Maximum acceleration up ramp

[diffusion]

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?

 

[Vuldoraq]

Hi,

Have you drawn a free body force diagram? Also we need to see your attempt at the solution before we can help you.

 

[diffusion]

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.

 

[Doc Al]

Not sure how to find the solution, but I'll wing it.

Looks like you knew more than you thought.

-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

 

[diffusion]

Looks like you knew more than you thought.


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?

 

[Doc Al]

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)

 

[diffusion]

No. Each term in the numerator has a single "m". It factors out like so:

mX + mY = m(X + Y)

Gotcha. Thanks.