lista6_solucao

8/18/2019 lista6_solucao

1/6

AA

1, 8x107kg

θ

24o

AA

θmax = µ = 0, 63 = 32o

θ

θmax

F

F + mg sin θ − f emax = 0

F e −mgcosθ = 0

f emax = µeF N

θ = 24o

m = 1, 8.107kg

F = mg(µe cos θ − sin θ) = 3, 0.107N

12N 5, 0N

0, 60

0, 40

F

F n

f

F n

f

µsF n f < µsF n f > µsF n

F −F n = 0

F n = F = 12N µsF n = (0, 60)(12N ) = 7, 2N

f − mg = 0

f = mg = 5, 0N

f < µsF n

f = 5, 0N

F n = 12N

F w = −F n î + f ̂ j = −(12, 0N )̂i + (5, 0N )ˆ j

5, 00

3, 00

30o

F = may F cos30, 0

o = mg

F = mgcos30,0o F sin30, 0o = mv

2

R

v =

Rg tan θ =

(5, 50m)(9, 8m/s2)tan30, 0o = 5, 58m/s


2/6

T = 2πRv = 2π(5,50)

5,58 = 6, 19s

F = ma

0.320m

40.0rev/min

0, 150m

60, 0rev/min

µs = v2

Rg

v = 2πRT µs = 4π2RT 2g

40, 0rev/min

1, 50s

µs = 4π2(0,150)(1,502)(9,8)

= 0, 269

(0, 150) 4060 = 0, 067m

0, 600metros

F y = may ⇒ n + mg = mv2

R

n → 0

n → 0

mg = m v2

R v =√

gR = 2, 42m/s

160km/h

310km/h

vt =

2mgcρA

A = 2mg

cρV 2 e a =

2mg

cρv2

V = 310km/h

v = 160km/h

A

a =

2mgcρV 2

2mgcρv2

= (V

v )2 = (

310

160)2 = 3, 75

1000Km/h 10Km

0.38Kg/m3

10Km 0.67Kg/m3

5.0Km

DaDh

=12

cρaAv2a

12

cρhAv2h=

ρaρh

.v2av2h

= 0, 38

0, 67.

10002

5002 = 2, 26

711N

0, 25 θ 30

o

T 1

T 2

θ = 30o)

F n P a (P a = ma.g) P b (P b = mb.g = 711N )


3/6

2a

4

T 1 − F smax = 0

F n − P b = 0

T 2 cos θ − T 1 = 0

T 2 sin θ − P a = 0

µs = 0, 25 P a = 103N =1, 0.102N

m1 = 40kg

m2 = 10kg

0, 60 0, 40

100N

100N

F Ns

F Nb ms mb

2a

−f = msas

F Ns − F Nb −msg = 0

f − F = mbab (2)

F Nb −mbg = 0

µsF Nb = µsmbg = (0, 60)(10kg)(9, 8m/s2) = 59N

as = ab a f

f = msF

ms + mb = (40kg)(100N )

40kg + 10kg = 80N

f smax

f = µkF Nb (2)

ab = µkmbg − F

mb=

(0, 40)(10kg)(9, 8, /s2)− 100N 10kg

= −6, 1m/s2

ab = (−6, 1m/s2

)̂i

as = µkmbg − F

ms=

(0, 40)(10kg)(9, 8, /s2)− 100N 40kg

= −0, 98m/s2

as = (−0, 98m/s2)̂i

800m 9, 80m/s2

3, 70m/s2

arad = g

T =2π

Rg

= 2π

4009,80

= 40, 1s

(60s/min)(40, 1s) = 1, 5rev/min

T = (1, 5rev/min)

3,709,8 =

0, 92rev/min

1125kg

2250kg

225metros

65mi/h

65, 0mi/h


4/6

tan β = (29,1m/s)2

(9,80,/s2)(225m) e β = 21, 0o

tan β

ncarro = (1125kg)(9,80m/s2)

cos21,0o = 1, 18.104 e npickup =

2ncarro = 2, 36.104N.mpickup = 2mcarro

θ = 40, 0o

µk = 0.400

m = 85.0Kg

A = 1.30m2

C = 0.150

1, 20Kg/m3

X :

F ext = ma Y :

F ext = ma

N − P cos θ = m(0)

N = P cos θ

−F at −D + P sin θ = m(0)

0

D = P sin θ − µkN

D = P (sin θ − µk cos θ)

1

2cρAv2 = P (sin θ − µk cos θ)

V t =

2P (sin θ − µk cos θ)

cρA

V t =

2(85.9, 8)(sin 40o − 0, 4cos40o)

0, 150.1, 20.1, 30 = 49m/s

D = 12

CρAv2 20Kg

0.040m2

0.80

1.21Kg/m3

0.80

10m

2.00

F = 156, 8N

D = F

F = 12cρAv2

V t =

2F

cρA =

2(156, 8)

(0, 80)(1, 21)(0, 040) = 90m/s = 3, 2.102m/s

V = 2V t = 2(3, 2.102) = 6, 4.102m/s

2, 6.102

D = 12

CρAv2

1300Km/h

60m/s

V t =

2F gcρA F g

cρA

cρA = 2mgV 2t

D = 12cρAv2 = 12 (2mgV 2t )v2 = mg( vV t )2

F ext = M a

|a| = D2m

= g

2(

v

V t)2 =

g

2(

360

60 )2 = 18g

0, 500m/s2

m1 m2 m3

T 23 T 12 2a

T 12 −m1g = m1a T 23 − µkm2g − T 12 = m2a


5/6

m3g − T 23 = m3a

m1 = M, m2 = 2M e m3 = 2M

2Mg − 2µkM g −M g = 5M a

a = 0, 500m/s2

µk = 0, 372

44N 22N

µs 0, 20

µk 0, 15

F N P ac P b

2a

X : T − f = 0 → f = T

Y : F N − P ac = 0 → F N = P ac

2a

P b − T = 0 T = P b f = T = P b

µF N P b <musP ac P ac

P ac = P bµs

= 22N

0, 20 = 110N

44N

(110− 44)N =66N

T − f = ( P ag

)a

F N − P a = 0

P b − T = ( P bg

)a

f = µkF k, e F N = P a f = µkP a

T = P b − (P bg )a

P b − ( P bg

)a− µkP a = ( P ag

)a

a = g(P b−µkP a)P a+P b = (9,8m/s2)(22N −(0,15)(44N ))

44N +22N = 2, 3m/s2

m = 16kg e M = 88kg

µs = 0, 38

f s f s = f smax = µsF , em que µs = 0, 38


6/6

2a

F = M totala → a = F m + M

2a

F − F

= ma → F

= F − m F

m + M

f s −mg = 0 → µsF −mg = 0

F = mg

µs(1− mm+M )

F = 4, 9.102

0, 40

2, 1m

49kg

F s −M G = m(0) F s µsF N µsF N − MG = 0 ⇒ F N = mgµs

−F N =

m(−v2)/ R F N

v =

(gR/µs) = 7, 2m/s

F N = mv2

R = 1200N

Documents

lista6_solucao