# calculating acceleration from coefficient of friction

• Sep 23rd 2008, 06:11 PM
winterwyrm
calculating acceleration from coefficient of friction
The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.40 with the floor. If the train is initially moving at a speed of 52 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?
• Sep 23rd 2008, 06:49 PM
skeeter
net force on the crates while stopping will be provided by $\displaystyle F_s$.

the max net force possible would then be $\displaystyle F_{smax} = \mu m g = ma_{max}$

so ... $\displaystyle a_{max}$ has magnitude $\displaystyle \mu g$

using the "no time" kinematics equation ...

$\displaystyle \Delta x = \frac{v_f^2 - v_0^2}{2(-\mu g)}$
• Sep 23rd 2008, 08:50 PM
winterwyrm
how would you use it in this situation though?
• Sep 24th 2008, 12:26 PM
topsquark
Quote:

Originally Posted by winterwyrm
how would you use it in this situation though?

The same way skeeter did it. You know the maximum force that friction can have from $\displaystyle f_s = \mu _s N = \mu _s mg$. (Igonre, for the moment that we don't know the mass.) Then using Newton's 2nd Law we get that the maximum acceleration the block can take without sliding is $\displaystyle \sum F = f_s = ma$. Now the mass cancels out so we don't have to worry about it. Thus we get $\displaystyle a = \mu _s g$.

-Dan