Respuesta :
Answer:[tex]\theta =\arctan \mu [/tex]
Step-by-step explanation:
we know force sin component would oppose the weight of object thus normal reaction will not be W rather it would be
[tex]N=W-Fsin\theta [/tex]
therefore force cos component will balance the friction force
F[tex]cos\theta[/tex] =[tex]\left ( \mu N\right )[/tex]
F[tex]cos\theta[/tex] =[tex]\left ( \mu \left ( W-Fsin\theta \right )\right )[/tex]
F=[tex]\frac{\mu W}{cos\theta +\mu sin\theta}[/tex]
F will be smallest when [tex]cos\theta +\mu sin\theta[/tex] will be maximum
and it will be maximum when we differentiate it to get
[tex]\theta =\arctan \mu[/tex]
The magnitude of the force, F, is varies with the angle the rope makes
with the plane according to the given equations.
F will be smallest when [tex]\underline{\theta \ is \ arctan (\mu)}[/tex].
Reason:
The given parameters are;
Angle the rope makes with the plane = θ
The magnitude of the force is, [tex]F = \dfrac{ \mu \cdot W}{\mu \cdot sin(\theta) +cos(\theta) }[/tex]
The value of θ for which the value of F is smallest.
Solution;
When, F is smallest, we have;
[tex]\dfrac{dF}{d \theta} = \dfrac{d}{d\theta} \left(\dfrac{ \mu \cdot W}{\mu \cdot sin(\theta) +cos(\theta) } \right) = \dfrac{-\mu \cdot W \cdot (\mu \cdot cos(\theta) -sin(\theta))}{\left( \mu \cdot sin(\theta) +cos(\theta) \right)^2} = 0[/tex]
Therefore;
-μ·W·(μ·cos(θ) - sin(θ))
μ·cos(θ) = sin(θ)
By symmetric property, we have;
sin(θ) = μ·cos(θ)
[tex]\mathbf{\dfrac{sin(\theta)}{cos(\theta)} = tan (\theta) = \mu}[/tex]
Which gives;
θ = arctan(μ)
Therefore;
F, will be smallest when [tex]\underline{\theta = arctan (\mu)}[/tex].
Learn more here:
https://brainly.com/question/14457114
Question; The given equation of the magnitude of the force in relation to the angle the rope makes with the plane, θ, is presented as follows;
[tex]F = \dfrac{ \mu \cdot W}{\mu \cdot sin(\theta) +cos(\theta) }[/tex]
