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]);