Maximum acceleration up ramp

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Homework Help Overview

The problem involves determining the maximum possible acceleration of a remote-controlled car on a ramp inclined at 20 degrees, considering the coefficients of static and kinetic friction.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of free body diagrams and the application of forces, including gravitational and frictional forces. There are attempts to derive an equation for acceleration, with some questioning the cancellation of mass in the calculations.

Discussion Status

The discussion is active, with participants providing feedback on each other's reasoning and calculations. There is a focus on clarifying the mathematical steps involved, particularly regarding the treatment of mass in the equations.

Contextual Notes

Participants express uncertainty about the solution process and the need for a clear understanding of the forces involved. There is an emphasis on ensuring that all steps are correctly followed and understood before proceeding.

diffusion
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You are trying to have your remote-controlled car leap off a ramp, which happens to be at a 20 degree incline to the ground. If the coefficients of static and kinetic friction are .70 and .50, repsectively, what is the maximum possible acceleration of the remote-controlled car up the ramp?
 
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Hi,

Have you drawn a free body force diagram? Also we need to see your attempt at the solution before we can help you.
 
Not sure how to find the solution, but I'll wing it.

Max acceleration = -mg(sin20) + max static friction / mass.

Max static friction = static coefficient x normal force = 0.70 x n

Find n (Net force in y-direction) = mgcos20.

Now you plug mgcos20 back into the original equation

-mg(sin20) + mgcos20 / mass.

Stuck here.
 
diffusion said:
Not sure how to find the solution, but I'll wing it.
Looks like you knew more than you thought. :wink:

-mg(sin20) + mgcos20 / mass.

Stuck here.
Looks good to me. Realize that mass = m, so it cancels. And use parentheses, to reduce mistakes. I'd write that as:

a = ( -mgsin20 + mgcos20 )/m
 
Doc Al said:
Looks like you knew more than you thought. :wink:


Looks good to me. Realize that mass = m, so it cancels. And use parentheses, to reduce mistakes. I'd write that as:

a = ( -mgsin20 + mgcos20 )/m

I realize this is going to be a stupid question, but how do they cancel? In the numerator you have two mg's, only one in the denominator. Isn't that going to leave you with another m?
 
diffusion said:
In the numerator you have two mg's, only one in the denominator. Isn't that going to leave you with another m?
No. Each term in the numerator has a single "m". It factors out like so:

mX + mY = m(X + Y)
 
Doc Al said:
No. Each term in the numerator has a single "m". It factors out like so:

mX + mY = m(X + Y)

Gotcha. Thanks.
 

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