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A crate is at rest on an inclined plane. As the slope increases the crate remains at rest until the incline reaches an angle of 32.7° from the horizontal. At this angle the crate begins to slidedown the ramp.If the coefficient of kinetic friction between the surface of the ramp and the crate is 0.33,what would be the acceleration of the moving crate at this angle?

Answer :

The free body diagram of the crate can be shown as,

According to free body diagram, the net force acting on the crate is,

[tex]F=mg\sin \theta-f\ldots\ldots\text{ (1)}[/tex]

The frictional force acting on the crate is,

[tex]f=\mu N[/tex]

The normal force acting on crate is,

[tex]N=mg\cos \theta[/tex]

Therefore, the frictional force becomes,

[tex]f=\mu mg\cos \theta[/tex]

According to Newton's second law of motion, the net force acting on the crate is,

[tex]F=ma[/tex]

Therefore, equation (1) becomes,

[tex]\begin{gathered} ma=mg\sin \theta-\mu mg\cos \theta \\ a=g(\sin \theta-\mu\cos \theta) \end{gathered}[/tex]

Substitute the known values,

[tex]\begin{gathered} a=(9.8m/s^2)(\sin 32.7^{\circ}-(0.33)\cos 32.7^{\circ}) \\ =(9.8m/s^2)(0.540-(0.33)(0.842)) \\ =(9.8m/s^2)(0.262) \\ \approx2.57m/s^2 \end{gathered}[/tex]

Thus, the acceleration of the moving crate is 2.57 m/s2.

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