Dyna Project

EXAMPLES/NOTES: UNIFORM RECTILINEAR MOTION 1. Car A at a gasoline station stays there for 10 minutes after Car B passes at a constant speed of 40 miles per hour. How long will it take Car A to overtake Car B if it accelerates at 4 m/s² Solution: Ta = Tb Sa = S10 + Sb

Consider B:

Consider A: (

For S10: (

)(

Substitute:

Answer: T = 77.85 seconds

)

)(

)

2. A stone is thrown up from the ground with a velocity of 300ft/s. How long must one wait before dropping a second stone from the top of 600ft tower if the two stones are to pass each other 200ft from the top of the tower?

Solution: Consider 1

)( )

(

(

)( )

Consider 2

)( )

(

How long must one wait? t = 17.19 seconds – 3.52 seconds

Answer: 13.67 seconds

VARIABLE ACCELERATION 1. The motion of a particle is given by the values of

and

when

where is in .

and in

. Compute

( ) ( )

( )

( );

( )

;

2. A particle moves in a straight line according to the law a. When

, compute .

b. Find the average velocity during the 3rd to 4th seconds. c. When the particle comes to stop, what is its acceleration?

a.

( )

;

(

b.

( )

( );

)

;

where is in

and in

.

( )

( );

c.

(

);

3. The rectilinear motion of a given particle is given by ,

, and

( )

. Determine

; ( )

( )

;

,

, and

where is in relations.

and

is in

. When

MOTION CURVES 1. A particle starting with an initial velocity of 60 ft /s has a rectilinear motion with the constant deceleration of 10 ft/ . Determine the velocity and displacement at t = 9 sec.

Solution: For 𝑨𝟏 and 𝑨𝟐 :

𝑨

𝐴 𝐴𝑟𝑒𝑎 ( 𝐴𝑟𝑒𝑎

𝑡

)( )

(

)(𝑥)

(

)(

𝐴

𝑡 𝐴

𝑥)

𝐴

𝑽

𝐴

𝑥

𝑥 𝑥

𝑡

𝐴

𝑥

𝑥

𝑠𝑒𝑐

𝑠

𝑉𝑜 𝑡

𝑎𝑡

𝑺 ( )

𝑡

(

𝒔

𝟏𝟑𝟓 𝒇𝒕

𝑉

𝑎𝑡 (

𝑽

𝑉𝑜 )( )

𝟑𝟎 𝒇𝒕 𝒔

)( )

2. An auto travelled 1800 ft in 40sec. The auto accelerates uniformly and decelerates uniformly at 6 ft/ , starting from rest at A and coming to stop at B. Find the maximum speed in fps.

Solution:

A 6x – 6(40 – x) = 0 𝐴

𝑥

6x – 0 = 6(40 – x)

𝑥 𝑥

𝐴

(

𝑡 𝑥)

𝑠𝑒𝑐

6x = 240 – 6x x = 20

V ( ) 120 𝐴

𝐴 𝑡

3. An Auto starts from rest and reaches a speed of 60 ft/s in 15 sec. The acceleration increases uniformly from zero for the first 9 sec after which the acceleration reduces uniformly to zero in the next 6 sec. Compute for the displacement in this 15 sec interval.

Solution: V:

𝑨 𝑎

𝑨𝟏

𝑓𝑡 𝑠𝑒𝑐

𝑨𝟏

𝑡 𝐴

𝑡

𝑠

𝑓𝑡 𝑠

𝑎

𝟔𝟎𝒇𝒕 𝒔𝒆𝒄

𝑨𝟐

For 𝑨𝟏 and 𝑨𝟐 : ( )(𝑎)

𝐴 𝐴

( ); 𝐴

𝑎

𝐴

( )(𝑎)

𝐴

𝑎

𝐴

𝐴 𝑡

𝚫𝑽

𝐴 𝑡

𝑽

𝑨𝟐

𝑡

𝑎𝑠𝑒𝑐

𝑎 𝑎

( ); 𝐴

𝑎 𝑓𝑡 𝑠𝑒𝑐 𝑭𝒐𝒓 𝑨𝟑 𝒂𝒏𝒅 𝑨𝟒 ∶

𝑺 (

𝐴

)( )

𝐴 (

𝐴 𝑡 𝑡

𝑎𝑠𝑒𝑐

)( )

(

)( )

;𝑺

𝟒𝟐𝟎

𝐴 𝑆

5

PROJECTILE MOTION 1. A golf ball is fired from the top of a cliff 50 m high with a velocity of 10 m/s directed at 45⁰ to the horizontal. Find the range of the projectile.

t1 = 3.99 s (checked) Vo cosө = =

t2 = -2.55 s For range:

10cos45⁰ = 10cos45⁰ = For t:

10cos45⁰ =

y = Vo sin t - gt2 -50 = 10sin45⁰t - ( 9.81) t2

Range = 28.21 m

2. In figure 9-6.10, a ball thrown down the incline strikes it at a distance s = 254.5 ft. If the ball rises to a maximum height h = 64.4 ft above the point of release. Compute its initial velocity and inclination ө.

(

Horizontal motion:

(

) )

Vo cosө = (

) = (2)(32.2)(64.4)

.4 Vertical Motion: y = Vo sin t - gt2

For -80.48 = Vo sin t - gt2 -80.48 = Vo sin (

For maximum height:

-80.48 = 241.44 tan -

) - (32.2)( (

)

(

)

H = 64.4 ft. (

) (

)

-80.48 = 241.44 tan -

)

tan

;

tan

53.06⁰ ;

-15.11⁰

therefore,

Vo = 80.57 ft/

3. Find the take-off velocity that is just enough to clear the gap. Using

Using horizontal motion formula: Vo cosө = Vo cos30⁰ =

t=

Using vertical motion formula: y = Vo sin t - gt2 -22.2= Vo sin

t - (32.2)t2

-22.2= Vo sin

(

) - (32.2)(

Vo = 14.14 ft/s

4. How high is the hill?

Using horizontal motion formula:

)2

Vo cosө = 100cos60⁰ = t=

5

5

t = 10 sec Using vertical motion formula: y = Vo sin t - gt2 y= 100sin

(10)- (32.2)(10)2

y = -743.97 ft.

KINETICS 1. Determine P that will give the body an acceleration of 6 ft/sec2 µ = 0.20.

(1)

(2)

5

5

Substituing 2 in 1 we get: P = 722.17 lbs.

2. Determine the acceleration of the system and tension in the chord. µ = 0.30

Consider 200 N Block FBD:

Consider 100 N Block FBD:

a = 5.56 m/sec2 T = 86.67 N

3. Find the acceleration of the system and tension in the block.

Consider 300 N Block FBD:

Consider 100 and 200 N Block FBD:

5

T = 208 N a = -3 m/sec2

5

5

5

4.

1. What is the tension in the card? 2. Acceleration of the blocks 3. Velocity of B after 2 seconds?

B 196.2 N A 981 N Consider B

Consider A

T

T

T

196.2 N a

a

W 981 N

For

:

For

:

⁄

⁄

⁄

⁄

1

2

Solve 1 and 2 simultaneously:

Therefore:

T = 327N

= 6.54

⁄

⁄

5

150 N aA

aC

aB

50 N

100 N

Determine the tension and acceleration of each blocks. Consider block A:

T1 aA 150 N 5

1

Consider block B: T2

aB

2

100 N

Consider block C: T2

aC

50N

5

3

For T1 and T2:

T1

A: 5

SA

B:

150N

(

)

(

)

Original C: 5

50N

100N

SA SB SB’

For SA:

For SB’: ⁄

For SB:

⁄

⁄

⁄

But

’ ⁄

⁄

⁄

⁄ (

(

)

)

Equation

2

(

will now be: )

1 5

T2 = 120 aA = 5.836 ⁄ aB’ = -7.848 ⁄

3 5

(

)

5

(

)

T = 70 N aA = -0.58

⁄

aB’ = 3.46

⁄

ASSIGNMENT NO. 1 9-3.6 How fast must an automobile of the previous problem move in the last 8 minutes to obtain an average speed of 35 mph? From the previous problem:

(

)(

)

(

)(

)

(

)(

)

;

5

9-3.8 On a certain stretch of track, trains run at 60 mph. How far back of a stopped train should a warning torpedo be placed to signal an oncoming train? Assume that the brakes are applied at once and retard the train at the uniform rate of 4 fps2.

( )

9-3.10 A ship being launched slides down the ways with a constant acceleration. She takes 4 seconds to slide the first foot. How long will she take to slide down the ways if their length is 900 feet? ;

( (

)

) ;

9-3.12 A stone is dropped down a well and 5 seconds later the sound of the splash is heard. If the velocity of sound is 1120 fps, what is the depth of the well?

For the stone,

For the sound,

( )

(

) (

)

5

(√

5

) + (√

)

9-3.14. A train moving with constant acceleration travels 24 ft during the 10th sec of its motion and 18 ft during the 12th sec of its motion. Find its initial velocity. Solution: S1oth = 24 ft @ t = 9 sec to 10 sec S12th = 18 ft @ t = 11 sec to 12 sec @ S1oth:

( ) ( S9-10

( ) )

( (

; )

;

) (eq. 1)

@ S12th:

(

) (

)

(

)

(

)

(

) (

(

S11-12

)

;

(

)

;

(

)

) (eq. 2)

Using the eq. 1 & 2: ;

9-3.16. An auto A is moving at 20 fps and accelerating at 5 fps2 to overtake an auto B which is 382 ft ahead. It auto B is moving at 60 fps and decelerating at 3 fps2, how soon will A pass B? Solution: @ Auto A:

(

)

@ Auto B:

( )

(eq. 2)

(

)

(

)

(eq. 2)

Subtract eq. 2 from eq. 1:

9-3.18. The rectilinear motion of a particle is governed by the equation s = r sin t where r and constants. Show that the acceleration is a = - 2s. Solution:

; ;

are

; ;

Since Therefore:

9-3.20 A ladder of length L moves with its ends in contact with a vertical wall and a horizontal floor. If a ladder starts from a vertical position and its lower end A moves along the floor with a constant velocity vA, show that the velocity of the upper end B is vB = – vA tan Ɵ where Ɵ is the angle between the ladder and the wall. What does the minus sign mean? Is it physically possible for the upper end B to remain in contact with the wall throughout the entire motion? Explain. Solution:

√ (

)

But Therefore: When Ө = 90°,

, which is impossible.

9-3.22 The velocity of a particle moving along the x-axis is defined by v=kx³ -4x² + 6x, where v is in fps, x is in feet, and k is a constant. If x = 1, compute the value of the acceleration when x = 2 feet. Solution: At x = 2 feet v = (1)(2)³ - 4(2)² + 6(2) = 4fps

Substituting v = 4 fps. ( )( ) ( ) Answer:

( )( )

( )

ASSIGNMENT NO. 2 Determine the acceleration of the 2 blocks after touching each other. Determine the time at which the block will touch the block.

;

(

)

(

);

(

);

(1)

;

(

)

(2)

Equating (1) and (2)

; Acceleration after touch: ;

ASSIGNMENT NO. 4 1044. An elevator weighing 3220 lb starts from rest and acquired an upward velocity of 600 ft per min in a distance of 20 ft. If the acceleration is constant. What is the tension in the elevator cable? Given: W = 3220 lb

T

Sol’n:

v = 600 ft/min = 10 ft/sec v

(

s = 20 ft

)

(

)

W

Req’d: T

(

)

1045. A man weighing 161 lb is in an elevator moving upward with an acceleration of 8 ft per sec2. (a) What pressure does he exert on the floor of the elevator? (b) What will the pressure be if the elevator is descending with the same acceleration?

Given: Sol’n: Wman = 161 lb a = 8 ft/sec2

(a) ( )

Req’d: (a) Pressure he exert (b) Pressure if the elevator Descends with the same acceleration

(b) ( )

1046. The block in Fig. P-1046 reaches a velocity of 40 ft per sec in 100 ft, starting from rest. Compute the coefficient of kinetic friction between the block and the ground.

P= 60 lb

161 lb Given: Sol’n: v = 40 ft/sec (

s = 100 ft

)

Req’d: Coefficient of kinetic Friction, µ ( )

(

)

1047. Determine the force P that will give the body in Fig. P-1047 an acceleration of 6 ft per sec2. The coefficient of kinetic friction is 0.20.

P 3

322 lb

Given: Sol’n: a = 6 ft/sec2 µ = 0.2

( )

(

5

Req’d: force,P

5

5

)

4

( )

5

(

)(

5

)

1053. Referring to Fig. P-1052, assume A weighs 200lb and B weighs 100lb. Determine the acceleration of the bodies if the coefficient of kinetics friction is 0.10 between the cable and the fixed drum.

B

Fig. P-1052

Given:

Solution: 1

2

( )

3

A

Substitute 3 to 1: 4

From 2:

Substitute 2 to 4: (

)

⁄

1055. If the pulleys in Fig. P-1055 are weightless and frictionless, find the acceleration of the body A.

200 lb

300 lb B.Fig. P-1055

A

For A:

aA 200 lb

1

For B: T

T ( )

aB

2

300 lb Equate 1 and 2: (

)

⁄

1057. The coefficient of kinetic friction under block A in Fig. P-1057 is 0.30 and under block B it is 0.20. Find the acceleration of the system and the tension in each cord.

B 200lb

A 100lb 30o

300lb

At C, 300 – T2 =

a -----1

At B, T2 – T1 – 200sin30o – 200cos30o (0.2) =

a

T2 – T1 – 134.64 = 6.21a -----2

At A, T1 – 100sin30o – 100cos30o (0.3) =

a

T1 – 75.98 = 3.11a -----3 T1 = 75.98 + 3.11a

Substitute T1 to 2, T2 – (75.98 + 3.11a) – 134.64 = 6.21a T2 – 210.62 = 9.32a T2 = 210.62 + 9.32a -----4

Substitute 4 to 1 300 – 210.62 + 9.32a =

a

89.38 = 18.64a

a= 4.8 ft/sec2  ans.

T1 = 75.98 + 3.11 (4.8) = 90.91 lb  ans. T2 = 210.62 + 9.32 (4.8) = 255.36 lb  ans.

1059. Compute the acceleration of body B and the tension in the cord supporting body A in Fig. P-1059.

300lb fh = 0.20

200lb

A 3 4

In block A, 200 – T =

aA

In block B, 2T - 5 (300) - 5 (300) (0.20) = 2T – 228 =

aB

aB

In getting the acceleration for B, Since aA = 2 aB

2 [ 200 – T =

aA ]

+ -228 + 2T = 400 – 228 = 550aA

172 = 500 (2aB)

(0.5) aA

aB / 32.2 = 172/1100

aB = 5.03 ft/sec2  ans.

1061. Compute the time required for the 100-lb body in Fig. P-1061 to move 10 ft starting from rest.

100 lb fh=0.20

80lb 3 4

For 100-lb block, T1 - (100) = 5

a1

For 800-lb block, 5

(80) - (80) (0.20) – T2 = 5

2T2 – T1 = ma 2T2 – T1 = 0 2T2 = T1

Since a2 = 2 a1

a2

a2 = 2.82 ft/sec2 Solving for t1 (100lb), S = a1t2 10 = (2.82) t2 t = 2.663 sec.  ans.

1063. Determine the acceleration of each weight in Fig. P-1063, assuming the pulleys to be weightless and frictionless.

A B

C 150 lb

B 480 lb

For A, T – 150 =

5

aA

For B, 2T - 480 =

For C,

aB

300 lb

300 – T =

aC

Since aB= aC - aA

For Tension, 5

aA = T -150

(480) ( aC - aA) = 2T - 480 aC = 300 – T

T = 218.7 lb

Solving for acceleration, 218.7 – 150 =

300 – 218.7 =

5

aA

; aA = 14.7476 ft/sec2  ans.

aC

; aC = 8.7262 ft/sec2  ans.

2(218.7) – 480 = 480/32.2aB aB = -2.85775 ft/sec2 or aB = 2.85775 ft/sec2 (downward) ans.

1065. Determine the maximum and minimum weights of the body C on Illustration Problem 1043 that will keep C stationary. All other data remain unchanged. Solution:

1000 T

2B F = 160 800 N = 800

2T B

aB = WA

800

W 8W

F

N =gW

F = 16W

6W

For A; 600 – 160 – T = For B; 2T – 800 =

(

)

Solving for T; T = 407 lb For up plane impending motion of C:

T = 407 = 6W + 16W W = 535 lb For down plane impending motion of C;

T = 407 = 6W - 16W W = 924 lb

1067. In the system of connected blocks in Fig 1067, the coefficient of kinetic friction is 0.20 under bodies B and C. determine the acceleration of each body and the tension in the cord. Solution:

1000lb C B fh = 0.20

fh = 0.20

800lb

400lb

A 3 4

Direction of motion: Assuming at rest T= 400 On B, Net force = 2T = 800 – 480 > F = 128 (B rises) On B, Net force = 600 – (T=400) > F = 160 ( C falls) With C at rest, Sa’

= 2Sb

With B at rest Sa “ = Sc Net motion = Sa’ - Sa” = Sa = 2Sb Differentiating : aA = 2aB – aC aB = (

For ; 400 – T =

)

- Sc

3 4

For B; 2T – 480 – 128 = Solving; T = 348.2lb aA = 4.18 fps2 aB = 3.57 fps2 aC = 2.96 fps2

1069. Two blocks A and B each weighing 96.6 lb and connected by a rigid bar of negligible weight move along the smooth surfaces shown in Fig 1069. They start from rest at the given position. Determine the acceleration of B at this instant. Hint: To relate aA to aB, use the method developed in Illus Prob on 258.

Solution: V=

a= Va + xaA + Vb + Yab = 0 At start,

Va = Vb = 0 aA = or if Ab is down + down aA =

=

For A; 6P =

= 4ab

For B; 96.6 – 8P =

ab = 3ab

Solving; aB = 11.6 fps2 aA = 15.47 fps2

1071. The pulleys in the preceding problem have been assumed to be frictionless and weightless. What changes would there be in the solutions of these problems if the pulleys (a) had friction (b) had appreciable weight? Solution: (a) with friction, the tensions on the opposite sides of the pulley would be unequal. (b) With appropriate weight, the supporting tension would not equal twice the outside tensions.

SEATWORK A ball is dropped from the tower of 80 ft. high at the same instant that a second ball is thrown upward from the ground with an initial velocity of 40 ft/sec. When and where do they pass, and with what velocities?

SOLUTION:

(

)

( )

( ) (

) ( )

( )

Equate (2) and (1): (

Substitute t to (1) and (2): ( )

)

An automobile starting from rest speeds up to 40 ft/sec with a constant acceleration of 4 ft/sec2 run at this speed for a time and finally comes to rest with a deceleration of 5 ft/sec2. If the total distance travelled is 1000 ft, find the total time required.

SOLUTION:

(

( )(

)

)( )

( )( )

The velocity of a particle moving along the x-axis is defined by v=kx3-4x2+6x where v is in m/s and x = meter and k = 1. Compute the acceleration when x = 2m.

SOLUTION:

When k =1 ; ( )

(

)(

)

When x = 2 m

( )

( )

( )

( )

( )

a=

; when t = 2 sec; v = 36 m/sec; s = 30 m. Determine s at t = 3 sec.

SOLUTION:

∫

∫

When v = 36 m/sec; t = 2 sec (

)

Therefore,

Or

( )

∫

∫

When s = 30 m; t = 2 sec ( )

Therefore,

When t = 3 sec ( )