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I have this (physics) question, but am missing something as to why the math works for it. The problem is as follows:

A 4- kg sphere rests on t he smooth parabolic surface. Determine the normal force it exerts on the surface and the mass of block B needed to hold it in the equilibrium position shown.

With the following image:

diagram

The problem i'm having is finding the angle $\theta$ that the horizontal makes with the surface. My book has it in the example as follows:

$$\tan(\theta)|_{x=.4m} = \frac{dy}{dx}|_{x=.4m} = 5.0x|_{x=.4m}=2.00$$ $$\arctan(2)=63.43^\circ$$

This supposedly gets the angle that the surface (I presume the curve?) is making with the the horizontal.

My question is: I know that $tangent$ is opposite over adjacent, but I'm not sure why are we finding the derivative of the function that represents the curve/ramp $y=2.5x^2$, or how plugging that into the $arctangent$ we get the angle between the curve and the horizontal? (Note, I can do the problem given the angle, that is I can find the forces, but why and how are we finding the angle this way?)

Edit: The image is cut off, the bottom distance marked is also $0.4 m$

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  • $\begingroup$ Geometrically, what is the derivative of $y=2.5x^2$ at the point $x=.4$? $\endgroup$ – john w. Jul 6 '14 at 16:18
  • $\begingroup$ @johnw. That is the derivative is $y=5.00x$ and plugging in we get $y=5.00*.4=2.00$ $\endgroup$ – Rivasa Jul 6 '14 at 16:20
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    $\begingroup$ the derivative of a curve at a point is the slope of the tangent line at that point. $\endgroup$ – john w. Jul 6 '14 at 16:23
  • $\begingroup$ @johnw. I understand that, but why do we put it into tan? How is that giving the angle? $\endgroup$ – Rivasa Jul 6 '14 at 16:26
  • $\begingroup$ Draw the tangent line. Then make a triangle with this tangent line, the horizontal axis and a vertical line from the tangent point to the horizontal axis. Now where is $\theta$ and how do we find it. From trig $\tan(\theta)$ is opp/adj, the problem is that you do not know what the length of the bottom of the triangle is. But opp/adj is rise/run which is the slope of the tangent line. $\endgroup$ – john w. Jul 6 '14 at 16:37
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The slope $m$ of the tangent line to a curve $y=f(x)$ is given by the derivative $m=f'(x_0)$ at the point $(x_0,f(x_0))$ of interest - essentially by definition of derivative. The slope is also the tangent of the angle between the horizontal line and the tangent line (not the two distinct meanings of the word "tangent" here), that is $\tan\theta = f'(x_0)$. You can see that by drawing a right triangle with a horizontal and a vertical leg and a segment of the tangent line as hypothenuse.

Remark: At first sight I thought your text has something wrong, because we are given $y=0.4$, not $x=0.4$. Luckily, this makes no difference as $2.5\cdot 0.4^2=0.4$

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  • $\begingroup$ The $x$ distance is also $0.4m$ The image just got cut off. $\endgroup$ – Rivasa Jul 6 '14 at 16:22

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