At equilibrium, the coefficient of static friction is given by the tangent of
the angle formed by the wire and the horizontal.
- The highest distance above the x-axis at which the particle will be in equilibrium is; [tex]\underline{\displaystyle y = \frac{a \cdot \mu^2}{a} }[/tex]
Reasons:
The given parabolic equation is; x² = a·y
The coefficient of friction = μ
At equilibrium, we have;
μ·[tex]F_N[/tex] = [tex]\mathbf{F_P}[/tex]
Where;
[tex]F_N[/tex] = The normal reaction = m·g·cos(θ)
[tex]F_P[/tex] = The component of the force of the weight of the bread acting along the plane = m·g·sin(θ)
Therefore;
[tex]\displaystyle \mu = \frac{m\cdot g \cdot sin(\theta)}{m \cdot g \cdot cos (\theta)} = \mathbf{tan (\theta)}[/tex]
Where;
θ = The angle the wire forms with the horizontal at the point the bread is in equilibrium
From trigonometric ratios, we have;
[tex]\displaystyle tan(\theta) = \mathbf{ \frac{\Delta y}{\Delta x}}[/tex]
Which for the curved wire gives;
[tex]\displaystyle tan(\theta) = \lim_{\Delta x \to0} \frac{\Delta y}{\Delta x} = \mathbf{\frac{dy}{dx}}[/tex]
[tex]\displaystyle y = \frac{x^2}{a}[/tex]
Therefore;
[tex]\displaystyle \frac{dy}{dx} = \frac{d}{dx} \left(\frac{x^2}{a} \right) = \mathbf{\frac{2 \cdot x}{a}}[/tex]
Which gives;
- [tex]\mu = tan(\theta) = \displaystyle \mathbf{\frac{dy}{dx}} =\frac{2 \cdot x}{a}[/tex]
Therefore;
[tex]\displaystyle x = \frac{a \cdot \mu }{2}[/tex]
Which gives;
[tex]\displaystyle y = \frac{ x^2}{a} = \frac{\left(\frac{a \cdot \mu}{2} \right)^2 }{a} = \mathbf{\frac{a \cdot \mu^2 }{4}}[/tex]
- [tex]Highest \ distance \ above \ the \ x-axis \ for \ equilibrium \ of\ particle \ is; \ \displaystyle \underline{y = \frac{a \cdot \mu^2 }{4}}[/tex]
Learn more about friction on an inclined plane here:
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