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## anonymous 5 years ago 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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1. anonymous

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)

2. anonymous

ok cool, thats what i ended up with, i just wasn't sure if it'd be that simple.

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