Ramirez King
03/10/2024 · High School
4. Jism qiya tekislikda 15 N kuch ta'sirida ko'tarildi. Jism og'irligi 16 N qiya tekislik balandligi 5 m , uzunligi \( 6,4 \mathrm{~m} \). Qiya tekislikning FIKini toping
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Qiya tekislikning FIK (friksion koeffitsienti) ni topish uchun avval jismning qiya tekislikda ko'tarilishida ta'sir etuvchi kuchlarni tahlil qilishimiz kerak.
1. **Jismning og'irligi**: \( W = 16 \, \text{N} \)
2. **Ko'tarilish kuchi**: \( F = 15 \, \text{N} \)
3. **Qiya tekislik balandligi**: \( h = 5 \, \text{m} \)
4. **Qiya tekislik uzunligi**: \( L = 6.4 \, \text{m} \)
Qiya tekislikning burchagini topish uchun Pifagor teoremasidan foydalanamiz:
\[
\sin(\theta) = \frac{h}{L} = \frac{5}{6.4}
\]
Burchakni hisoblash:
\[
\theta = \arcsin\left(\frac{5}{6.4}\right) \approx 0.6947 \, \text{rad} \approx 39.8^\circ
\]
Jismning og'irligi qiya tekislikda quyidagi komponentlarga bo'linadi:
- **Tenglama**: \( W_{\perp} = W \cdot \cos(\theta) \)
- **Qiya tekislik bo'ylab og'irlik kuchi**: \( W_{\parallel} = W \cdot \sin(\theta) \)
Bu yerda \( W_{\perp} \) normal kuch, \( W_{\parallel} \) esa qiya tekislik bo'ylab og'irlik kuchidir.
Hisoblaymiz:
\[
W_{\perp} = 16 \cdot \cos(39.8^\circ) \approx 16 \cdot 0.7771 \approx 12.43 \, \text{N}
\]
\[
W_{\parallel} = 16 \cdot \sin(39.8^\circ) \approx 16 \cdot 0.6340 \approx 10.14 \, \text{N}
\]
Endi, qiya tekislikda jismni ko'tarish uchun ta'sir etuvchi kuchlar:
\[
F - W_{\parallel} - F_{friks} = 0
\]
Bu yerda \( F_{friks} \) friksion kuchdir. Shunday qilib, friksion kuchni topamiz:
\[
F_{friks} = F - W_{\parallel} = 15 - 10.14 \approx 4.86 \, \text{N}
\]
Friksion kuchni topgach, FIK ni hisoblash uchun normal kuchni bilishimiz kerak:
\[
FIK = \frac{F_{friks}}{W_{\perp}} = \frac{4.86}{12.43} \approx 0.39
\]
Natijada, qiya tekislikning friksion koeffitsienti (FIK) taxminan \( 0.39 \) ga teng.
Quick Answer
Qiya tekislikning friksion koeffitsienti (FIK) taxminan 0.39 ga teng.
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