Math Problem Statement

minimum force 알려줘

minimum force 알려줘

Solution

To find the minimum force FF, let’s analyze the given equation:

F=mgμsinθ+μcosθF = \frac{mg\mu}{\sin\theta + \mu \cos\theta}

Given:

  • μ=0.54\mu = 0.54
  • Range for θ\theta: 0θπ20 \leq \theta \leq \frac{\pi}{2}

Steps to find the minimum force:

  1. Behavior of the denominator: The denominator (sinθ+μcosθ)(\sin\theta + \mu \cos\theta) determines the minimum force. The force FF is inversely proportional to the denominator.

  2. Maximizing the denominator: Maximize sinθ+μcosθ\text{Maximize } \sin\theta + \mu \cos\theta Differentiate with respect to θ\theta and find the critical points: ddθ(sinθ+μcosθ)=cosθμsinθ\frac{d}{d\theta} (\sin\theta + \mu \cos\theta) = \cos\theta - \mu \sin\theta Set the derivative equal to zero: cosθ=μsinθtanθ=1μ\cos\theta = \mu \sin\theta \quad \Rightarrow \quad \tan\theta = \frac{1}{\mu} Substitute μ=0.54\mu = 0.54: tanθ=10.54θ=arctan(10.54)61.3\tan\theta = \frac{1}{0.54} \quad \Rightarrow \quad \theta = \arctan\left(\frac{1}{0.54}\right) \approx 61.3^\circ

  3. Evaluate the denominator at θ=61.3\theta = 61.3^\circ, θ=0\theta = 0, and θ=π2\theta = \frac{\pi}{2}:

    • At θ=0\theta = 0: sin0+0.54cos0=0+0.54=0.54\sin 0 + 0.54 \cos 0 = 0 + 0.54 = 0.54
    • At θ=π2\theta = \frac{\pi}{2}: sinπ2+0.54cosπ2=1+0=1\sin \frac{\pi}{2} + 0.54 \cos \frac{\pi}{2} = 1 + 0 = 1
    • At θ=61.3\theta = 61.3^\circ: Use sin61.30.874\sin 61.3^\circ \approx 0.874 and cos61.30.485\cos 61.3^\circ \approx 0.485: sin61.3+0.54cos61.3=0.874+0.54(0.485)1.135\sin 61.3^\circ + 0.54 \cos 61.3^\circ = 0.874 + 0.54(0.485) \approx 1.135

    The maximum denominator occurs at θ61.3\theta \approx 61.3^\circ.

  4. Calculate minimum force: Substitute the maximum denominator into the equation for FF: Fmin=mgμMax(sinθ+μcosθ)F_{\text{min}} = \frac{mg\mu}{\text{Max}(\sin\theta + \mu \cos\theta)} With sin61.3+0.54cos61.31.135\sin 61.3^\circ + 0.54 \cos 61.3^\circ \approx 1.135: Fmin=mg(0.54)1.1350.475mgF_{\text{min}} = \frac{mg(0.54)}{1.135} \approx 0.475 \, mg

Final Answer:

Fmin0.475mgNewtons\boxed{F_{\text{min}} \approx 0.475 \, mg \, \text{Newtons}}

Would you like me to expand further on the derivation or provide additional examples?

Related Questions:

  1. What happens to the force FF if μ\mu decreases to 0.3?
  2. Why is the minimum force not at θ=0\theta = 0 or θ=π2\theta = \frac{\pi}{2}?
  3. Can this result be applied if μ\mu is larger than 1? Why or why not?
  4. How does gravity gg impact the minimum and maximum forces?
  5. What if the angle of pulling exceeds π/2\pi/2? Would the formula still work?

Tip:

For any optimization problem involving trigonometric functions, analyze the behavior of sinθ\sin\theta and cosθ\cos\theta carefully over the given range.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Optimization
Forces and Friction

Formulas

F = (mgμ) / (sinθ + μ cosθ)
tanθ = 1 / μ

Theorems

Trigonometric Optimization

Suitable Grade Level

Grades 11-12