2000, W. E. HaislerFriction (R&S 9)1
Friction
Two contacting bodies, which are being moved relative to each other (sliding), will generate distributed contact forces between them. These can be broken down into components normal and parallel to their surfaces as shown below. The quantity R is the resultant of the distributed force (pressure) that acts on the surface of the block.
Coulomb found that the normal and parallel components are related depending on the properties of the contact surface and whether motion is impending or on-going.
Coulomb's Law of Dry Friction (RS 9.2)
Two types of frictional forces are "observed or deduced" from experiment. These are referred to as static and kinectic (or dynamic) friction. Consider a block of mass "m" on a flat plane (either horizontal or inclined). We desire to know the force required to move the block, or to keep it moving. Due to friction, a reaction R will develop at the contact surface as shown on the free-body. This reaction can be broken up into components normal (N) and parallel to the surface (F) as shown below.
As the force P is increased, the resisting force F increases approximately linearly (consequence of COLM in direction parallel to contact surface) without any motion. However, Coulomb found that when the force P reaches a critical value, motion of the block occurs and the force required to maintain motion decreases. Graphically, this experimental observation becomes:
is the maximum frictional force that can develop before motion begins and is equal to (). Upon further application of force P to the block, the block begins to move and the frictional force drops to (). Thus Coulomb observed that the frictional force depends upon whether motion of the body is impending, or on-going.
We define two frictional "constants" depending upon these two situations:
Impending Motion:
where
is the frictional force that must be overcome (i.e., force P applied) to initiate motion. If , no motion occurs.
Continuing Motion:
where
is the frictional force for continued motion (and equal to force P applied).
A table of selected experimental coefficients of friction is provided in Table 9-1 of R&S. It should be noted that in general by roughly 15-25%. The coefficient friction depends primarily on the material the surface of each material is made of and their roughness. Some examples of :
Steel on steel (clean)0.6Diamond on diamond0.1
Aluminum on aluminum1.9Aluminum on ice0.4
Wood on wood0.25-0.5Glass on metal0.5-0.7
Rubber on solids1-4
The force is oftened referred as the force needed to move an object on a plane. For a force less then , no motion occurs. As soon as is applied to the body, then the required force to continue motion will drop to . For continuing motion, any force greater than will accelerate the body.
It should be that the frictional and normal forces are resultants of distributed contact loads (pressure). These forces may not always act uniformly over the contact area, nor at the center of the contact area. This is particularly the case for a sliding body on an inclined plane when the incline is sufficient so that the block is on the verge of tipping over. In this case, the normal force will be close to corner or point about which the block would tend to pivot.
In most cases, however, we will take the distributed load and replace it by a normal force which acts at the "center of the distributed load."
Example
A 20-kg triangular block sits on top of a 10-kg rectangular block as shown. The coefficient of static friction is 0.40 at all surfaces. Determine the maximum horizontal force P for which motion will not occur.
Three types of motion are possible:
- The triangular block can slip, or
- the triangular block can tip, or
- the rectangular block can slip.
We have to consider all three posibilities in the solution.
If the triangular block tries to slip:
The weight of the block is 20(9.81)=196.2 N. Let =frictional force and =normal force on block. The location (line of action) of the force vector A must be between the front corner of the block and the vertical certerline of the block. Applying COLM gives
For impending motion,
If the triangular block tries to tip:
For tipping, the line of action of the force vector due to friction and normal force must intersect the left corner of the triangular block as shown.
Applying COAM about point B gives
Since, , the triangular block will not tip; i.e., it takes less force to make the block slip then to make it tip.
If the rectangular block tries to slip:
Applying COLM gives:
For impending motion,
Since, , the rectangular block will not slip; i.e., it takes less force to make the triangular block slip then to make the rectangular block slip.
Hence, when P=78.48 N, the triangular block has impending motion (is about to slip).
Example
The block has a mass of 100 kg. The coefficient of static friction between the block and the inclined surface is 0.20. Determine
a)If the system is in equilibrium when P=600N.
b)The minimum force P to prevent motion.
c)The minimum force P for which the system is in equilibrium.
Part a): Construct the free-body diagram of the block. The weight of the block is W=100(9.81)=981 N. Let =frictional force and =normal force on block; and A is the resultant of these. The three forces () must be concurrent if the block is to be in equilibrium.
Now apply COLM parallel (x) and perpendicular (y) to the inclined surface.
Thus, for equilibrium, . For impending motion, .
Since , the block will not slide and hence the block is in equilibrium.
Part b): The minimum force will be required when motion of the block down the incline is impending. For this condition, the frictional force will tend to resist this motion as shown in the free-body. Equilibrium exists when
Solving simultaneously yields
Part c): The maximum force will occur when motion of the block up the incline is impending. For this condition, the frictional force will tend to resist this motion as shown in the free-body. COLM provides the equilibium condition:
Solving simultaneously yields
Recitation Exercise:
Two blocks with masses and are connected with a flexible cable that passes over a frictionless pulley as shown. the coefficient of static friction between the blocks is 0.25. If motion of the blocks is impending, determine
a) the coefficient of friction between block B and the inclined surface and b) the tension in the cable between the two blocks.
Hint: you have to consider two cases: one where the the frictional force between A and B is less then the maximum (), and one where motion is impending and the frictional force is equal to the maximum ().
Example
Two blocks A and B with masses and are connected by a rope that passes over a pulley as shown. Assume that the rope and pulley have negligible mass and that the length of the rope remian constant. The kinetic coefficient of friction between block A and the inclined surface is 0.25. Determine the tension in the rope, and the acceleration of block A after the blocks are released from rest.
A quick check shows that if block A were on a frictionless surface, a force of 294 N acting down the incline would be necessary to hold the system in equilibrium (not moving).
Why? Tension in cable is 60(9.81)=588.6 N. The force needed parallel to inclined surface to keep A in equilibrium is
F = 588.6 - 50(9.81)(3/5) = 294N
acting down the plane.
Since no such force is present, block A must be pulled up the incline and friction must act down the incline as shown in the free-body below. Assuming that block B is heavy enough to cause motion, the magnitude of the friction force will be . Block A has no motion in the direction perpendicular to the surface so that . The acceleration of block A must be in the x-direction (up the incline) and is equal to . We now apply COLM to block A in the x and y directions to obtain:
x:
y:
With , the x-equation gives (a)
Now consider a free-body of block B. Since the rope and pulley are massless and the pulley is frictionless, the tension in the rope is the same on each side of the pulley. Since the rope does not stretch, kinematics requires that .
Applying COLM to block B gives:
x:
y:
Substituting into the y-equation gives
(b)
Solving equations (a) and (b) simultaneously gives