- #1

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A velocity-time or acceleration-time graph which models a particle going up a ramp and down a ramp.

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- Thread starter nehcrow
- Start date

- #1

- 15

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A velocity-time or acceleration-time graph which models a particle going up a ramp and down a ramp.

- #2

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Even if there are no external forces but gravity on the block, and your acceleration-time graph is just a constant you would still be unable to find the coefficient of friction without at least knowing the angle the ramp makes with the horizontal.

- #3

- 462

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You would need at least some kind of additional information.

No that's not right

Even if there are no external forces but gravity on the block, and your acceleration-time graph is just a constant you would still be unable to find the coefficient of friction without at least knowing the angle the ramp makes with the horizontal.

You do not know inclination as well. So you have got two equations and two variables. What else do you need?

- #4

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- #5

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- #6

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There should only be a single acceleration graph.. unless there are other forces acting on the block other than gravity. The gravitational acceleration on the block does not care if the block is going up or down, the acceleration is constant. Therefore the acceleration-time graph is just a straight line, and fully describes the block moving up and down the ramp together with the initial velocity and position, no second graph is required.

Well the acceleration experienced by the block regardless of whether it is going up or down the ramp and regardless of whatever its velocity is will be found by:

*in general*

[tex]F_{friction} = \mu F_{normal}[/tex]

[tex]F_{friction} = \mu mgcos\theta[/tex]

the component of gravity down the ramp will be:

[tex]F_{g_{x}} = mgsin\theta[/tex]

So the total force is:

[tex]F = F_{g_{x}} - F_{friction}[/tex]

[tex]F = mgsin\theta - \mu mgcos\theta[/tex]

So the acceleration is in general:

[tex]a = g(sin\theta - \mu cos\theta)[/tex]

Then from this the velocity is simply the integral, we find:

[tex]v = g(sin\theta - \mu cos\theta)t + v_{0}[/tex]

So for example, with a ramp angle of pi/4 and a coefficient of friction of 0.2, the acceleration is directed down the ramp with a magnitude of 5.544m/s^2

However, it is easy to verify that a ramp angle of pi/3 and a coefficient of friction of 0.601**also** has an acceleration down the ramp of 5.544m/s^2

There should actually be an infinite amount of combinations of ramp angle and coefficients of friction that will give the same acceleration.

If the acceleration is the same, then the velocity and position graphs are also the same. (provided that all situations have the same starting point and initial velocity).

Well the acceleration experienced by the block regardless of whether it is going up or down the ramp and regardless of whatever its velocity is will be found by:

[tex]F_{friction} = \mu F_{normal}[/tex]

[tex]F_{friction} = \mu mgcos\theta[/tex]

the component of gravity down the ramp will be:

[tex]F_{g_{x}} = mgsin\theta[/tex]

So the total force is:

[tex]F = F_{g_{x}} - F_{friction}[/tex]

[tex]F = mgsin\theta - \mu mgcos\theta[/tex]

So the acceleration is in general:

[tex]a = g(sin\theta - \mu cos\theta)[/tex]

Then from this the velocity is simply the integral, we find:

[tex]v = g(sin\theta - \mu cos\theta)t + v_{0}[/tex]

So for example, with a ramp angle of pi/4 and a coefficient of friction of 0.2, the acceleration is directed down the ramp with a magnitude of 5.544m/s^2

However, it is easy to verify that a ramp angle of pi/3 and a coefficient of friction of 0.601

There should actually be an infinite amount of combinations of ramp angle and coefficients of friction that will give the same acceleration.

If the acceleration is the same, then the velocity and position graphs are also the same. (provided that all situations have the same starting point and initial velocity).

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