The minimum coefficient of static friction required for this claim to be made possible is 0.7
In an inclined plane system, the coefficient of static friction is the angle at which an object slide over another.
The gravitational force component exceeds the static friction force when the angle rises. Hence, the object begins to slide.
According to Newton second law;
[tex]\sum F_x = \sum F_y = 0[/tex]
[tex]\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }[/tex]
[tex]\mathbf{mg sin \theta =f_s}[/tex]
[tex]\mathbf{mg sin \theta =\mu_s N}[/tex]
N = mg cos θ
Equating both force components on the L.H.S and R.H.S, we have the following
[tex]\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}[/tex]
[tex]\mathbf{sin \theta =\mu_s \ \ cos \theta}[/tex]
[tex]\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}[/tex]
According to the trigonometry rule:
[tex]\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}[/tex]
∴
[tex]\mathbf{\mu_s =\tan \theta}}[/tex]
[tex]\mathbf{\mu_s =\tan 35^0}}[/tex]
[tex]\mathbf{\mu_s = 0.700}}[/tex]
Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7
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