How are these derived for an inclined plane?

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Homework Help Overview

The discussion revolves around the derivation of equations related to mechanical advantage and efficiency for an inclined plane, specifically involving the angle theta and the coefficient of kinetic friction, mu_k.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss drawing a free-body diagram of the inclined plane and analyzing the forces acting on a mass on the incline. There are attempts to express mechanical advantage in terms of forces and to relate it to the efficiency equation.

Discussion Status

Some participants are sharing their interpretations and calculations regarding mechanical advantage, while others are clarifying the relationships between the forces involved. There is an ongoing exploration of how to express these concepts mathematically without reaching a definitive conclusion.

Contextual Notes

The original poster notes that the angle theta is unspecified, which may affect the derivation process. There is also mention of the coefficient of kinetic friction, mu_k, which is integral to the equations being discussed.

catzmeow
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I'm utterly lost on this and would appreciate any help!

I'm given mechanical advantage= 1/(sin(theta)+mu_k*cos(theta))

And efficiency = 1/(1+mu_k*cot(theta))

These are for an inclined plane and theta is not specified, I'm just supposed to show how those equations were derived.
Thanks!
 
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You might start by drawing an incline plane at an angle of theta with a mass on it, call it m as it doesn't matter in the end. Then draw the forces required to push the mass up the ramp. Compare this with the force required to raise the mass straight up. This should be the mechanical advantage.
 
Thanks barryj! I did that and now I have MA= (Mgsin(theta))/(input force-mgcos(theta)...

Is this close(er)?
 
Oops I meant all that I have equal to mu_k
 
Well, I think the MA = force required to lift the mass with no incline divided by the force required to push the mass up the incline. So, the force required to lift the mass with no incline would be mg, now you figure out the force required to push the mass up the incline and do the division.
 
Oh ok- that makes since. Thanks Barryj!
 

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