#

**Static Friction**

**The force of friction always opposes
the motion -- or opposes the motion that would occur if the
friction force were not present.**

**
**

**
Example: The weight is 100 newtons; what is the tension in the
cord supporting it? The coefficent of friction between the block
and the plane is 0.35.**
**Always start with a free body diagram
showing all the forces acting on the body.**

**The x- and y-axes have been chosen
parallel and perpendicular to the plane. From that diagram, we
find the components of the weight are **

**W**_{x} = W sin 25° =
42.3 N
**W**_{y} = - W cos 25° = -
90.6 N

**The normal force F**_{N} lies in
the positive y-direction; the tension T and friction force
F_{f} lie along the negative x-axis. Notice that the
friction force is its maximum value F_{N}
if the rope is required to hold the block in equilibrium. Now we
have all the components required for the First Condition of
Equilibrium.

**The First Condition of Equilibrium,
for the x-components of all the forces, is**

** F**_{x}
= W _{x} - F_{f} - T = 0
**W**_{x} - F_{f,max} - T
= 0

**W**_{x} - F_{N}
- T = 0

**42.3 N - 0.35 F**_{N} - T =
0

**The First Condition of Equilibrium,
for the y-components of all the forces, is**

** F**_{y}
= F _{N} - W _{y} = 0
**F**_{N} - 90.6 N =
0

**F**_{N} = 90.6
newtons

**This value of the normal force can be substituted
back in the F**_{x}
equation to find

**42.3 N - 0.35 (90.6 N) - T =
0**
**T = 10.6
newtons**

**If we wanted to know the friction
force, we would find that from**

**F**_{f} = F_{f,max} = F_{N}
= (0.35)(90.6 N) = 31.7 N
** **

**Notice that the normal force F**_{N} is not
equal to the weight in this case.

**There are just enough special cases
where the normal force does happen to be equal to the weight that
it is sometimes (and wrongly) assumed that the normal force and
weight are always the same. Be aware of that for it simply is not
true!**

(c) Doug Davis, 2001; all rights reserved