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Unformatted text preview: uu ‘ 
‘  “I. w "15 Uulluuav u: viableI‘ll] uu. aw.” i
, i ii Dete i _its speed when! = 4 5 if the horizontal traction
F var s 1th time as shown in the graph. 2 o' .1:
(w mlu.).+2J F,d1=m(u. )2
l ,, :0+j:(40.01:1)(10’)dt= 28(103)u i { ;v=0.564mls An, 1(5) .
l s g
1 l 1515. I The 20lb cabinet is subjected to the force F =
(3 + 2: lb, where t is in seconds if the cabinet is initially movin down the plane with a velocity of 6 ft/s, determine
how long it will take the force to bring the cabinet to rest.
F always acts parallel to the plane. Neglect the size of the rollers. i .i ' ? Principle of Linear Impulse and Momentum : Applying Eq.154, we have
'2 m(u,.),+zj Edt=m(u,.)z
'i ' ; 2 («33%)(4) +J; (3+ 2t)dt— zosin 20°:= (3%)(0)
!= 4.64 s Ans 1523. Detemiine the velocities of blocks A and B 2 s
_ after they are released from 'rest. Neglect the mass of the
; pulleys and cables. r 7"
t ‘3 2‘] + 255 3 l
a 1
3' VA = 'VB ‘ 5 A
l
l
(+l) moltajmxwwyh
3 Solving.
1 2
o—r(2)+2(2) = (3.272)" r= 2.67 lb BlockB: v, =21.sn/sL An: (44,) md'y); {.sz dtgmwyh VA =_—2L5ftls=21.5ft/ST AM 1 0 +40) — 7(2) = (3%», 1535.» ‘lLe two blocks A and B each h 5 and are su‘spended from parallel cords :fPZTas: “,5 kg
‘ stiffness of k = 60 N/m, is attached to B and is can avlng a
g 0.3 m agai st A and B as shown. Determine the m Pressed
' 808165 '9 a d ¢ of the cords when the blocks are 3:22:31! from rest nd the spring becomes unstretched [—4) th‘l 3me 3'
‘5
'3
‘. ‘5 i
3‘ O+O=SVA+5VB : t'Ai = vs = v
ForA orB°
Just b fore eblocks ‘ '  ‘
Pl begin to nse . Dammmowest point
Tl + K = 75 + K T: + W = 7}1 + l5
1
5(5"°'7343)’ +0 = 0+5(9.8l)(2)(l —c059) 1 (0+0) 60 21 1 +2“ no.3) 5(5)<v)’+§(5>(v)'+o
9=¢=9.52° Ans v = 0.7348 lm/s is held at rest on the smooth
block at A. If the 10g bullet is t becomes embedded in the
he distan 10kg bl, ki determine t ce the block will slide
up along the plane before momentarily stopping. 1547. ei 10—kg block
inclined lane by the stop
travelin at 300 m/s 'when i : If we consider the block and the bullet
impact is Ltive force F caused by the
ut. Also. the weight of the bullet momentum is Linear Momentum
the impu
ill cancel 0
As the reSult. linear Conservation of
as a syStem. then from the FBD. internal to the system. Therefore. it w
and the block are nonimpulst've forces.
conserved along the x' axis. )’ =(mb+ml)vx ’"b ( ”I! 1
001000005 30°) = (0.01 + l0) 1) (+ l 1
v = 0.2595 m/s oclt' 5 initial position. When
y are it _abave the
l)h = 98.l98lh. Energy : The datum is set at the bl
is at their highest point the
is (10+0.0l) (9.8 Cans erv atilm of
the block and the embedded bullet datum. Them gravitational potential energy
Applying 311.14  21. we have n+w=n+w l
0+i(10+°'°1)(°‘25952) =0+98.1981h
h = 0.003433 m = 3.43 mm d = 3.43 / sin 30° = 6.87 mm Ans ...
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 Fall '08
 Hudyma
 Dynamics

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