CE 201 - Statics

Chapter 8 – Lecture 2

Problems Involving Dry Friction

A body subjected to a system of forces including effect of friction

The body is in equilibrium, then the body has to satisfy:

Fx = 0

Fy = 0

Mo = 0

Fs s N

Fk = k N

Types of Friction Problems

Generally, there are three types Equilibrium Impending Motion at all points Tipping or impending motion at some points

Equilibrium

In equilibrium problems, the total number of unknowns should be equal to the available number of equilibrium equations to be solved.

In these cases, once frictional forces are found ( F and N ), the inequality F N should be checked.

If the values of F and N do not satisfy the inequality, then slipping will take place and the body will not be in equilibrium.

Example

ABC is a frame composed of two members (AB and BC). Each member has a weight of 100 N. Check if the membranes are in equilibrium.

A

B

C

A = 0.3 C = 0.5

100 N

Bx

NA

FC

100 N

By

Bx

By

FA

NC

Here we have six unknowns (Bx, By, FA, NA, FC, NC)We also have six equilibrium equations (3 for each member)

( Fx = 0; Fy = 0; Mo = 0)Solving the equations, FA, NA, FC, NC can be determinedCheck if

FA 0.3 NA

and FC 0.5 NC

If inequalities are satisfied OKIf inequalities are not satisfied frame is not in equilibrium

100 N

Bx

NA

FC

100 N

By

Bx

By

FA

NC

Impending Motion at all Points

In this case, the total number of unknowns will be equal to the total number of available equilibrium equations plus the total number of frictional equations (F = N).

If body is on the verge of moving, then Fs = s N will be used.

If body is moving, then Fk = k N will be used.

Example

Bar AB is placed against the wall. Find smallest so that the bar will not slip. The bar has a weight of 100 N.

Here we have five unknowns (FA, NA, FB, NB, )Also we have three equilibrium equations ( Fx = 0; Fy = 0; Mo = 0) and two static frictional equations (FA = 0.3 NA and FB = 0.4 NB)

NOTEIf the bar impends to move, then it will slip at both points (A) and (B) at the same time.

A

B

A = 0.3

B = 0.4

FA

100 N

NA

NB

FB

Tipping or Impending Motion at Some Points

In this case, the total number of unknowns is less than the total number of equilibrium equations plus the total number of frictional equations.

So, several possibilities of motion or impending motion will exist, and the actual situation needs to be determined.

Example 1

ABC is a two-member frame. Each member has a weight of 100 N. Find P needed to cause movement of the frame

A

B

C

A = 0.3 C = 0.5

100 N

Bx

NA

FC

100 N

By

Bx

By

FA

NC

PP

Here we have seven unknowns (Bx, By, FA, NA, FC, NC, P). We also have six equilibrium equations (three for each member) and one static frictional equation.

If A slips, then:

FA =A NA FC C NC

If C slips, then:

FC =C NC FA A NA

To solve the problem, find P for each case and choose the smallest P. If P was found the same for both cases, then slipping will occur at both points at the same time. That means seven unknowns should satisfy eight equations.

100 N

Bx

NA

FC

100 N

By

Bx

By

FA

NC

P

Example 2

A block having a weight W width b and height h is resting on rough surface. Find P needed to cause motion.

s

P

b

h

Here we have 4 unknowns (P, F, N, X). We also have 3 equilibrium equations and 1 static friction equation or conditional equation.

Two Possibilities of Motion

1. Block will slip

Where F = s N and 0 X (b/2) have to be satisfied

N

P

b/2

h

F

X

2. Block will tip

Where X = (b/2) and F s N have to be satisfied.

Solve for both cases and choose the smallest P. If P is the same for both cases, then slipping and tipping of the block will take place at the same time. In such cases, four unknowns should satisfy five equations.

N

P

b/2

h

F

X

Equilibrium versus Frictional Equations

Frictional forces act to oppose motion or impede the motion of a body over its contacting surface.

Frictional forces must always be shown acting with its correct sense on the free-body diagram whenever the frictional equation is used for the solution of the problem (F = N ). This is because frictional equation relates only the magnitude of two perpendicular vectors (F and N).

Procedure for Analysis

Draw the free-body diagram Determine the number of unknowns Always state frictional forces as unknowns unless stated There are 3 equilibrium equations for each body If there are more unknowns than equilibrium equations,

apply frictional equations at some or all points of contact Apply equilibrium equations and frictional equations to

solve for unknowns If the problem involves three-dimensional force system,

apply the equations of equilibrium using Cartesian vectors.