Let's say a man is pushing a box over a ramp with angle $\theta$ with the horizontal. By opposite adjacent angles, I think the normal force would be $mg\sin\theta$ why is the normal force $N = mg\cos\theta$?
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$\begingroup$ Show us what you mean by “opposite adjacent angles “. Also, the normal force depends on whether the man is pushing parallel to the ramp or at an angle with the ramp $\endgroup$– Bob DCommented Jun 4 at 15:59
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$\begingroup$ the man is pushing parallel to the ramp $\endgroup$– samsamradasCommented Jun 4 at 16:13
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1$\begingroup$ It's always best to start by drawing a picture. $\endgroup$– Kyle KanosCommented Jun 4 at 16:13
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1 Answer
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Simply, the normal force is at 90 degrees to the ramp's surface, so the component of gravity it resists is also at a 90 degree angle from the ramp*. This means the force is $mgsin(90-\theta)$. However, $sin(90-\theta)=cos(\theta)$, so we simplify to $mgcos(\theta)$. Try drawing out the triangles of the forces if this doesn't make sense.
*this assumes the box is being pushed parallel to the ramp and so the normal force is only resisting the perpendicular component of gravity.
