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An object with a mass of 5.10 kg rests on a plane inclined 18.59° from horizontal. What is the component of the weight force that is parallel to the incline?

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ANSWER

15.93 N

EXPLANATION

First let's draw a free body diagram:

As we can see, the component of the weigth force Fg that is parallel to the incline is the x-component, according with the coordinate plane chosen. This component we can translate it to be between the arrow of the y-component and the arrow of the actual force, so it is the opposite side to the given angle θ. Therefore, we can find it with Fg and the sine of the angle:

[tex]F_{gx}=F_g\cdot\sin \theta[/tex]

Where Fg = m*g, m = 5.10kg, g = 9.8 m/s² and θ = 18.59°:

[tex]\begin{gathered} F_{gx}=5.10\operatorname{kg}\cdot9.8m/s^2\cdot\sin 18.59\degree \\ F_{gx}\approx15.93N \end{gathered}[/tex]

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