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A rock of mass 10 kg sits on a slope which makes an angle of 25 degrees with the horizontal. If it does not slip down what is the magnitude of the frictional force?
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The weight of the rock is 98 N, so the component that is parallel to the slope would be \[98*\sin(25degrees)\approx41.4N\] Thus the friction force must also be 41.4 N (or maybe just 41, because of significant figures)
ok cool, thats what i ended up with, i just wasn't sure if it'd be that simple.
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