Physicsiunit02_4p


    (Dynamics)


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SUT Physics I unit02

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SUT Physics I unit02

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ΣF = ma
ΣFx = ma x ΣF !' 8BC  >ก ΣFy = ma y m
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I (newton, N) khanchai

N = kg m / s 2

SUT Physics I unit02

   

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4

+%$ (a)

F =

G m1 m2 r2

=

(6.67 × 10-11 N.m2 /kg 2 )(7.3 kg)(7.3 kg)

= 8.4 × 10

(b)

F =

=

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(0.65 m) 2 −9

N

G m1 m2 r2

(6.67 × 10 −11 N ⋅ m 2 /kg 2 )(5.98 × 10 24 kg )(7.36 × 10 22 ) (3.82 × 108 m) 2

= 2.01 × 1020 N khanchai

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µ s        

SUT Physics I unit02

W =

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F = 2 × 1.40 × 105 N

ΣFy = 0 90 N30


fs W

2 ×1.40 ×105 N=m(2.3 m/s 2 ) ΣFx = 0

m = 1.22 × 105 kg ∴ W = mg = (1.22 × 105 kg )(9.8 m/s 2 ) = 1.19 × 10 6 N khanchai

f s = 90 N cos 30
3 µs R = 90 N × 2

R + (90 N ) sin 30
− W = 0 1 R + (90 N )  − (20 kg )(9.8 m/s 2 ) = 0 2 R = 196 − 45 N = 151 N

∴ µs =

90 N × 3 151 N × 2

= 0.52 khanchai

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F = ma  
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R



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Fx = ma x

W

Fx = 5.2 N m = 40 kg

ax =

SUT Physics I unit02


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5.2 N = 0.13 m/s 2 40 kg

(c) 15m − x

x

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6

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ax =

t2 15m − x

x


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1 1 x = (0)t + (0.13m / s 2 )t 2 15m − x = (0)t + (0.62m / s 2 )t 2 2 1 2 2 2 1 x = + (0.13m / s )t 15m − x = + (0.62m / s 2 )t 2 2 2

R

Fx = ma x Fx = 5.2 N m = 8.4 kg

t1

W

5.2 N = 0.62 m/s 2 8.4 kg khanchai

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v 2 = v02 + 2 as

SUT Physics I unit02

1 (1) (0.62m / s 2 )t 2 2 1 (2) x = (0.13m / s 2 )t 2 2 1 1 15m = (0.62m / s 2 )t 2 + (0.13m / s 2 )t 2 2 2 1 2 2 15m = (0.75m / s )t 2 1 15m × 2 x = + (0.13m / s 2 )t 2 2 t = 2 1 0.75m / s 2 x = (0.13m / s 2 )(6.32s ) 2 30 2 2 2 t = s 0.75 x = 2.59m t = ±6.32s t = 6.32s

15m − x =

ΣF = ma

a

m1 g − T = m1a m1g (110 kg )(9.8 m/s ) − T = (110 kg )a 2

T − m2 g = m2 a

T

T − (74 kg )(9.8 m/s 2 ) = (74 kg ) a m2g

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7

(110 kg )(9.8 m/s 2 ) − T = (110 kg )a

(1)

T − (74 kg )(9.8 m/s 2 ) = (74 kg ) a

(2)

(1) + (2) ;

(1) (2)

(110 kg − 74 kg )(9.8 m/s 2 ) = (184 kg ) a

a =

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1 15m − x = (0.62m / s 2 )t 2 2 1 x = (0.13m / s 2 )t 2 2 1 (0.62m / s 2 )t 2 15m − x 2 = 1 x (0.13m / s 2 )t 2 2

36 × 9.8 = 1.92 m/s 2 184

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v 2 = v02 + 2 as = 0 2 + 2(1.92 m/s 2 )(12 m) = 46.08 m 2 /s 2 v = 6.79 m/s (b)&
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x = 2.59m khanchai

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T − (5 kg) g sin 40
= (5 kg)a

(1) (2)

(1) + (2) ; (10kg) g − (5 kg) g sin 40
= (15 kg ) a SUT Physics I unit02

10 kg

(10 kg ) g − T = (10 kg )a

(10 kg)(9.8 m/s 2 ) − (5 kg)(9.8 m/s 2 )(0.643) 15 kg = 4.43 m/s 2

∴ a =

5 kg 40

o

(10 kg )(9.8 m/s 2 ) − T = (10 kg )(4.43 m/s 2 )

T = 53.7 N khanchai

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8


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ΣF = ma

T

W1 y

T 5 kg 40°

w2 cos 40

°

W2

N

x

w2 sin 40°

N

N cosθ

ΣFx = m2 a x T − m2 g sin θ = m2 a T − (5 kg) g sin 40
= (5 kg)a

N cos θ = mg θ

mg

N sin θ

ΣFy = m2 a y

ΣFy = 0 ΣFr = mar v2 r v2 (8.7 m/s) 2 tan θ = = = 0.31 gr (9.8 m/s 2 )(25 m) θ = arctan 0.31 N sin θ = m

= 17.2


N − m2 g cos θ = 0 N − (5 kg) g cos 40
= 0 khanchai

N cos (17.2
) = (85 kg )(9.8 m/s 2 ) = 833 N 833 N 833 N = 871.9 N N = = cos 17.2 0.955

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m1 g − T = m1a (10 kg) g − T = (10 kg )a

10 kg


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SUT Physics I unit02

mg


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ΣF = ma 7 kg

mg − R = ma (0.25kg )(9.8m / s 2 ) − R = (0.25kg )(9.2m / s 2 ) R = (0.25kg )(9.8 − 9.2)(m / s 2 )

R = 0.15 N khanchai

khanchai

9


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ΣF = ma

N T

4 kg

y

mg sin θ mg cos θ

θ

θ

x

ΣF = ma

m1 g

fs

mg mg sin θ = µs mg cos θ

ΣFy = 0

N = mg cos θ

7 kg

f s = µs N

θ = tan −1 0.04 = 2.3


(1)

ΣFy = m2 a m2 g − T = m2 a

(2)

(1)+(2) m2 g − µ k m1 g = (m1 + m2 )a

= 0.04

ΣFx = 0

mg sin θ = f s

µ s = tan θ

T

T − µ k m1 g = m1a

ΣFy = 0 N = m1 g Khanchai Khosonthongkee

fk

N

ΣFx = m1a T − f k = m1a T − µ k N = m1a

khanchai

m2 g

a=

( m2 − µ k m1 ) g (m1 + m2 ) khanchai

( m2 − µ k m1 ) g (m1 + m2 )

a=

(7 kg − (0.30)4kg )9.8m / s 2 (7 kg + 4kg )

SUT Physics I unit02

a=

a = 5.17 m / s 2 m2 g − T = m2 a

(2)

T = m2 ( g − a )

T = 7 kg (9.8 − 5.17)m / s 2

T = 32.41N

khanchai

10 Khanchai Khosonthongkee