Solution - Mecânica Vetorial Para Engenheiros, Dinâmica - 11ª edição Solucionário Beer PDF

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Solucionário do Mecânica Vetorial Para Engenheiro da 11ª edição , BEER. Solucionário em inglês....

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CHAPTER 11

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PROBLEM 11.1 A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS coordinates are used to determine his displacement as a function of time: x= 0.5t3 + t2 + 2t where x and t are expressed in ft and seconds, respectively. Determine the position, velocity, and acceleration of the boarder when t = 5 seconds.

SOLUTION x  0.5t 3  t 2  2t

Position:

v

Velocity:

dx  1.5t 2  2t  2 dt

a

Acceleration:

dv  3t  2 dt

x  0.5  5  5 2  2 5

x  97.5 ft 

v  1.5 5  2 5  2

v  49.5 ft/s 

a  3  5  2 

a  17 ft/s2 

3

At t  5 s,

2

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PROBLEM 11.2 The motion of a particle is defined by the relation x  2 t 3  9 t 2 12 t 10, where x and t are expressed in feet and seconds, respectively. Determine the time, the position, and the acceleration of the particle when v  0.

SOLUTION x  2t 3  9t 2  12t  10

Differentiating,

v

dx  6t 2  18t  12  6( t 2  3t  2) dt  6(t  2)(t  1)

a

dv  12t  18 dt

So v  0 at t  1 s and t  2 s. At t  1 s,

x1  2  9  12  10  15

t  1.000 s 

a1  12  18  6

x1  15.00 ft 

a1  6.00 ft/s2  At t  2 s, 3

2

x2  2(2) 9(2) 12(2) 10  14

t  2.00 s 

x2  14.00 ft 

a2  6.00 ft/s2 

a2  (12)(2)  18  6

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PROBLEM 11.3 The vertical motion of mass A is defined by the relation x  10 sin 2t  15cos 2t  100, where x and t are expressed in mm and seconds, respectively. Determine (a) the position, velocity and acceleration of A when t  1 s, (b) the maximum velocity and acceleration of A.

SOLUTION x  10sin 2t  15cos 2t  100

v

dx  20cos 2t  30sin 2t dt

a

dv  40sin 2t  60cos 2t dt

For trigonometric functions set calculator to radians: (a) At t  1 s.

x1  10sin 2  15cos 2  100  102.9

x1  102.9 mm  v1  35.6 mm/s 

v1  20cos 2  30sin 2  35.6

2

a1  40sin 2  60cos 2  11.40

a1  11.40 mm/s 

(b) Maximum velocity occurs when a  0. 40sin 2 t  60 cos 2t  0

tan 2t  

60  1.5 40

2 t  tan 1 (1.5)  0.9828 and  0.9828   Reject the negative value. 2t 2.1588

t  1.0794 s t  1.0794 s for vmax vmax  20cos(2.1588)  30sin(2.1588)   36.056

so

vmax  36.1 mm/s 

Note that we could have also used 2

2

vmax  20  30  36.056 by combining the sine and cosine terms. For amax we can take the derivative and set equal to zero or just combine the sine and cosine terms. 2

amax  402  602  72.1 mm/s2

amax  72.1 mm/s 

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PROBLEM 11.4 A loaded railroad car is rolling at a constant velocity when it couples with a spring and dashpot bumper system. After the coupling, the motion of the car is 4.8 t defined by the relation x  60e  sin16t where x and t are expressed in mm and seconds, respectively. Determine the position, the velocity and the acceleration of the railroad car when (a) t  0, (b) t  0.3 s.

SOLUTION x  60e4.8t sin16 t dx  60( 4.8) e4.8t sin16 t  60(16) e4.8t cos16 t dt   v  288 e 4.8t sin16 t  960 e 4.8 t cos16 t v

a

dv    1382.4e 4.8t sin16t  4608e 4.8t cos16t dt    4608e 4.8t cos16t  15360e 4.8t sin16t

a  13977.6e (a) At t  0,

4.8t

sin16t  9216e

4.8

cos16t

x0  0

x0  0 mm 

v0  960 mm/s

v0  960 mm/s

a0  9216 mm/s2 (b) At t  0.3 s,

a0  9220 mm/s2

 

e 4.8t  e  1.44  0.23692 sin16t  sin 4.8   0.99616 cos16 t  cos 4.8  0.08750

x0.3  (60)(0.23692)(0.99616)  14.16

x0.3  14.16 mm



v0.3  87.9 mm/s



v 0.3   (288)(0.23692)( 0.99616) (960)(0.23692)(0.08750) 87.9

a0.3  (13977.6)(0.23692)( 0.99616)  (9216)(0.23692)(0.08750)  3108

a0.3  3110 mm/s2  or 3.11 m/s2

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

PROBLEM 11.5 The motion of a particle is defined by the relation x  6 t 4 2 t 3 12 t 2  3t  3, where x and t are expressed in meters and seconds, respectively. Determine the time, the position, and the velocity when a  0.

SOLUTION We have

4 3 2 x  6t  2t  12t  3t  3

Then

v

dx 3 2  24 t  6 t  24 t  3 dt

and

a

dv  72t 2  12t  24 dt

When a  0:

72 t2 12 t  24 12(6 t2  t  2)  0 (3t  2)(2t  1)  0

or t

or At t 

2 1 s and t   s (Reject) 3 2 4

3

t  0.667 s 

2

 2  2  2  2 x 2/3  6    2    12    3    3 3 3 3        3

2 s: 3

 v 2/3  24  

3

2    6 3 

2

2    24  3 

or

2 3 3

x2/3  0.259 m 

or v2/3  8.56 m/s 

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PROBLEM 11.6 The motion of a particle is defined by the relation x  t3  9 t2  24 t  8, where x and t are expressed in inches and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero.

SOLUTION 3

2

We have

x  t  9t  24 t  8

Then

v

dx 2  3 t  18 t  24 dt

and

a

dv  6 t  18 dt

3 t2 18 t  24  3( t2  6 t  8)  0

When v  0:

(a)

(t  2)(t  4)  0 t  2.00 s and t  4.00 s  When a  0:

(b)

6t  18  0 or t  3 s

At t  3 s:

x3  (3) 3  9(3) 2  24(3)  8

First observe that 0  t  2 s:

v0

2 s  t  3 s:

v0

or

x3  10.00 in. 

Now

At t  0:

x0  8 in.

At t  2 s:

x2  (2)3 9(2)2  24(2) 8 12 in.

x2  x0  12  ( 8)  20 in.

Then

| x3  x2 |  |10 12|  2 in. Total distance  22.0 in. 

Total distance traveled  (20  2) in.

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PROBLEM 11.7 A girl operates a radio-controlled model car in a vacant parking lot. The girl’s position is at the origin of the xy coordinate axes, and the surface of the parking lot lies in the x-y plane. She drives the car in a straight line so that the x coordinate is defined by the relation x(t) = 0.5t3 - 3t2 + 3t + 2, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and total distance travelled when the acceleration is zero.

SOLUTION Position:

x  t   0.5 t3  3 t2  3 t  2

Velocity:

v t  

dx dt

v t   1.5t 2  6 t  3 2 0 1.5t  6t  3

(a) Time when v=0

t

6  62  4 1.5 3

a t  

Acceleration:

t  0.586 s and t  3.414 s 

2 1.5 dv dt

a t   3 t  6 0  3t  6

Time when a=0

t2 s

x 2   0.5  2  3 2  3  2  2 3

(b) Position at t=2 s

2

x 2  0 m  To find total distance note that car changes direction at t=0.586 s

x 0   0.5 0  3 0  3 0  2 3

Position at t=0 s

2

x 0   2 x 0.586  0.5 0.586  3 0.586  3 0.586 2 3

Position at t=0.586 s

2

x 0.586  2.828 m Distances traveled: From t=0 to t=0.586 s:

x 0.586  x 0  0.828 m

From t=0.586 to t=2 s:

x 2  x 0.586  2.828 m Total distance 3.656 m 

Total distance traveled  0.828 m  2.828 m

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PROBLEM 11.8 The motion of a particle is defined by the relation x  t   t  2 , where x and t are expressed in feet and seconds, respectively. Determine (a) the two positions at which the velocity is zero, (b) the total distance traveled by the particle from t  0 to t  4 s.. 3

2

SOLUTION Position:

x  t   t 2   t  2

Velocity:

v t  

3

dx dt

v t   2t  3 t  2 

2



v t   2 t  3 t 2  4t  4



2

 3t 14t 12

0  3t 2  14t  12

Time when v(t)=0

t

14  142  4  3  12

2  3  t  1.131 s and t  3.535 s

(a) Position at t=1.131 s

x1.131   1.131   1.1.31  2

3

Position at t=3.535 s

x 3.535   3.535   3.535  2

3

2

x1.131  1.935 ft.  2

x 3.531  8.879 ft.  To find total distance traveled note that the particle changes direction at t=1.131 s and again at t=3.535 s.

x  0    0   0  2 2

Position at t=0 s

3

x  0  8 ft

x  4    4   4  2 2

Position at t=4 s

3

x  4   8 ft (b) Distances traveled: From t=0 to t=1.131 s:

x1.131  x 0   6.065 ft.

From t=1.131 to t=3.531 s:

x 3.535   x 1.131  6.944 ft.

From t=3.531 to t=4 s:

x 4   x 3.535  0.879 ft.

Total distance traveled 6.065 ft  6.944 ft  0.879 ft

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Total distance  13.888 ft. 

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PROBLEM 11.9 The brakes of a car are applied, causing it to slow down at a rate of 10 ft/s2. Knowing that the car stops in 100 ft, determine (a) how fast the car was traveling immediately before the brakes were applied, (b) the time required for the car to stop.

SOLUTION a  10 ft/s2 (a)

Velocity at x  0.

v



0 v0

0

dv  a   10 dx xf



vdv  

0

(10) dx

v02  10 x f  (10)(300) 2

v0  77.5 ft/s 

v 02  6000 (b)

Time to stop. dv  a  10 dx



0

dv  

v0



tf

10 dt

0

0  v0  10 t f tf 

v0 77.5  10 10

t f  7.75 s 

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PROBLEM 11.10 The acceleration of a particle is defined by the relation a  3e  0.2t , where a and t are expressed in ft/s 2 and seconds, respectively. Knowing that x 0 and v 0 at t 0, determine the velocity and position of the particle when t 0.5 s.

SOLUTION Acceleration:

a  3e 0.2t ft/s 2

Given :

v0  0 ft/s, a

Velocity:

dv  dv  adt dt

v



x0  0 ft

t

dv 

v0

 adt 0 t

v  vo 



0

0



t

3e 0.2t dt  v  0=  15 e0.2t



v  15 1  e 0.2t ft/s v

Position:

x



dx  dx  vdt dt t



dx  vdt

x0

0 t

 





x  xo  15 1  e 0.2 t dt  x  0=15 t  5 e 0.2 t 0

 v  15 1  e



t 0

  ft/s

 x 15 t  5 e 0.2t  75 ft  0.2 0.5

Velocity at t=0.5 s

v 0.5   1.427 ft/s 





x  15 0.5  5 e0.2 0.5  75 ft

Position at t=0.5 s

x 0.5  0.363 ft. 

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PROBLEM 11.11 The acceleration of a particle is directly proportional to the square of the time t. When t  0, the particle is at x  24 m. Knowing that at t  6 s, x  96 m and v  18 m/s, express x and v in terms of t.

SOLUTION a  kt2

We have

k  constant

dv  a  kt 2 dt

Now v



or

1 v  18  k (t 3  216) 3

18

dv 



t

At t  6 s, v  18 m/s:

6

kt 2 dt

1 3 v  18  k (t  216)(m/s) 3

or

dx 1 3  v 18  k( t  216) dt 3

Also



At t  0, x  24 m:

x 24

dx 

t

0

x  24  18t 

or

1

 18  3 k (t

3

  216) dt 

1 1 4  k t  216 t  3  4 

Now 1 1  96  24 18(6)  k  (6) 4  216(6)  3 4 

At t  6 s, x  96 m:

k

or

1 m/s4 9

1  1  1 4  x  24  18 t     t  216 t  3  9  4 

Then

x (t ) 

or

1 4 t  10t  24 108



11 v  18    ( t 3  216) 39

and

v (t ) 

or

1 3 t  10 27



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PROBLEM 11.12 The acceleration of a particle is defined by the relation a  kt 2 . (a) Knowing that v   8 m/s when t  0 and that v  8 m/s when t  2 s, determine the constant k. (b) Write the equations of motion, knowing also that x  0 when t  2 s.

SOLUTION a  kt 2

(1)

dv  a  kt 2 dt t  0, v  8 m/s and t  2 s, v  8 ft/s



(a)

8 8

dv 



2 0

2

kt dt

1 k (2) 3 3

k  6.00 m/s 

dv  a  6t 2 dt

a  6t 2 

8  ( 8) 

4

Substituting k  6 m/s4 into (1)

(b)



t  0, v  8 m/s:

v 8

dv 

v  (8) 



t

0

6t 2 dt

1 3 6(t ) 3

3

v  2t  8 

dx 3  v  2t  8 dt



t  2 s, x  0:

x 0

dx 



t 2

(2t 3  8)dt ; x 

1 4 t  8t 2

t

2

1  1  x   t4  8t   (2)4  8(2) 2  2  1 4 x  t  8t  8 16 2

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1 x  t 4  8t  8  2

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PROBLEM 11.13 A Scotch yoke is a mechanism that transforms the circular motion of a crank into the reciprocating motion of a shaft (or vice versa). It has been used in a number of different internal combustion engines and in control valves. In the Scotch yoke shown, the acceleration of Point A is defined by the relation a 1.8sin kt, where a and t are expressed in m/s2 and seconds, respectively, and k 3 rad/s. Knowing that x  0 and v 0.6 m/s when t 0, determine the velocity and position of Point A when t 0.5 s.

SOLUTION Accel...