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Compute the moment MO of the 1.2-kN force about the axis O-O.

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እነዚህን ጥያቄዎች እንፍታ።

4) የ1.2-kN ኃይል በ O-O ዘንግ ዙሪያ የሚያመነጨውን ሞመንት MO ያሰሉ።

  1. መጋጠሚያ ስርዓት (coordinate system) እና የኃይሉን አቀማመጥ ቬክተር (position vector) ይግለጹ።

    የ O-O ዘንግን እንደ x-ዘንግ እንውሰድ፣ መነሻው (origin) በቀኝ በኩል ባለው O ላይ ይሁን።
    የኃይሉ አተገባበር ነጥብ P መጋጠሚያዎች:
    xP=200 mm=0.2 mx_P = 200 \text{ mm} = 0.2 \text{ m}
    yP=105 mm=0.105 my_P = 105 \text{ mm} = 0.105 \text{ m}
    zP=300 mm=0.3 mz_P = 300 \text{ mm} = 0.3 \text{ m}
    ስለዚህ፣ የአቀማመጥ ቬክተር r⃗=(0.2i^+0.105j^+0.3k^)\vec{r} = (0.2 \hat{i} + 0.105 \hat{j} + 0.3 \hat{k}) m ይሆናል።

  2. ኃይሉን ወደ ክፍሎቹ (components) ይከፋፍሉ።

    የኃይሉ መጠን F=1.2 kNF = 1.2 \text{ kN} ነው።
    ከ y-ዘንግ ጋር 60° እና በ x-z አውሮፕላን ውስጥ ካለው x-ዘንግ ጋር 40° ይሰራል።
    Fx=Fsin⁡(60∘)cos⁡(40∘)=1.2sin⁡(60∘)cos⁡(40∘)=1.2×0.8660×0.7660=0.795 kNF_x = F \sin(60^\circ) \cos(40^\circ) = 1.2 \sin(60^\circ) \cos(40^\circ) = 1.2 \times 0.8660 \times 0.7660 = 0.795 \text{ kN}
    Fy=Fcos⁡(60∘)=1.2×0.5=0.6 kNF_y = F \cos(60^\circ) = 1.2 \times 0.5 = 0.6 \text{ kN}
    Fz=Fsin⁡(60∘)sin⁡(40∘)=1.2sin⁡(60∘)sin⁡(40∘)=1.2×0.8660×0.6428=0.669 kNF_z = F \sin(60^\circ) \sin(40^\circ) = 1.2 \sin(60^\circ) \sin(40^\circ) = 1.2 \times 0.8660 \times 0.6428 = 0.669 \text{ kN}
    ስለዚህ፣ የኃይል ቬክተር F⃗=(0.795i^+0.6j^+0.669k^)\vec{F} = (0.795 \hat{i} + 0.6 \hat{j} + 0.669 \hat{k}) kN ይሆናል።

  3. በ O-O ዘንግ ዙሪያ ያለውን ሞመንት ያሰሉ።

    በ O-O ዘንግ (x-ዘንግ) ዙሪያ ያለው ሞመንት MO−O=(r⃗×F⃗)xM_{O-O} = (\vec{r} \times \vec{F})_x ነው።

    r⃗×F⃗=∣i^j^k^0.20.1050.30.7950.60.669∣\vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0.2 & 0.105 & 0.3 \\ 0.795 & 0.6 & 0.669 \end{vmatrix}

    የ x-ክፍል (x-component) የሚከተለው ነው፦

    MO−O=(0.105)(0.669)−(0.3)(0.6)M_{O-O} = (0.105)(0.669) - (0.3)(0.6) MO−O=0.070245−0.18M_{O-O} = 0.070245 - 0.18 MO−O=−0.109755 kN⋅mM_{O-O} = -0.109755 \text{ kN}\cdot\text{m}

  4. የመጨረሻውን መልስ ይግለጹ።

    MO−O=−0.110 kN⋅mM_{O-O} = \boxed{-0.110 \text{ kN}\cdot\text{m}}

    5) በ O ላይ ያለውን የኃይል-ጥንድ ስርዓት (force-couple system) ያግኙ። R ለ Mo ቀጥ ያለ ነው?

  5. መጋጠሚያ ስርዓት እና የኃይሎቹን አቀማመጥ ቬክተሮች ይግለጹ።

    መነሻው በ O ላይ ነው። x-ዘንግ በ AOB ዘንግ ላይ ነው።
    ለ F1=600 NF_1 = 600 \text{ N} በ A ላይ: r⃗A=(−80i^) mm=(−0.08i^) m\vec{r}_A = (-80 \hat{i}) \text{ mm} = (-0.08 \hat{i}) \text{ m}
    ለ F2=800 NF_2 = 800 \text{ N} በ B ላይ: r⃗B=(160i^) mm=(0.16i^) m\vec{r}_B = (160 \hat{i}) \text{ mm} = (0.16 \hat{i}) \text{ m}

  6. ኃይሎቹን ወደ ክፍሎቻቸው ይከፋፍሉ።

    ለ F1=600 NF_1 = 600 \text{ N}: በ x-z አውሮፕላን ውስጥ ከ x-ዘንግ ጋር 30° ይሰራል።
    F1x=−600cos⁡(30∘)=−600×32=−519.62 NF_{1x} = -600 \cos(30^\circ) = -600 \times \frac{\sqrt{3}}{2} = -519.62 \text{ N}
    F1y=0 NF_{1y} = 0 \text{ N}
    F1z=600sin⁡(30∘)=600×12=300 NF_{1z} = 600 \sin(30^\circ) = 600 \times \frac{1}{2} = 300 \text{ N}
    F⃗1=(−519.62i^+300k^)\vec{F}_1 = (-519.62 \hat{i} + 300 \hat{k}) N

    ለ F2=800 NF_2 = 800 \text{ N}: በ x-y አውሮፕላን ውስጥ ከ x-ዘንግ ጋር 45° ይሰራል።
    F2x=800cos⁡(45∘)=800×22=565.69 NF_{2x} = 800 \cos(45^\circ) = 800 \times \frac{\sqrt{2}}{2} = 565.69 \text{ N}
    F2y=−800sin⁡(45∘)=−800×22=−565.69 NF_{2y} = -800 \sin(45^\circ) = -800 \times \frac{\sqrt{2}}{2} = -565.69 \text{ N}
    F2z=0 NF_{2z} = 0 \text{ N}
    F⃗2=(565.69i^−565.69j^)\vec{F}_2 = (565.69 \hat{i} - 565.69 \hat{j}) N

  7. አጠቃላይ ኃይል (resultant force) R⃗\vec{R} ያሰሉ።

    R⃗=F⃗1+F⃗2\vec{R} = \vec{F}_1 + \vec{F}_2 R⃗=(−519.62i^+300k^)+(565.69i^−565.69j^)\vec{R} = (-519.62 \hat{i} + 300 \hat{k}) + (565.69 \hat{i} - 565.69 \hat{j}) R⃗=(46.07i^−565.69j^+300k^) N\vec{R} = (46.07 \hat{i} - 565.69 \hat{j} + 300 \hat{k}) \text{ N}

  8. በእያንዳንዱ ኃይል በ O ዙሪያ የሚፈጠረውን ሞመንት ያሰሉ።

    ለ r⃗A×F⃗1\vec{r}_A \times \vec{F}_1:

    r⃗A×F⃗1=(−0.08i^)×(−519.62i^+300k^)\vec{r}_A \times \vec{F}_1 = (-0.08 \hat{i}) \times (-519.62 \hat{i} + 300 \hat{k}) =(−0.08)(300)(i^×k^)=−24(−j^)=24j^ N⋅m= (-0.08)(300) (\hat{i} \times \hat{k}) = -24 (-\hat{j}) = 24 \hat{j} \text{ N}\cdot\text{m}

    ለ r⃗B×F⃗2\vec{r}_B \times \vec{F}_2:

    r⃗B×F⃗2=(0.16i^)×(565.69i^−565.69j^)\vec{r}_B \times \vec{F}_2 = (0.16 \hat{i}) \times (565.69 \hat{i} - 565.69 \hat{j}) =(0.16)(−565.69)(i^×j^)=−90.51k^ N⋅m= (0.16)(-565.69) (\hat{i} \times \hat{j}) = -90.51 \hat{k} \text{ N}\cdot\text{m}

  9. አጠቃላይ ሞመንት (resultant moment) M⃗O\vec{M}_O ያሰሉ።

    M⃗O=(r⃗A×F⃗1)+(r⃗B×F⃗2)\vec{M}_O = (\vec{r}_A \times \vec{F}_1) + (\vec{r}_B \times \vec{F}_2) M⃗O=(24j^−90.51k^) N⋅m\vec{M}_O = (24 \hat{j} - 90.51 \hat{k}) \text{ N}\cdot\text{m}

  10. R⃗\vec{R} ለ M⃗O\vec{M}_O ቀጥ ያለ መሆኑን ያረጋግጡ።

    ሁለት ቬክተሮች ቀጥ ያሉ ከሆኑ፣ የነሱ የነጥብ ብዜት (dot product) ዜሮ መሆን አለበት።

    R⃗⋅M⃗O=(46.07i^−565.69j^+300k^)⋅(0i^+24j^−90.51k^)\vec{R} \cdot \vec{M}_O = (46.07 \hat{i} - 565.69 \hat{j} + 300 \hat{k}) \cdot (0 \hat{i} + 24 \hat{j} - 90.51 \hat{k}) R⃗⋅M⃗O=(46.07)(0)+(−565.69)(24)+(300)(−90.51)\vec{R} \cdot \vec{M}_O = (46.07)(0) + (-565.69)(24) + (300)(-90.51) R⃗⋅M⃗O=0−13576.56−27153\vec{R} \cdot \vec{M}_O = 0 - 13576.56 - 27153 R⃗⋅M⃗O=−40729.56 N⋅m2\vec{R} \cdot \vec{M}_O = -40729.56 \text{ N}\cdot\text{m}^2

    R⃗⋅M⃗O≠0\vec{R} \cdot \vec{M}_O \neq 0 ስለሆነ፣ R⃗\vec{R} ለ M⃗O\vec{M}_O ቀጥ ያለ አይደለም።

  11. የመጨረሻውን መልስ ይግለጹ።

    የኃይል-ጥንድ ስርዓት በ O ላይ:

    • የተጣመረ ኃይል: R⃗=(46.1i^−566j^+300k^) N\boxed{\vec{R} = (46.1 \hat{i} - 566 \hat{j} + 300 \hat{k}) \text{ N}}
    • የተጣመረ ሞመንት: M⃗O=(24.0j^−90.5k^) N⋅m\boxed{\vec{M}_O = (24.0 \hat{j} - 90.5 \hat{k}) \text{ N}\cdot\text{m}}
    • R⃗\vec{R} ለ M⃗O\vec{M}_O ቀጥ ያለ ነው? አይደለም\boxed{\text{አይደለም}}
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