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.

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

A unit vector in this direction:

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

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

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

[itex] 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) [/itex]

I don't think this is right

 

[Gregg]

This can't be right because then

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

 

[Gregg]

Could someone have a look at this

 

[Redbelly98]

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.

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

A unit vector in this direction:

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

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

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

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.

[itex] 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) [/itex]

I don't think this is right

test

 

[rwisz]

because there isn't a force perpendicular to the direction of motion

I would like to clarify that parallel and perpendicular vectors are two types of vectors that can exist in a two or three-dimensional space. Parallel vectors have the same direction, while perpendicular vectors have a 90 degree angle between them. In the context of the given scenario, the force of gravity acting on the mass is a parallel vector to the direction of motion, while the normal force (if present) would be a perpendicular vector.

In order to resolve the parallel and perpendicular forces acting on the mass, we can use vector decomposition techniques. This involves breaking down the force vector into its parallel and perpendicular components. In this case, the parallel component would be the force of gravity, while the perpendicular component would be the normal force (if present).

However, in the given equation, there is no mention of a normal force or any other perpendicular force. Therefore, it is not possible to resolve the forces into parallel and perpendicular components. The equation only describes the path of the mass and the force of gravity acting on it.

In summary, in order to resolve the parallel and perpendicular forces, we need to know all the forces acting on the mass, not just the force of gravity. Without this information, it is not possible to accurately calculate the parallel and perpendicular components of the force.