y

=

1

≤t ≤

md

t 2

δ

= =

eVt12 2md eV � mdv0

+

eVt12 2md

y

=

y1

+ (v y )1 ( t − t 1 )

y2

=

δ

=

md

y1

(t2 − t1 ) =

 �  v + 2


=

0

eVt1

0

y1

and

L v0

−

1 � 2 v0

+ (v y )1 ( t2 − t 1 ) eVt1   t2 md 

  

−

1 2

 

t1 

or δ

=

eV � L mdv02

�


PROBLEM 12.65

In Prob. 12.64, determine the smallest allowable value of the ratio d / � in terms of e, m, v0 , and V if at x � the minimum permissible distance between the path of the electrons and the positive plate is 0.075 d . =

Problem 12.64: In

the cathode-ray tube shown, electrons emitted by the cathode and attracted by the anode pass through a small hole in the anode and then travel in a straight line with a speed v0 until they strike the screen at A. However, if a difference of potential V is established between the two parallel plates, the electrons will be subjected to a force F perpendicular to the plates while they travel between the plates and will strike the screen at point B, which is at a distance δ from A. The magnitude of the force F is F eV /d , where – e is the charge of an electron and d is the distance between the plates. Neglecting the effects of gravity, derive an expression for the deflection δ in terms of V , v0 , the charge – e and the mass m of an electron, and the dimensions d , �, and L. =

SOLUTION

Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x = � at the right edge of the plate. At the screen, x

�

+ L 2 Horizontal motion: There are no horizontal forces acting on the electron so that a x = 0.

Let t 1

=

=

0 when the electron passes the left edge of the plate, t

when it passes the right edge, and t

=

= t 1

t 2 when it impacts on the screen.

For uniform horizontal motion, x

=

v0t ,

t1

so that

=

�

t 2

and

v0

=

�

2v0

+

L v0

.

Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection produced by the electric force. While the electron is between the plates (0 ≤ t ≤ t 1 ) , the vertical force on the electron is F y

=

eV / d . After it passes the plates ( t1

t 2 ) ,

≤

t

+

eVt 2 2md

≤

it is zero. For 0

≤

t

≤ t 1,

Σ F y =


ma y : a y

v y

=

(v )

y

=

y0

y

0

+

=

F y m

a yt

=

(v y ) t

+

+

0

0 1 2

=

+

eV md eVt md

a yt 2

=

0

+

0



At t

=

But y

So that

�

t1,

<

v0

d 2

, y

−

0.075d

eV �2 2mdv02 d2 �

2

=

>

<

eV � 2 2mdv02 =

0.425d

0.425d

1

eV 2 0

0.425 2mv


=

1.176

eV 2 0

mv

d �

>

1.085

eV mv02

t


PROBLEM 12.66

A 0.5-kg block B slides without friction inside a slot cut in arm OA which rotates in a vertical plane at a constant rate, θ � = 2 rad/s. At the instant when θ = 30°, r = 0.6 m and the force exerted on the block by the arm is zero. Determine, at this instant, ( a) the relative velocity of the block with respect to the arm, (b) the relative acceleration of the block with respect to the arm.

SOLUTION θ =

θ� =

30°, r

2 rad/s, θ ��

0.6 m, W

=

=

=

0

mg

Block B: Only force is weight weight. Fr (a) Fθ

=

maθ

2r�θ�

� = r

−

=

W cos30°,

=

(

Fθ

= −W sin30°

)

�θ � : m rθ�� + 2r

=

Fθ

− rθ�� = −

m

g sin30°

�� + r θ

mg sin30° m

(9.81) sin 30° + ( 0.6)( 0) (2) (2)

= −

2θ �

− rθ�� = − g sin 30° − rθ��

v B / rod =

(b) Fr

=

mar

�� r

=

(

= −1.226

m/s 60° �

1.226 m/s

)

m �� r − r θ� 2 : Fr

=

rθ�

=

(0.6 )( 2)

+

=

m 2

+

rθ� 2

+

mg cos30° m

(9.81) cos 30°

=

=

rθ � 2

g cos30°

10.90 m/s 2

a B / rod =


+

10.90 m/s 2

60° t


PROBLEM 12.67

A 0.5-kg block B slides without friction inside a slot cut in arm OA which rotates in a vertical plane. The motion of the rod is defined by the relation �� = 10 rad/s2 , constant. At the instant when θ = 45°, r = 0.8 m and the θ velocity of the block is zero. Determine, at this instant, ( a) the force exerted on the block by the arm, (b) the relative acceleration of the block with respect to the arm.

SOLUTION θ

= 45°,

r = 0.8 m,

vr = r� = 0, (a)

�� θ

= 10 rad/s2

vθ = rθ � = 0,

W = mg

�θ � ) Σ Fθ = maθ : N − W cos 45° = m ( rθ�� + 2r

(

�θ � N = mg cos 45° + m rθ�� + 2r

)

= (0.5)(9.81) cos 45° + 0.5 ( 0.8)(10 ) + 0 = 7.468 N = 7.47 N (b)

45° t

Σ Fr = mar : mg sin 45° = m (�� r − r θ� 2 ) �� r =

mg m

sin 45° + rθ� 2 = g sin 45° + rθ �2

= (9.81) sin 45° + 0 = 6.937 m/s 2 a B / rod


= 6.94 m/s 2

45° t


PROBLEM 12.68

1.8 kg block B in a horizontal plane is defined by the The motion of a 1.8-kg block B in 22 33 and 33tt �t t and

relations rr

= =

�

22 wherer r is isexpressed expressedininmeters, meter , t t in 2t 2t , ,where in

==

seconds, and θ in radians. Determine the radial and transverse components of the force exerted on the block when (a) ( a) t t = 0, (b) (b) t t = 1 s.

SOLUTION

Use radial and transverse components of acceleration. r

=

3t 2

r�

=

6t

t 3 m, m

θ

=

2t 2 rad

3t 2 m/s m/s,

θ �

=

4t rad/s

�� θ

=

4 rad/s2

−

−

�� r = 6 − 6t m/s, m/s a

r

=

�� r − rθ � 2 = 6 =

aθ = rθ�� + 2r�θ �

Mass: m

=

(a) t = 0.

W

=

g a

r

=

−

6t

−

(3t2 − t 3 )(16t2 )

16t 5 − 48t 4 − 6t + 6 m/s2 =

(3t 2 − t 3 )(4)

=

60t 2 − 28t 3 m/s2

17.658 9.81

=

+

(2)(6t

−

3t 2 )(4t )

1.8 kg

6 m/s 2

aθ = 0 Apply Newton’s second law. Fr = mar = (1.8 )(6)

F r = 10.8 N t

Fθ = maθ = (1.8)(0)

F θ = 0 t

(b) t = 1 s.

a

r

= −

32 m/s2

aθ = 32 m/s2 Apply Newton’s second law.

Fr = mar = (1.8)(−32)

F r =

Fθ

F θ =

=

maθ = (1.8 )(32)


−

57.6 N t 57.6 N t


PROBLEM 12.69

.45 kg block B in The motion of a .45-kg block B in aa horizontal horizontal plane plane is is defined defined by by the the cos2π and θθ == 22π where r r is is expressed expressed in in meters, relations rr == 66((11 ++ cos2 meter, π t t )) and π t t ,,where t in seconds, and θ in t in radians. radians. Determine Determine the the radial radial and and transverse transverse components of the force force exerted exerted on on the the block block when when ( a) ( a) t t == 0,0, t = 0.75 s. s. b) t ((b)

SOLUTION

Use radial and transverse components of acceleration. r

=

6(1 + cos 2π t ) m m,

θ

=

2π t rad

r�

= −

m/s 12π sin 2π t m/s,

θ�

=

2π rad/s

�� θ

=

0

+

6 cos 2π t )(2π )2

�� r = − 24π 2 cos 2π t m/s2 , a

r

=

r�� − rθ� 2

= −

24π 2 cos 2π t

= −

24π 2 − 48π 2 cos 2π t m/s2

aθ = rθ�� + 2r�θ�

=

0

= −

Mass: m

=

(a) t = 0.

W g a

r

=

4.4 9.81

= −

=

+

−

(6

(2)(−12π sin 2π t )(2π )

48π 2 sin 2π t

.45 kg

710.61 m/s2

aθ = 0 Apply Newton’s second law. Fr = mar = (.45 )(710.61) Fθ

=

F r =

−

maθ = 0

319.7 N t F θ

(b) t = 0.75 s. a

r

= −

=

0 t

236.87 m/s2

aθ = 473.74 m/s2 Apply Newton’s second law. Fr = mar = (.45 )(−236.87)

Fθ

=

F r =

−

106.6 N t

F θ = 213.2 N t

maθ = ( .45 )(473.74)



PROBLEM 12.70

The 2.7-kg 2.7 kg collar B slides on the frictionless arm AA′. The arm is attached to drum D and rotates about O in a horizontal plane at the rate θ � = 0.8 t , where θ � and t are expressed in rad/s and seconds, respectively. As the arm-drum assembly rotates, a mechanism within the drum releases cord so that the collar moves outward from O with a constant speed of .457 m/s. Knowing that at t = 0, r = 0, determine the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on B by arm AA′.

SOLUTION

dr

Kinematics:

∫

dt r

0

r = 0 = r� = .457 m/s , �� t

dr =∫ 0 r dt θ�

r = .457 t m

or

�� θ

= 0.8t rad/s,

= 0.8 rad/s 2 2

ar = r�� − rθ � 2 = 0 − (.457)( 0.8t ) = −.292 t 3m/s2

�� + 2r� � = (.457)( 0.8) + ( 2)(.457)( 0.8t ) = 1.0968 m/s aθ = rθθ 2 Kinetics: Sketch the free body diagrams for the collar.

Σ Fr = mar : − T = mar Σ Fθ = maθ : Q = maθ Set T = Q to obtain the required time.

−mar = maθ

− ar = aθ

or

expressions, Using the calculated expressions

.292t 3 = 1.0968,

t 2 =

0968

.292

= 3.75 s 2 t = 1.936 s t



PROBLEM 12.71

The horizontal rod OA rotates about a vertical shaft according to the relation θ � = 10t , where θ � and t are expressed in rad/s and seconds, respectively. A .23-kg .23 kg collar B is held by a cord with a breaking strength of 1.8 kg. Neglecting friction, determine, immediately after the cord breaks, (a) the relative acceleration of the collar with respect to the rod, (b) the magnitude of the horizontal force exerted on the collar by the rod.

SOLUTION θ�

=

m

=

Before cable breaks: Fr Fr mrθ� 2 = mr�� + T or

�� θ

10t rad/s,

2.25 9.81

T and �� r

mar :

=

θ �

2

=

−

T

mr�� + T

θ �

mr

=

10 rad/s2

.23 kg. kg 2

=

= −

=

=

0.

=

m �� r − rθ � 2

(

0

=

+

17. 658

(.23)(.457)

) =

168 rad 2 / s 2

13 rad/s

Immediately after the cable breaks: Fr

=

0, r�

=

0

rod: (a) Acceleration of B relative to the rod.

(

2 m �� r − rθ�

)

=

0

or

�� r

=

rθ �

2

=

a B / rod

(.457)(13) =

2

=

77.2 m / s2

77.2 m/s 2 radially outward �

(b) Transverse component of the force: force. Fθ F θ

=

=

maθ : Fθ

=

(

m rθ�� + 2r�θ �

(.23)[(.457)(10) + (2)(0)(12 )] = 1.05

) F θ

=

1.05 N

�



315

PROBLEM 12.72

Disk A rotates in a horizontal plane about a vertical axis at the constant rate of θ �0 = 15 rad/s. Slider B has a mass of 230 g and moves in a frictionless slot cut in the disk. The slider is attached to a spring of constant k = 60 N/m, which is undeformed when r = 0 . Knowing that at a given instant the acceleration of the slider relative to the disk is �� r = −12 m/s 2 and that the horizontal force exerted on the slider by the disk is 9 N, determine at that instant ( a) the distance r , (b) the radial component of the velocity of the s lider.

SOLUTION θ� =

15 rad/s, m

=

230 g

Due to the spring, Fr

=

0.230 kg, θ ��

= −kr,

Σ Fr =

Fr

=

0, Fθ

=

9 N, �� r

= −12

m/s 2

k = 60 N/m =

mar :

− kr =

(k − mθ �2 ) r = −mr��

(

m �� r − r θ� 2

)

(a) Radial coordinate. coordinate: r =

=

−

�� mr k

−

mθ � 2

−

mr θ ��

2mθ �

=

=

(

m rθ��

=

F θ m 9

�θ � 2r

+

)

�� − r θ

−

0

( 2)(0.230)(15)

(b) Radial component of velocity: velocity.


2

r = 335 mm

maθ : Fθ 2r�θ�

Fθ

(0.230 )( −12 ) 60 − ( 0.230 )(15)

0.33455 m, m

Σ Fθ =

�= r

= −

vr

=

=

�, r

1.304 m/s vr

=

1.304 m/s t


PROBLEM 12.73

A 1.5-kg collar is attached to a spring and slides without friction along a circular rod in a vertical plane. Knowing that the tension in the spring is 70 N and the speed of the collar is 3.8 m/s as it passes through point A, determine, at that instant, the radial and transverse components of acceleration of the collar.

SOLUTION

an =

At point A.

tan θ

v2 ρ

= −

=

(3.8)2

= 115.52

0.125 125

θ =

175 + 125

m/s 2

− 22.62°

Draw the free body diagram of the collar. + Σ F =

at =

ma :

F s cos 22.6° m

F s cos 22.6°

=

70 cos 22.62° 1.5

=

=

mat

43.077 m/s 2

continued




Acceleration vector. vector: a

=

an2 + at 2

=

115.522

tan φ =

115.52 43.077

+

43.077 2

,

= 123.29

m/s 2

φ = 69.55°

φ − 22.62° = 46.93° ar = a cos46.93° 9co os46 s46..93° = = 123.29c

84.2 m/s 2 in negative r -direction -direction ar =

− 84. 84.2 2

m/ss 2 � m/

aθ = a sin 46.93° 123.29sin 29sin 46.9 46.93 3° = 123. aθ = 90.1 m/ m/ss 2 �



PROBLEM 12.74

The two blocks are released from rest when r = .73 m an and d θ = 30° . Neglecting the mass of the pulley and the effect of friction in the pulley and between block A A and and the horizontal surface, determine (a ( a) the initial A,, (c tension in the cable, ( b) the initial acceleration of block A ( c) the initial acceleration of block B block B..

SOLUTION

Let r and and

θ

be polar coordinates of block A block A as as shown, and let y B be the

position coordinate (positive downward, downward, origin at the pulley) for the rectilinear motion of block B block B.. r


r� For block A block A,,

y B

+

+ Σ F x =

=

constant,

+ v B =

�r� + aB ��

0,

mAaA : T cos θ

For block B block B,,

=

Σ F y =

Adding Eq. (1) to Eq. (2) to eliminate T , Radial and transverse components of

W A g W B g W B

=

0

aA or T

=

aB : WB

−

W A

=

g

�r� = −aB (1)

or W A g T

aA secθ (2) W B

=

aA secθ

g

+

W B g

aB (3) aB (4)

a A.

Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components.

�r� − rθ� 2

=

Noting that initially initially θ � signs gives

=

ar

= aA ⋅ e r = −a A cosθ

(5)

0, using Eq. (1) to eliminat eliminatee �r �, and changing a B

=

a A cosθ

(6)

Substituting Eq. (6) into Eq. (4) and solving for a A, a A

=

W B g W A secθ

From Eq. (6), a B

+

W B cos θ

=

= 5.5 cos 30º

(a) From Eq. (2), T = (176.5 / A.: A. (b) Acceleration of block block A B.: (c) Acceleration of block block B B.


222 2.6)(918 . ) (22 176.5 sec 30º =

4.76 m / s

)

+ 222.6

cos 30º

=

5.5 m / s

2

2

9.81 81 ( 5. 5) sec 30º

= 114. 26

T = = 114.26 N �

a A = 5.5 m/s

2

� 2

a B = 4.76 m/s

�


PROBLEM 12.75

The velocity of block A block A is is 1.8 m/s to the right at the instant when r = = .73 m and an d θ = 30°. Neglecting the mass of the pulley and the effect of friction in the pulley and between block A A and and the horizontal surface, determine, A,, at this instant, (a ( a) the tension in the cable, (b ( b) the acceleration of block A (c) the acceleration of block B block B..

SOLUTION

Let r and and

θ

be polar coordinates of block A block A as as shown, and let y B be the

position coordinate coordinate (positive downward, downward, origin at the the pulley) for the rectilinear motion of block B block B.. Radial and transverse components of

v A.

Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components. r�

vr

=

= v A ⋅ e r = −v A cos30°

= −1.8 cos30° = −1.56 m/s

rθ �

=

vθ

= v A ⋅ eθ = −v A sin30°

= 1.8 sin30° = .9 m/s

vθ

θ � =


r

r� For block A block A,, +

Σ F x =

r

+

.9

=

y B

.73

=

= 1.232 rad/s

constant,

+ v B =

0,

�r� + aB ��

mA aA : T cos θ

=

W A g

Σ F y =

For block B block B,,

Adding Eq. (1) to Eq. (2) to eliminate T , Radial and transverse components of

0

=

aA or T

W B g

W B

aB : WB

=

�r� = −aB (1)

or

W A g

=

−

W A g T

a A secθ

a A secθ (2)

=

+

W B g W B g

aB (3) aB (4)

a A.

Use a method similar to that used for the components of velocity.

�r� − rθ� 2

=

ar

= aA ⋅ e r = −a A cosθ

(5)

�� and changing signs gives Using Eq. (1) to eliminat eliminatee r a B


=

aA cosθ

− r θ �

2

(6)


CH 12 P1-75.pdf

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