Quiz03 Div1 Soln
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ME 274 – Spring 2009 Quiz No. 3 – Div 1 Monday, February 2
SOLUTION
The mechanism shown below is made up of links OA, AB and BC. Links OA and BC are pinned to ground at point O and C, respectively. Link AB joins links OA and BC at pin joints A and B, respectively. Pins O and C are on the same horizontal line. At the instant shown, • link AB is horizontal and BC is vertical. • link BC is rotating in the CCW sense at a constant rate of ωBC = 12 rad/sec. At the instant shown, a) find the angular velocities of links OA and AB. Write your answers as vectors. b) find the angular accelerations of links OA and AB. Write your answers as vectors. c) 5 BONUS POINTS -- make an accurate sketch of the direction for the velocity of point D on link AB (HINT: consider the location of the instant center for link AB.)
ICAB
VA
VD D
A 300 mm
y
B
VB
500 mm
ωBC
x O
C
Velocity analysis
( ) OA : v A = vO + ! OA " r A/O = 0 + (! OA k ) " ( 400i + 300 j ) = #300! OA i + 400! OA j BC : v B = vC + ! BC " r B/C = 0 + (! BC k ) " 300 j = #300! BC i
AB : v B = v A + ! AB " r B/ A
#
(
)
!300" BC i = !300" OA i + 400" OA j + (" AB k ) # ( 300i ) = ( !300" OA ) i + ( 400" OA + 300" AB ) j
i:
! 300" BC = !300" OA
#
j:
0 = 400" OA + 300" AB
#
300 mm
$
" OA = " BC = 12 rad / sec
" AB = ! ( 4 / 3) " OA = !16 rad / sec
Acceleration analysis
(
)
2 2 2 BC : a B = aC + ! BC " r B/C # $ BC r B/C = 0 + 0 # $ BC 300 j = #300$ BC j 2 OA : a A = aO + ! OA " r A/O # $ OA r A/O
(
)
(
2 = 0 + (! OA k ) " 400i + 300 j # $ OA 400i + 300 j
(
) (
)
)
2 2 = #300! OA # 400$ OA i + 400! OA # 300$ OA j
2 AB : a B = a A + ! AB " r B/ A # $ AB r B/ A
( = ( !300#
) (
%
)
2 2 2 2 !300" BC j = !300# OA ! 400" OA i + 400# OA ! 300" OA j + (# AB k ) $ ( 300i ) ! 300" AB i
i:
OA
) (
)
2 2 2 ! 400" OA ! 300" AB i + 400# OA ! 300" OA + 300# AB j
2 2 0 = !300" OA ! 400# OA ! 300# AB
%
2 2 $ " OA = ! ( 4 / 3)# OA ! # AB = !448 rad / sec 2
2 2 2 2 j : ! 300# BC = 400" OA ! 300# OA + 300" AB $ " AB = !# BC + # OA ! ( 4 / 3)" OA = 597.3 rad / sec 2
Answers:
! OA = ! OA k = (12k ) rad / sec
! AB = ! AB k = ( "16k ) rad / sec
# OA = # OA k = ( "448 k ) rad / sec 2
# AB = # AB k = ( 597.3k ) rad / sec 2
SOLUTION
The mechanism shown below is made up of links OA, AB and BC. Links OA and BC are pinned to ground at point O and C, respectively. Link AB joins links OA and BC at pin joints A and B, respectively. Pins O and C are on the same horizontal line. At the instant shown, • link AB is horizontal and BC is vertical. • link BC is rotating in the CCW sense at a constant rate of ωBC = 12 rad/sec. At the instant shown, a) find the angular velocities of links OA and AB. Write your answers as vectors. b) find the angular accelerations of links OA and AB. Write your answers as vectors. c) 5 BONUS POINTS -- make an accurate sketch of the direction for the velocity of point D on link AB (HINT: consider the location of the instant center for link AB.)
ICAB
VA
VD D
A 300 mm
y
B
VB
500 mm
ωBC
x O
C
Velocity analysis
( ) OA : v A = vO + ! OA " r A/O = 0 + (! OA k ) " ( 400i + 300 j ) = #300! OA i + 400! OA j BC : v B = vC + ! BC " r B/C = 0 + (! BC k ) " 300 j = #300! BC i
AB : v B = v A + ! AB " r B/ A
#
(
)
!300" BC i = !300" OA i + 400" OA j + (" AB k ) # ( 300i ) = ( !300" OA ) i + ( 400" OA + 300" AB ) j
i:
! 300" BC = !300" OA
#
j:
0 = 400" OA + 300" AB
#
300 mm
$
" OA = " BC = 12 rad / sec
" AB = ! ( 4 / 3) " OA = !16 rad / sec
Acceleration analysis
(
)
2 2 2 BC : a B = aC + ! BC " r B/C # $ BC r B/C = 0 + 0 # $ BC 300 j = #300$ BC j 2 OA : a A = aO + ! OA " r A/O # $ OA r A/O
(
)
(
2 = 0 + (! OA k ) " 400i + 300 j # $ OA 400i + 300 j
(
) (
)
)
2 2 = #300! OA # 400$ OA i + 400! OA # 300$ OA j
2 AB : a B = a A + ! AB " r B/ A # $ AB r B/ A
( = ( !300#
) (
%
)
2 2 2 2 !300" BC j = !300# OA ! 400" OA i + 400# OA ! 300" OA j + (# AB k ) $ ( 300i ) ! 300" AB i
i:
OA
) (
)
2 2 2 ! 400" OA ! 300" AB i + 400# OA ! 300" OA + 300# AB j
2 2 0 = !300" OA ! 400# OA ! 300# AB
%
2 2 $ " OA = ! ( 4 / 3)# OA ! # AB = !448 rad / sec 2
2 2 2 2 j : ! 300# BC = 400" OA ! 300# OA + 300" AB $ " AB = !# BC + # OA ! ( 4 / 3)" OA = 597.3 rad / sec 2
Answers:
! OA = ! OA k = (12k ) rad / sec
! AB = ! AB k = ( "16k ) rad / sec
# OA = # OA k = ( "448 k ) rad / sec 2
# AB = # AB k = ( 597.3k ) rad / sec 2
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