Parallel and perpendicular vectors

Gregg

If a mass takes a path down a slope of a parabola. The force downward is its weight. I need to resolve the perpendicular and parallel (to direction of motion) forces.

y=ax^2+bx+c, it has a y intercept of h and a repeated root l. a>0. The mass slides from x=0 to x=l with no friction.

[latex] d\vec{r} = \left(
\begin{array}{c}
1 \\
2ax+b
\end{array}
\right)\text{dx} [/latex]

A unit vector in this direction:

[latex] \hat{u} = \frac{1}{\sqrt{1+(2ax+b)^2}}\left(
\begin{array}{c}
1 \\
2ax+b
\end{array}
\right) [/latex]

To get the size of the component of the force parallel to the curve:

[latex] mg \cos (\theta) = mg\frac{(2ax+b)}{\sqrt{1+(2ax+b)^2}} [/latex]

[latex] mg \cos (\theta) \hat{u} = mg\frac{(2ax+b)}{\sqrt{1+(2ax+b)^2}} \frac{1}{\sqrt{1+(2ax+b)^2}}\left(
\begin{array}{c}
1 \\
2ax+b
\end{array}
\right) = mg \frac{(2ax+b)}{1+(2ax+b)^2}\left(
\begin{array}{c}
1 \\
2ax+b
\end{array}
\right) [/latex]

I don't think this is right

Gregg

This can't be right because then

[latex] \int \vec{F}\cdot d\vec{r} = Mg(al^2 + bl) \ne Mgh [/latex]

Gregg

Could someone have a look at this

Redbelly98

There is a subtle mistake at play here. You want the magnitude of the parallel force component, i.e., its absolute value. So there should be a big absolute value sign on the r.h.s. in that last equation.

The absolute value is not an issue for m, g, and the square-root expression since those are all positive. But what about the 2ax+b term, is that negative or positive?

Hints:
1. We are only considering 0<x<l.
2. Express a and b each in terms of h and l, if you have not already done so.
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