Friction

- Introduction

-Types of friction

-Dry friction

•Mechanism of Dry Friction

•Static Friction

•Kinetic friction

-Types of Frictional Problems

Introduction

Friction may be defined as a force of resistance acting on a body whichprevents or retards slipping of the body relative to a second body orsurface with which it is in contact.The frictional force always acts tangent to the surface of contactbetween two bodies. This force opposes sliding motion between thebodies.

People could not walk or drive automobiles without the beneficialeffects of friction to make attractive force possible. Belt drives andbrakes all require frictional forces in order to function.

Types of friction

a) Dry Friction: Dry friction occurs when the unlubricated surfaces oftwo solids are in contact under a condition of sliding or a tendency toslide.

b) Fluid Friction: Fluid friction occurs when the contacting surfaces areseparated by a film of fluid (gas or liquid).The nature of fluid friction isstudied in fluid mechanics.

In general, two types of friction can occur between surfaces.

Dry Friction

Consider a solid block of weight W resting on a rough horizontal surface, this block are pulled by horizontal force P (fig.1).

The floor exerts a distribution of both normal force ∆Nn

and frictional force ∆Fn along the contacting surface(fig.2).

Close examination of the contacting surfaces betweenthe floor and block reveals how this frictional andnormal forces develop, (fig.3). It can be seen that manymicroscopic irregularities exist between the twosurfaces and, as a result, reactive forces ∆Rn aredeveloped at each of the protuberances. Each reactiveforce contributes both a frictional component ∆Fn anda normal component ∆Nn.

(fig.1)

(fig.3)

(fig.2)

• Mechanism of Dry Friction

For simplicity:As shown in the FBD (fig.4)

N : resultant of distributed normal forceF : resultant of distributed frictional force

F always acts tangent to the contacting surface, opposite to the direction of P.

N is directed upward to balance the blocksweight W. Notice that N acts at a distance x tothe right of the line of action of W.

(fig.4)

If we perform the experiment andrecord the friction force F as a functionof P, we obtain the relation shown inthe (fig.5).

(fig.5)

As P is increased, the friction force mustbe equal and opposite to P as long asthe block does not slip. During thisperiod the block is in equilibrium, and allforces acting on the block must satisfythe equilibrium equations. Finally, wereach a value of P which causes theblock to slip and to move in the directionof the applied force.

At the same time the friction force decreases slightly and abruptly. It thenremains essentially constant for a time but then decreases still more as thevelocity increases.

• Static Friction

(fig.6)

The region in (fig.5) up to the point of slippage or impending motion is calledthe range of static friction, and in this range the value of friction force isdetermined by the equation of equilibrium. Experimentally, it has beendetermined that the limiting (maximum) static frictional force Fs is directlyproportional to the resultant normal force N .

μs is called the coefficient of static friction

If Rs is the resultant of Fs and N as shown in (fig.6), and making an angle Фs with the N.

tan Ø = Fs/N

• Kinetic friction

After slippage occurs, a condition of kinetic friction accompanies theensuing motion. Kinetic friction force is usually somewhat less than themaximum static friction force. The kinetic friction force Fk is alsoproportional to the normal force.

(fig.7)

μk is called the coefficient of kinetic friction

50(9.81)

• Determine the maximum force P that can be applied to block A in Fig. (a) without

causing either block to move.

• Solution• The problem statement indicates that we are to find P that would cause impending

motion. However, there are two possible ways in which motion can impend:

• impending sliding at surface 1, or impending sliding at surface 2

• The free-body diagrams of the entire system and each block are shown in

• Figs. (b) and (c), respectively. Note that the equilibrium of each block yields

Tipping (Overturning) :

- If x < b/2 no tipping - If x = b/2 verge to tipping

- If x > b/2 tipping occurs

Note:- Greater height h or smaller width b leads to greater chance for tipping.

Impending Tipping

• In the preceding article, we restricted our attention to sliding; the possibility of tipping was

neglected. We now discuss problems that include both sliding and tipping as possible motions.

• Consider again a homogeneous block on a friction surface being pushed by a force P, as shown

in Fig. 7.4 (a). We assume that the weight W of the block, and the dimensions b, h, and d are

known. We wish to determine the magnitude of P that will cause impending motion of the block,

either impending sliding or impending tipping.

• We can gain insight into the solution by comparing the number of unknowns with the number of

available equilibrium equations. From the free-body diagram of the block, Fig. 7.4(b), we see that

there are four unknowns: the applied force P, the resultant normal force N, the friction force F, and

the distance x that locates the line of action of N. Because there are only three independent

equilibrium equations, an additional equation must be found before all unknowns can be

calculated. If we assume impending sliding, the additional equation is F = Fmax =μsN. On the other

hand, if impending tipping about corner A is assumed, the additional equation is x =b/2, becauseN acts at the corner of the block when tipping impends.

• This classification can be easily reworded to include the possibility of impending tipping.

• Type I The problem statement does not specify impending motion (sliding or tipping).

• Type II The problem statement implies impending motion, and the type of motion (sliding at known

surfaces, or tipping) is known.

• Type III The problem statement implies impending motion, but the type of motion (sliding or

tipping) and/or the surfaces where sliding impends are not known.

• Sample problem

• The man in Fig. (a) is trying to move a packing crate across the floor by applying a horizontal force

P. The center of gravity of the 250-N crate is located at its geometric center. Does the crate moveif P =60 N? The coefficient of static friction between the crate and the floor is 0.3.

Free Body Diagram

• Solution• This is a Type I problem because the problem statement does not specify impending motion. To

determine if the crate moves for the conditions stated, we first assume equilibrium and then check

the assumption. However, the check must answer two questions—(1) does the crate slide and (2)

does the crate tip?

• The free-body diagram of the crate is shown in Fig. (b). If the block is assumed to remain in

equilibrium, the three equilibrium equations can be used to calculate the three unknowns: the

normal force N1, the friction force F1, and the distance x locating the line of action of N1, as shownin the following.

• Sample problem

• Calculate the force P required to cause tipping of the packing crate in previous Sample Problem

7.8. Also determine the minimum coefficient of static friction that permits tipping.

Angle of friction

• Figure below shows a block on a friction surface subjected to the horizontal force P. As seen inthe free-body diagram, we let φ be the angle between the contact force R and the normal n to thecontact surface. The angle φ is given by tan φ = F/N, where N and F are the normal and frictionforces, respectively. The upper limit of φ, denoted by φs , is reached at impending sliding when

• F = Fmax =μsN. Therefore, we have

• tan φs = μs

• The angle φs is called the angle of static friction. Note that φ ≤ φs signifies equilibrium and thatφ = φs indicates impending sliding. Therefore, the direction of the contact force R is known at allsurfaces where sliding impends. This knowledge can be frequently utilized to gain insight intoproblems involving two- and three-force bodies.

• In Fig. above, the friction force F opposes the tendency of P to slide the block to the right. If the directionof P is reversed, the direction of F would also be reversed. This leads to the conclusion that the block canbe in equilibrium only if the line of action of R stays within the sector AOB (bounded by ±φs ), as shown

• in Fig. 7.6. For more general loadings, the line of action of R must lie within the cone, called the cone ofstatic friction, that is formed by rotating sector AOB about the normal n. Observe that the vertex angle ofthe cone of static friction is 2φs .

Angle of Repose :

From equilibrium:F = W sin α

N = W cos α

tan α = F/N

- If α < Øs no motion.

- If α = Øs motion impend.

- If α > Øs motion occurs

Note:- When α = Øs this angle called the angle of repose.