mg x - ecourse2.ccu.edu.tw
TRANSCRIPT
HW4-1sol: Two blocks 1 and 2 rest on a frictionless horizontal surface. They are connected by three massless strings and two frictionless, massless pulleys as shown below. .
(a) Draw the free-body diagram for each block and each pulley.(b) Find the acceleration of the each block
Fig. 1(a)
2m g
2N T
2T
2T F1N
1m g
(b)
3T = m2a2!!!(1)F − 2T = m1a1!(2)2a1 = 3a2!!!(3)
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(1),(3)® 3T =
23
m2a1!!(4)
(2),(4)®
a1 =9F
9m1 + 4m2
a2 =6F
9m1 + 4m2
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��m1 ������m2�������2/3��m1�� m2�� �3/2�
x2 x1 x0 x ʹp
xp
xp − x2( ) + xp − x ʹp( ) + x0 − x ʹp( ) = ℓx1 − xp = a = constant → xp = x1 − a
x ʹp − x2 = b = constant → x ʹp = x2 +b
⇒ 2x1 − 3x2 = constant or 2!!x1 = 3!!x2
When the small block will leave the surface of the wedge. Also
−F − N sinθ = MaN g − N cosθ − Mg = 0 (no use)
N cosθ −mg = may
N sinθ = max
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M = 3m, θ = 37° ,cosθ = 4
5,sinθ = 3
5
F = 2mg →
a = −3142
g
ax =3
14g
ay = −57
g
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F ≥ 4mg
HW4-2sol: (a)
M
q
m
q
N
NF
Mgmg
gN
With the reference point O, the displacement of the wedge is atand the displacement of the small block is at . The free body diagrams for the two objects are: (x, y)
(X , 0)
O
!F
(X , 0) (x, y)
x
y
X − x
y
Note:
tanθ = yX − x
, and a = !!X ,ax = !!x,ay = !!y
such that tanθ =ay
a − ax
−F − N 35= 3ma→ a = − 1
3Fm−
315
Nm
N 45−mg = may → ay =
45
Nm− g
N 35= max → ax =
35
Nm
ay = tanθ (a − ax ) = 34
(a − ax ) = 45
Nm− g
N =
57
mg − 14
F⎛
⎝⎜
⎞
⎠⎟ ≥ 0 to keep the small block on the wedge.
(b)
F = 5mgN = 0
→
a = −53
g
ax = 0ay = −g
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Am Bm Cm
Am gBm g Cm g
TAF FT TCF
TAF TBF
TCF
FTA = FTB = FT
FTC = 2FTA = 2FT �pulley���
xA − xP + xB − xP = ℓ A
xP − x ʹP + xC − x ʹP = ℓ B
Constraint:
xA
aA = !!xA
xB
aB = !!xB
xC
aC = !!xC
FT −mAg = mAaA
FT −mB g = mBaB
2FT −mC g = mCaC
xP
x ʹP : fixed
→ xA + xB + 2xC = ℓ A − 2ℓ B + 4x ʹP = constant
or ""xA + ""xB + 2""xC = aA + aB + 2aC = 0
HW4-3sol:
→
aA = −9
49g
aB =1149
g
aC = −149
g
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(a)
(1) FT −mAg = mAaA
(2) FT −mB g = mBaB
(3) 2FT −mC g = mCaC
aA + aB + 2aC = 0
4 equations and 4 unknown: FT ,aA ,aB ,aC
(1)− (2)→ mB −mA( ) g = mAaA −mBaB
(1)+ (2)− (3)→ mC −mA −mB( ) g = mBaB +mAaA −mCaC
3aA − 2aB = −g3aA + 2aB − 5aC = 0aA + aB + 2aC = 0
FTC = 2FTA = 5(g + aC )
FTA =12049
g , FTC =24049
g
(b)
For numerical solution, you may just type>> syms a1 a2 a3 g>> S = solve([2*a2-3*a1==g,2*a2+3*a1-5*a3==0,a1+a2+2*a3==0], [a1,a2,a3]);