Plan for Today:
Dry Friction
Belt Friction
Friction
▪ Depends on the characteristics of the contacting surfaces.
▪ Acts in a direction that opposes motion.
▪ In Statics we further specify that friction opposes
impending motion
▪ Acts parallel to the contacting surfaces.
CHARACTERISTICS OF DRY FRICTION
Experiments show that the friction force F is
directly proportional to force (F = P) until a
certain point at which slip occurs.
That point is the point of maximum static
fraction and is given as:
Fs = s N
where s is the coefficient of static friction and
N is the normal force.
After the point of maximum static friction is
surpassed, motion occurs and friction force is
given as F = k N where k is the coefficient of
kinetic friction.
DETERMING s EXPERIMENTALLY
A block with weight W is placed on an inclined plane. The
plane is slowly tilted until the block just begins to slip.
The inclination, s, is noted. Analysis of the block just
before it begins to move gives (using Fs = s N):
+  Fy = N – W cos s
= 0
+  FX = S N – W sin s = 0
Using these two equations, we get
s = (W sin s ) / (W cos s ) = tan s
This simple experiment allows us to find the S between two
materials in contact.
Will the body tip or slip?
At rest (A), the normal force is uniformly
distributed.
As a pushing force is applied (B), the
distributed normal force is redistributed,
moving the equivalent point load to the right.
If pushing force becomes large enough (C),
the moment couple exerted by gravity and
normal force will be unable to counter the
couple exerted by the pushing force and
friction force
If the moment couple formed by gravity and the normal force is less than the moment
couple formed by the friction force and pushing force, tipping will occur.
In this, class we will only be dealing with problems for which tipping does not occur.
Decision Process for Solving Dry Friction Problems (Is friction µN?)
PROBLEMS INVOLVING DRY FRICTION
Steps for solving equilibrium problems involving dry friction:
1. Draw necessary free body diagrams. Make sure that you
show the friction force in the correct direction (it always
opposes the motion or impending motion).
2. Determine the number of unknowns. Do not assume that
F = Fs = S N unless the motion is impending.
3. Apply the equations of equilibrium and appropriate
frictional equations to solve for the unknowns.
Note: If you are not told that motion is impending, solve for the friction force with equilibrium equations
and then check to see if it is greater than Fs. If it is, the body is slipping and F is actually k N . If it isn’t,
then F is just equal to the F that you found.
BELT FRICTION
When there is no friction, T1 = T2 as we have
previously discussed.
However, now we need to consider what happens when
friction is of concern.
Consider a flat belt passing over a fixed curved surface
with the total angle of contact equal to  radians.
If the belt slips or is just about to slip, then T2 must be larger
than T1 and the direction of motion resisting friction forces.
Hence, T2 must be greater than T1.
Detailed analysis (see textbook) shows that T2 = T1 e  
where  is the coefficient of static friction between the belt
and the surface. Be sure to use radians when using this
formula!!
Example
The coefficient of static friction between the 50-lb box and the inclined surface is 0.10. The
coefficient of static friction between the rope and the fixed cylinder is 0.05. Determine the
force the woman must exert on the rope to cause the box to start moving up the inclined
surface.
Example:
The maximum tension that can be developed in the cord is 500 N. If the pulley at A is free to
rotate and the coefficient of static friction at the fixed drums B and C is 0.25, determine the
largest mass of the cylinder that can be lifted by the cord.
Start by looking at drum B:
When trying to determine wrap angle, one helpful
fact to remember is that the angle between the
line tangent to the point where the rope contacts
the drum and radial line is a right angle.
Example: A 180-lb farmer tries to restrain the cow from escaping by wrapping the rope two
turns around the tree trunk as shown. If the cow exerts a force of 250 lb on the rope, determine
if the farmer can successfully restrain the cow. The coefficient of static friction between the
rope and the tree trunk is µs = 0.15, and between the farmer’s shoes and the ground is µs = 0.3.