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# 請幫我解兩題動力學題目

有兩題動力學課本上的題目我不會解

希望聰明的各位大大

能夠幫我解題

希望能夠列出詳細的運算方式

以便我能學習

拜託啦

有關題目如下:

第一題 圖片是一架飛機

The jet aircraft has a total mass of 22 Mg and a center of amss at G .

Initially at take-off engines provide a thrust 2T=4 kN and T= 1.5 kN.

Determine the acceleration of the plane and the normal reactions on the nose wheel and each of the two wing wheels located at B . Neglect the mass of the wheels and , due to low velocity , neglect any lift caused by the wings.

第二題 圖片是一台卡車

The pipe has a mass of 460 kg and is held in place on the truck bed using the two boards A and B Determine the greatest acceleration of the truck so that the ;I;e begins to lose contact at A and the bed of the truck and starts to pivot about B Assume board B will not slip on the bed of the truck, and the pipe is smooth. Also, what force does board B exert on the pipe during the acceleration ?

而圖片我會貼在我的部落格

會放在至頂的文章

標題是動力學的題目圖片

我的部落格網址如下:

http://tw.myblog.yahoo.com/k90187566

麻煩各位啦!!!

### 1 Answer

- LILv 71 decade agoFavorite Answer
Q1：

F = T’+2T = ma

1.5*103 + 4*103 = 22*103(kg)*a

a = 0.25

因為飛機不轉動

所以

ΣMB = 0

- mg*3 + 9*NA – T’(2.5-1.2) – 2T(2.3-1.2) = 0

NA= 72572 (N)

垂直方向上：

mg= NA + 2NB →兩翼，所以一邊有一個輪B

22000*9.8 = 72572+ 2NB

NB = 71514 (N)

Q2：(畢業很久了，沒把握)

先解決轉動慣量

管的面積 =π(R2 – r2)

密度ρ= 質量/面積 = m/[π(R2 – r2)]

IO = ∫r2 dm = ∫∫r2 (ρdr * rdθ) = ∫2π0∫0.5 0.4 r2 (ρdr * rdθ)

=ρ∫2π0∫R r r3 dr dθ

= 2πρ*1/4 *r4 (r=r~r=R)

=πρ/2*(R4 – r4)

=π/2*(R4 – r4) * m/π(R2 – r2)

= 1/2* m (R2 + r2)

= 1/2* m (0.52 + 0.42)

= 0.205m

依平行軸定理

IB = IO + (r)BO2*m = 0.205m + 0.52*m = 0.455m

圖片參考：http://tw.myblog.yahoo.com/jw!Y.4_oUuBAAV5T3qgsG.h...

http://tw.myblog.yahoo.com/jw!Y.4_oUuBAAV5T3qgsG.h...

自由體圖(左圖)

→A點恰好無施力

(1)ΣM0 = 0.4*Fx –0.3*Fy = Ioα=0.205mα

(2)ΣMB = - 0.3*mg + IBα=0

→α=0.3*mg/( 0.455m ) = 6.46

(3)水平方向上：Fx =m*(aO)x

(4)垂直方向上：Fy – mg =m*(aO)y

Fy = 9.8m + m*(aO)y

※ (aO) 為重心的加速度；下標x、y表示水平、垂直方向

自由體圖右圖 ，aB 是車子的加速度

自由體圖中圖，ao/B 是環的重心相對於 B點的加速度

aO = aB + ao/B →有底線者為向量

= ai + αk 外積r

= ai + 6.46k 外積(0.3i + 0.4j)

= ai - 2.58 i + 1.94j

= (a-2.58) i + 1.94j

= (aO)x i + (aO)y j

(aO)x = a-2.58 →代入(3)→Fx =m(a-2.58)

(aO)y = 1.94 →代入(4) → Fy = 9.8m + 1.94m = 11.74m

再代回(1)式

0.4* m(a-2.58) – 0.3* 11.74m = 0.205m *6.46

a = 14.7

Fx =m*(aO)x = 460 * (14.7-2.58) = 5575.2

Fy = 11.74m = 5400.4

F = √(Fx2 + Fy2 ) = 7762 (N)

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