What Is the Maximum Tangential Acceleration Before a Car's Wheels Start to Spin?

[Roushrsh]

A car goes around a 40m radius turn at a speed of 15m/s . If the coefficient
of static friction μs between the tires and the road is 0.7, what is the
maximum tangential acceleration the car could have before the wheels
start to spin?

OK IT'S MY LAST QUESTION AND I THINK I'M TOO TIRED TO FIGURE IT OUT!
What I've concluded is that tangential acceleration is v/t not v^2/r
and that Ffr = umg = 6.86m
So basically I need an equation with m and a on the other side
what first comes to mind is 6.86m= ma, then a = 6.86m/s^2 which seems totally wrong for some reason.
Then I tried Calculating time
so 2pi r = 251.33
t=d/v 251.33/15 = 16.76
Then concluded that that was a waste of time due to v/t is really of no use to me...
HELP PLEASE! library closes in 30 minutes :/

 

[collinsmark]

Hello Roushrsh,

Welcome to Physics Forums!

Have you tried the Pythagorean theorem?

The trick to solving this problem is to realize that the radial component and the tangential component of acceleration are perpendicular to one another. The resulting acceleration vector's magnitude is the maximum acceleration due to friction.

 

[Roushrsh]

Yes but doesn't it ask for only the maximum tangential acceleration?
Also what would I do with the 0.7mg then?
If I were to do that I'd have the sqrt((15/16.76)^2 + (15^2/40)^2), how would that be equal to the 0.7mg? :S
Thanks

 

[collinsmark]

Yes but doesn't it ask for only the maximum tangential acceleration?

Hmm. Consider [tex] c = \sqrt{a^2 + b^2}. [/tex] Suppose b = 4 and the maximum possible c is 5. Given that information, what is the maximum possible a?

(Hint: Regarding problem about the car. You don't know the magnitude of the maximum tangential acceleration -- that's what you trying to find. And you can find it if you know the magnitude of the max overall acceleration, and the magnitude of the radial component of acceleration. )

Also what would I do with the 0.7mg then?
If I were to do that I'd have the sqrt((15/16.76)^2 + (15^2/40)^2), how would that be equal to the 0.7mg? :S
Thanks

0.7mg is maximum force of friction the car can/track handle without sliding. The important point here is that it's a force. Newton's second law will help you find the associated maximum acceleration.

 

[Roushrsh]

Yes... but I don't have the mass overall acceleration.
If it's the friction (since that's the max it can go before sliding)
then I'd have:
(6.86m) = sqrt ((ma)^2 + ((15^2 /40)m)^2)
= *after a few steps*
15.45 = -0.47m +2ma +a^2
Which means I can't do anything :/
Edit: unless of course you want me to replace the a in ma by 15/16.25... which shouldn't make sense since the 16.25 is time in one period :/
Edit: Oh wait I think I may have it hmm...
Edit: nvm, I was thinking of (6.86m)^2-(ma)^2-(5.625m)^2 = 0 then dividing both side by m, but then realized that I can't just remove the m due to the ^2

 

[collinsmark]

Yes... but I don't have the mass overall acceleration.
If it's the friction (since that's the max it can go before sliding)

Yes, it relates to the friction. The maximum overall acceleration is the maximum friction force divided by the mass.

F = ma → a = F/m

then I'd have:
(6.86m) = sqrt ((ma)^2 + ((15^2 /40)m)^2)
= *after a few steps*
15.45 = -0.47m +2ma +a^2
Which means I can't do anything :/
Edit: unless of course you want me to replace the a in ma by 15/16.25... which shouldn't make sense since the 16.25 is time in one period :/
Edit: Oh wait I think I may have it hmm...
Edit: nvm, I was thinking of (6.86m)^2-(ma)^2-(5.625m)^2 = 0 then dividing both side by m, but then realized that I can't just remove the m due to the ^2

You can remove the mass. Keep going...

[tex] \mu m g = \sqrt{(m a_{\|})^2 + (m a_{\bot})^2} [/tex]

[tex] = \sqrt{m^2 a_{\|}^2 + m^2 a_{\bot}^2} [/tex]

[tex] = \sqrt{m^2 (a_{\|}^2 + a_{\bot}^2)} [/tex]

[tex] = m \sqrt{a_{\|}^2 + a_{\bot}^2} [/tex]

thus,

[tex] \mu g = \sqrt{a_{\|}^2 + a_{\bot}^2} [/tex]

You should be able to take it from there.

 

[Roushrsh]

Argh, how did I keep missing that! Thanks a ton! I never connected that A(T) and A(C) using Pythagoras would give the max acc. Definitely something worthwhile to know.