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Ans.a = 2a2t + a2n = 2(4)2 + (7.622)2 = 8.61 m>s2

an =yB

2

r=

82

8.396= 7.622 m>s2

r =C1 + (dydx)2 D3>2

`d 2y

dx2 `

4x=2 m

=C1 + (1.6)2 D3>2

|0.8|= 8.396 m

d2y

dx22x=2 m

= 0.8

dy

dx2x=2 m

= 0.8x 2x=2 m

= 1.6

y = 0.4 x2

12147. The box of negligible size is sliding down along acurved path defined by the parabola .When it is atA ( , ), the speed is and theincrease in speed is . Determine themagnitude of the acceleration of the box at this instant.

dvB>dt = 4 m>s2vB = 8 m>syA = 1.6 mxA = 2 m

y = 0.4x2y

x

2 m

y 0.4x2

A

Ans.a = 2a2t + a2n = 20 + (1.81)2 = 1.81 ft>s2

an =y2

r=

402

885.7= 1.81 ft>s2

r 4x=600 ft

=[1 + (dydx)

2 D3>2

`d2y

dx2 `

4x=600 ft

=[1 + (1.08)2]3>2

|3.6(10)-3|= 885.7 ft

d2y

dx22x=600 ft

= 6(10)-6x 2x=600 ft

= 3.6(10)-3

dy

dx2x=600 ft

= 3(10)-6x2 2x=600 ft

= 1.08

y = (10)-6x3

*12148. A spiral transition curve is used on railroads toconnect a straight portion of the track with a curvedportion. If the spiral is defined by the equation

, where x and y are in feet, determine themagnitude of the acceleration of a train engine moving witha constant speed of 40 when it is at point .x = 600 ftft>s

y = (10-6)x3

600 ft

v 40 ft/s

x

y

y (106)x3

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Time Derivatives:

When ,

Velocity:

Thus, the magnitude of the pins velocity is

Ans.

Acceleration:

Thus, the magnitude of the pins acceleration is

Ans. = 2a44u# 4 + u

$2

a = 2ar 2 + au 2 = 4C -22a A2u#2 + u

$B D2 + C22a Au

$- 2u

#2 B D2

au = ru$

+ 2r# u#= 22au

$+ 2 A -22au

#B Au

#B = 22a Au

$- 2u

#2 B

ar = r$ - ru

#2 = -22a Au

#2 + u

$B - 22au

#2 = -22a A2u

#2 + u

$B

v = 2vr 2 + vu 2 = 4A -22au#B2 + A22au

#B2 = 2au

#

vr = r# = -22au

# vu = ru# = 22au#r$ u = p4 = -2a 1

22 u#2 +

1

22 u$ = -22a Au# 2 + u$ B

r# u = p4 = -2a 1

22u# = -22au#

r u = p4 = 2a 122 = 22a

u =p

4 rad

r$ = -2a Ccos uu

#u#+ sin uu

#D = -2a Ccos uu

#2 + sin uu

#D

r# = -2a sin uu

#r = 2a cos u

12167. The slotted arm OA rotates counterclockwiseabout O such that when , arm OA is rotating withan angular velocity of and an angular acceleration of .Determine the magnitudes of the velocity and accelerationof pin B at this instant. The motion of pin B is constrainedsuch that it moves on the fixed circular surface and alongthe slot in OA.

u$

u# u = p>4

O

BA

a

u

r 2 a cos u

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Position Coordinates: By referring to Fig. a, the length of the two ropes written interms of the position coordinates , , and are

(1)

and

(2)

Eliminating from Eqs. (1) and (2) yields

Time Derivative: Taking the time derivative of the above equation,

Here, . Thus,

Ans.vB = -20 m>s = 20 m>s c

vB + 4(5) = 0

vA = 5 m>s

A + T B vB + 4vA = 0sB + 4sA = l1 - 2a + 2l2

sC

2sA - sC = l2

sA + (sA - sC) = l2

sB + 2sC = l1 - 2a

sB + 2a + 2sC = l1

sCsBsA

12198. If end A of the rope moves downward with a speedof , determine the speed of cylinder B.5 m>s

5 m/s

A

B

Position Coordinates: By referring to Fig. a, the length of the two cables written interms of the position coordinates are

(1)

and

(2)

Eliminating from Eqs. (1) and (2) yields

Time Derivative: Taking the time derivative of the above equation,

Here, . Thus,

Ans.vE = -2.14 m>s = 2.14 m>s c

7vE - C2 A -5 B D - A -5 B = 0

vA = vB = -5 m>s

A + T B 7vE - 2vA - vB = 07sE - 2sA - sB = 2l1 + l2

sC

3sE - sB - 2sC = l2

(sE - sB) + 2(sE - sC) = l2

2sE - sA + sC = l1

sE + (sE - sA) + sC = l1

12199. Determine the speed of the elevator if eachmotor draws in the cable with a constant speed of .5 m>s

B C

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Solution I

Vector Analysis: The velocity of the smoke as observed from the ship is equal to thevelocity of the wind relative to the ship. Here, the velocity of the ship and windexpressed in Cartesian vector form are

and .Applying the relative velocity equation,

Thus, the magnitude of is given by

Ans.

and the direction angle that makes with the x axis is

Ans.

Solution II

Scalar Analysis: Applying the law of cosines by referring to the velocity diagramshown in Fig. a,

Ans.

Using the result of and applying the law of sines,

Thus,

Ans.u = 45 + f = 74.0

sin f

10=

sin 7519.91 f = 29.02

vw/s

= 19.91 m>s = 19.9 m>s

vw>s = 2202 + 102 - 2(20)(10) cos 75

u = tan-1a19.145.482

b = 74.0

vw/su

vw = 2(-5.482)2 + (-19.14)2 = 19.9m>s

vw/s

vw>s = [-5.482i - 19.14j] m>s

8.660i - 5j = 14.14i + 14.14j + vw>s

vw = vs + vw>s

= [8.660i - 5j] m>svw = [10 cos 30 i - 10 sin 30 j]= [14.14i + 14.14j] m>svs = [20 cos 45 i + 20 sin 45 j] m>s

12218. The ship travels at a constant speed of and the wind is blowing at a speed of , as shown.Determine the magnitude and direction of the horizontalcomponent of velocity of the smoke coming from the smokestack as it appears to a passenger on the ship.

vw = 10 m>svs = 20 m>s

vs 20 m/s

vw 10 m/sy

x

3045

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