If a crate weighing 350. N is at rest on an inclined plane whose angle provides the crate with 150. N of normal force, at what angle to the horizontal would the ramp be?

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cryssatemp cryssatemp · Answer 1 · 2019-12-22T23:42:43+01:00

Answer: [tex]25.37\°[/tex]

Explanation:

If we draw a free body diagram of the crate we will have the following related to the net force in the y-axis:

[tex]N- W sin \theta=0[/tex] (1)

Hence:

[tex]N=W sin \theta[/tex] (2)

Where:

[tex]N=150 N[/tex] is the normal force

[tex]W=350 N[/tex] is the weight of the crate

[tex]\theta[/tex] is the angle of the inclined plane where the crate is

Findind [tex]\theta[/tex]:

[tex]\theta=sin^{-1} (\frac{N}{W})[/tex] (3)

[tex]\theta=sin^{-1} (\frac{150 N}{350 N})[/tex] (4)

Finally:

[tex]\theta=25.37\°[/tex]