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Análise Dinâmica de Dois Blocos em Inclinação, Exercícios de Engenharia Elétrica

Faculdade Satc (FASATC)Engenharia Elétrica

O análise dinâmica de dois blocos colocados em uma pendícula. Os diagramas de equilíbrio livre para cada bloco são fornecidos, juntamente com as massas, forças de fricção e tensão. Aplicando as leis de newton no sentido das direções x e y de cada bloco, obtemos quatro equações que descrevem completamente o sistema dinâmico, desde que as acelerações dos blocos sejam as mesmas e t > 0. As equações resultam em uma aceleração de 3.5 m/s² e em uma tensão de 0.21 n.

Tipologia: Exercícios

Antes de 2010

Compartilhado em 08/10/2007

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bg1
23. The free-body diagrams for the two blocks are shown below. Tis the magnitude of the tension force of
the string,
NAis the normal force on block A(the leading block),
NBis the the normal force on block
B,
fAis kinetic friction force on block A,
fBis kinetic friction force on block B. Also, mAis the mass
of block A(where mA=WA/g and WA=3.6N),andmBis the mass of block B(where mB=WB/g
and WB=7.2 N). The angle of the incline is θ=30
◦.
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•
mAg
T
fA
NA
θ
A.
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•
mBg
T
fB
NB
θ
B
For e ac h bl oc k we take + xdownhill (which is toward the lower-left in these diagrams) and +yin the
direction of the normal force. Applying Newton’s second law to the xand ydirections of first block A
and next block B, we arrive at four equations:
WAsin θ−fA−T=mAa
NA−WAcos θ=0
WBsin θ−fB+T=mBa
NB−WBcos θ=0.
which, when combined with Eq. 6-2 (fA=µkA
NAwhere µkA =0.10 and fB=µkB
NBwhere µkB =
0.20), fully describe the dynamics of the system so long as the blocks have the same acceleration and
T>0.
(a) These equations lead to an acceleration equal to
a=gsin θ−µkA
WA+µkB
WB
WA+WBcos θ=3.5m/s2.
(b) We solve the above equations for the tension and obtain
T=WAWB
WA+WB(µkB −µkA
)cosθ=0.21 N .
Simply returning the value for afound in part (a) into one of the above equations is certainly fine,
and probably easier than solving for Talgebraically as we have done, but the algebraic form does
illustrate the µkB −µkA factor which aids in the understanding of the next part.
(c) Reversing the blocks is equivalent to switching the labels (so Ais now the block of weight 7.2Nand
µkA is now the 0.20 value). We see from our algebraic result in part (b) that this gives a negative
value fo r T, which is impossible. We conclude that the above set of four equations are not valid in
this circumstance (specifically, afor one block is not equal to afor the other block). The blocks
move independently of each other.

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Baixe Análise Dinâmica de Dois Blocos em Inclinação e outras Exercícios em PDF para Engenharia Elétrica, somente na Docsity!

23. The free-body diagrams for the two blocks are shown below. T is the magnitude of the tension force of

the string,

N

A

is the normal force on block A (the leading block),

N

B

is the the normal force on block

B,

f

A

is kinetic friction force on block A,

f

B

is kinetic friction force on block B. Also, m

A

is the mass

of block A (where m

A

= W

A

/g and W

A

= 3.6 N ), andm

B

is the mass of block B (where m

B

= W

B

/g

and W

B

= 7.2 N). The angle of the incline is θ = 30

m

A

g

T

f

A

N

A

A

m

B

g

T

f

B

N

B

B

For each block we take +x downhill (which is toward the lower-left in these diagrams) and +y in the

direction of the normal force. Applying Newton’s second law to the x and y directions of first block A

and next block B, we arrive at four equations:

W

A

sin θ − f

A

− T = m

A

a

N

A

− W

A

cos θ = 0

W

B

sin θ − f

B

+ T = m

B

a

N

B

− W

B

cos θ = 0.

which, when combined with Eq. 6-2 (f

A

k A

N

A

where μ

k A

= 0.10 and f

B

k B

N

B

where μ

k B

0 .20), fully describe the dynamics of the system so long as the blocks have the same acceleration and

T > 0.

(a) These equations lead to an acceleration equal to

a = g

sin θ −

k A

W

A

k B

W

B

W

A

+ W

B

cos θ

= 3.5 m/s

(b) We solve the above equations for the tension and obtain

T =

W

A

W

B

W

A

+ W

B

k B

k A

) cos θ = 0.21 N.

Simply returning the value for a found in part (a) into one of the above equations is certainly fine,

and probably easier than solving for T algebraically as we have done, but the algebraic form does

illustrate the μ

k B

k A

factor which aids in the understanding of the next part.

(c) Reversing the blocks is equivalent to switching the labels (so A is now the block of weight 7.2 N and

k A

is now the 0.20 value). We see from our algebraic result in part (b) that this gives a negative

value for T , which is impossible. We conclude that the above set of four equations are not valid in

this circumstance (specifically, a for one block is not equal to a for the other block). The blocks

move independently of each other.