1. Friction
Lecture 8
Friction: is defined as a force of resistance acting on a body which prevents or retards slipping of the body relative to a second body.
Experiments show that frictional forces act tangent (parallel) to the contacting surface in a direction opposing the relative motion or tendency for motion.
Types of Friction
Dry Friction: When the unlubricated surfaces of two solids are in contact under a condition of sliding or tendency to slide.
Fluid Friction: When adjacent layers in a fluid are moving at different velocities.
Internal Friction: When solid materials are subjected to cyclical loading.

2. Friction The Laws of Dry Friction. Coefficients of Friction Lecture 8
Block of weight (W )placed on horizontal surface. Forces acting on block are its weight and reaction of surface ( N ).
Small horizontal force ( P )applied to block. For block to remain stationary, in equilibrium, a horizontal component ( F ) of the surface reaction is required. ( F ) is a static-friction force.
As ( P )increases, the static-friction force ( F ) increases as well until it reaches a maximum value ( Fm ).
Further increase in ( P ) causes the block to begin to move as ( F ) drops to a smaller kinetic-friction force ( Fk ).

3. Characteristics Of Dry Friction
Lecture 8
Characteristics Of Dry Friction
For a body to be in equilibrium, the following must be true:
To study the characteristics of the friction force ( F ), let us assume that tipping does not occur (i.e., “h” is small or “a” is large).
Then we gradually increase the magnitude of the force ( P ). Typically, experiments show that the friction force ( F ) varies with ( P ).
The maximum friction force is attained just before the block begins to move (a situation that is called “impending motion”).
The value of the force is found using ( Fs = s N ) where ( s ) is called the coefficient of static friction. The value of ( s ) depends on the materials in contact.
Once the block begins to move, the frictional force typically drops and is given by Fk = k N. The value of ( k ) coefficient of kinetic friction is less than ( s ).

4. Characteristics Of Dry Friction
Lecture 8
Characteristics Of Dry Friction
It is also very important to note that the friction force may be less than the maximum friction force.
Because the object is not moving, do not assume the friction force is at it’s maximum of ( Fs = s N ) unless you are told or know motion is impending!
Thus, when the block is on the verge of sliding, the normal force ( N ) and frictional force ( Fs ) combine to create a resultant ( Rs ).
From the figure:
Steps for solving equilibrium problems involving dry friction:
1. Draw the 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 ( F = S N ) unless the impending motion condition is given.
3. Apply the equations of equilibrium and appropriate frictional equations to solve for the unknowns.

5. Motion impending, (Px = Fm)
Four Situations can occur in Friction System:
Lecture 8
No friction, (Px = 0)
No motion, (Px < Fm)
Motion impending, (Px = Fm)
Motion, (Px > Fm) F Fm Fk
Impending Tipping versus Slipping:
For a given W and h, how can we determine if the block will slide first or tip first?
In this case, we have four unknowns ( F, N, x, and P ) and only three E-of-E. F N x W

6. Assumption: Or: Impending Tipping versus Slipping:
Lecture 8
Impending Tipping versus Slipping:
Hence, we have to make an assumption to give us another equation.
Then we can solve for the unknowns using the three E-of-E.
Finally, we need to check if our assumption was correct.
Assumption:
Assume: Slipping occurs
Known: F = s N
Solve: x, P, and N
Check: 0  x  b/2
Or:
Assume: Tipping occurs
Known: x = b/2
Solve: P, N, and F
Check: F  s N
Slipping
Tipping

7. EXAMPLES of Friction: Lecture 8
Example 1: Drum weight = 100 lb, s = 0.5 , a = 3 ft and b = 4 ft.
Find: The smallest magnitude of ( P ) that will cause impending motion (tipping or slipping) of the drum.
Solution
Plan:
a) Draw a FBD of the drum.
b) Determine the unknowns.
c) Make friction assumptions, as necessary.
d) Apply E-of-E (and friction equation as appropriate) to solve for the unknowns.
e) Check assumptions, as required. X 3 4 5
1.5 ft
100 lb
4 ft F P
There are four unknowns: P, N, F and x.
First, let’s assume the drum slips. Then the friction equation is : ( F = s N = 0.5 N ).
A FBD of the drum
+   FX = (4 / 5) P – 0.5 N = 0
+   FY = N – (3 / 5) P – = 0
These two equations give: N

8. Lecture 8
Continue Example 1:
P = 100 lb and N = 160 lb
+  MO = (3 /5) 100 (1.5) – (4 / 5) 100 (4) (X) = 0
Check: X =  so, it is OK!
Drum slips as assumed at P = 100 lb
Example 2: A uniform ladder weighs 20 lb. The vertical wall is smooth. The floor is rough and s = 0.8.
Find: The smallest magnitude of ( P ) needed to move (tip or slide) the ladder.
Solution
There are four unknowns: NA, FA, NB, and P. Let us assume that the ladder will tip first. Hence, NB = 0
  FY = NA – 20 = 0 ; so, NA = 20 lb
+  MA = 20 ( 3 ) – P( 4 ) = 0 ; so, P = 15 lb
+   FX = 15 – FA = 0 ; so, FA = 15 lb
Now check the assumption:
Fmax = s NA = 0.8 * 20 lb = 16 lb
Is FA = 15 lb  Fmax = 16 lb? Yes, hence our assumption of tipping is correct.

9. NB 4 ft P FA NA 3 ft Lecture 8 Continue Example 2: A FBD of the ladder
20 lb NB
4 ft
3 ft NA FA
A FBD of the ladder
Remember:
That if you assume that the ladder slips first ( F = s . NA ), and find ( NB and P ).
Then your ( P ) would be (17 lb)
And the situation at which ( P = 15 lb ) happens before ( P = 17 lb ).
So the ladder tips first.
Example 3: Determine the maximum weight ( W ) the man can lift using the pulley system, without and then with the ( leading pulley at A ). The man has a weight of ( 200 lb ) and the coefficient of friction between his feet and the ground is (  = 0.6 ).
Solution
We start with a FBD of the pulley ( C ). It show that weight is suspended with three ropes.
The tension in each rope can be calculated to be equal to:

10. Without the ( Pulley A ), Examine a FBD of the man:
Continue Example 2:
Lecture 8 T W N F T
The man is using the pulleys to gain ( 3 : 1 ), which is the mechanical advantage.
Without the ( Pulley A ), Examine a FBD of the man:
200 lb
N = 200 – 1/3 W. sin 45
We can solve this last equation to find that:
W = 318 lb
b) With the ( Pulley A ), the analysis is exactly the same except that the angle is ( 0 o ) instead of ( 45 o ). In this case we found that:
N = 200 lb, F = 0.6 * 200 = 120 lb
T = F = 120 lb, W = 3 T = 3 * 120 = 360 lb