Datasheet

Chapter 1: Mechanical Adjustments · Page 33
Next, convert to the corresponding units of force (newtons):
N86.6
s/m8.9kg7.0f
N58.20
s/m8.9kg1.2N
mgW
2
S
2
=
×=
=
×=
=
Finally, plug these forces into the coefficient of kinetic friction equation and calculate the
result:
...333.0
N58.20
N86.6
N
f
K
S
=
=
=µ
Why couldn't I have just used the ratio of the masses? For this particular problem on the
earth's surface, you can use this shortcut. The ratio of the masses does give you the correct
answer, 700/2100 = 0.333... However, you should always keep in mind that the µ
s
and µ
k
are ratios of forces. While you won't encounter it in this book, the problems get more
complex when they incorporate things like thrust and a body's tendency to rotate. In
general, when the problems get more complex, you will have to be strict about the difference
between mass and force.
Coefficients of friction make predicting how much force it will take to get something to
slide really easy. The same applies for sustained sliding and the coefficient of kinetic
friction. In either case, all you have to do is multiply the normal force by the coefficient
of friction.
NfandNf
KKSS
×=×=
µ
µ
When the surface is tilting, the normal force takes some extra calculating, but when it's
horizontal to the ground, like in Figure 1-8, the normal force is just the object's weight on
the surface. The force to start it sliding is the coefficient of friction multiplied by the
object's weight, and the force to keep it sliding is the coefficient of kinetic friction
multiplied by the object's weight.
WfandWf
KKSS
×=×=
µ
µ