Turnbull O'Quinn
03/11/2023 · Primary 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
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
Qiya tekislikning FIK (friksion koeffitsienti) ni topish uchun avval jismning qiya tekislikda ko'tarilishi uchun zarur bo'lgan kuchlarni tahlil qilishimiz kerak.
1. **Jismning og'irligi**: \( W = 16 \, \text{N} \)
2. **Ko'tarish kuchi**: \( F = 15 \, \text{N} \)
3. **Qiya tekislikning balandligi**: \( h = 5 \, \text{m} \)
4. **Qiya tekislikning uzunligi**: \( L = 6.4 \, \text{m} \)
Qiya tekislikning burchagini topish uchun Pifagor teoremasidan foydalanamiz:
\[
\sin(\theta) = \frac{h}{L} = \frac{5}{6.4}
\]
Burchakni topish uchun:
\[
\theta = \arcsin\left(\frac{5}{6.4}\right)
\]
Endi, jismning og'irligi qiya tekislikda qanday ta'sir qilishini hisoblaymiz. Jismning og'irligi qiya tekislikda quyidagi komponentlarga bo'linadi:
- **Tenglama**: \( W_{\parallel} = W \cdot \sin(\theta) \) (qiya tekislik bo'ylab)
- **Tenglama**: \( W_{\perpendicular} = W \cdot \cos(\theta) \) (qiya tekislikka perpendikulyar)
Friksion kuch \( F_{friction} \) quyidagi tenglama bilan ifodalanadi:
\[
F_{friction} = \mu \cdot W_{\perpendicular}
\]
Bu yerda \( \mu \) - friksion koeffitsient.
Jismni ko'tarish uchun zarur bo'lgan kuch:
\[
F = W_{\parallel} + F_{friction}
\]
Bu tenglamani quyidagicha yozishimiz mumkin:
\[
F = W \cdot \sin(\theta) + \mu \cdot W \cdot \cos(\theta)
\]
Endi \( \mu \) ni topish uchun tenglamani qayta yozamiz:
\[
\mu = \frac{F - W \cdot \sin(\theta)}{W \cdot \cos(\theta)}
\]
Burchakni hisoblagandan so'ng, \( \sin(\theta) \) va \( \cos(\theta) \) ni topamiz:
\[
\sin(\theta) = \frac{5}{6.4} \approx 0.78125
\]
\[
\cos(\theta) = \sqrt{1 - \sin^2(\theta)} \approx \sqrt{1 - (0.78125)^2} \approx 0.623
\]
Endi \( \mu \) ni hisoblaymiz:
\[
\mu = \frac{15 - 16 \cdot 0.78125}{16 \cdot 0.623}
\]
Hisoblaymiz:
\[
\mu = \frac{15 - 12.5}{9.968} \approx \frac{2.5}{9.968} \approx 0.250
\]
Shunday qilib, qiya tekislikning friksion koeffitsienti \( \mu \approx 0.25 \) ga teng.
Quick Answer
Qiya tekislikning friksion koeffitsienti \( \mu \approx 0.25 \) ga teng.
Answered by UpStudy AI and reviewed by a Professional Tutor
UpStudy ThothAI
Self-Developed and Ever-Improving
Thoth AI product is constantly being upgraded and optimized.
Covers All Major Subjects
Capable of handling homework in math, chemistry, biology, physics, and more.
Instant and Accurate
Provides immediate and precise solutions and guidance.
Try Now
Ask Tutors
Ask AI
10x
Fastest way to Get Answers & Solutions
By text
Enter your question here…
By image
Re-Upload
Submit