CE 201 Statics

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CE 201 - Statics

• Chapter 8 Lecture 1

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FRICTION

• We used to assume that surface of contact between
  two bodies are smooth
• Reaction is assumed normal to surface of contact
• In reality, surfaces are rough

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Characteristics of Dry Friction

• What is friction?
• Friction is a force of resistance acting on a
  body preventing it from slipping. The friction
  force is always tangent to the surface at point
  of contact and its direction is opposite to the
  possible motion of the body.

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Theory of Friction

• Consider block of weight W on a rough surface
• Surface of contact of the block is non-rigid or
  deformable
• N is upward to balance W and Ff is to the left to
  prevent P from moving the block.
• Close look at the contacting surfaces

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Equilibrium

• For simplicity, the resultants N and F will be
  used.
• F is tangent to the surface and opposite to P
• N is found from the distribution of ?Nn and is
  upward to balance W
• N acts at a distance X right to W line of action
• X is necessary for the tipping effect
• Take moment about O
• WX Ph ? X Ph / W
• The block will be on the verge of tipping if X
  a/2

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Impending Motion

• As P increase, F will increase till it reaches a
  maximum value Fs called the limiting static
  frictional force
• Any further increase in P will cause deformation
  at the points of contact and consequently the
  block will move. Fs was found to be directly
  proportional to N
• Fs ?s N
• Where ?s is the coefficient of static friction
  and is dimensionless.

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Motion

• If P was increased to be greater than Fs, the
  frictional force slightly drops to Fk called the
  kinetic frictional force.
• Since P gt Fk , the block will move.
• Fk was found to be
• Fk ?k N
• Where ?k is the coefficient of kinetic friction
  and is approximately 25 less than ?s

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Characteristics of Dry Friction

• Frictional force is tangent to the surface and
  opposite to motion
• Maximum Fs is independent of the area of contact
• Maximum Fs is greater than maximum Fk
• If one body is moving with very low velocity,
  then Fs Fk or ?s ?k
• When body is about to move, then Fs ?s N
• When body is moving, then Fk ?k N

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Angle of Friction

• If block is stationary
• Fs ? ?s N
• If block is on the verge of moving
• Fs ?s N
• If block is moving
• Fk ?k N

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At equilibrium, Fs and N combine to have a
resultant Rs, the angle of static friction ?s
is?s tan-1 ( Fs / N) tan-1 (?s N / N)
tan-1 ?sIf the block is in motion, then the
angle of kinetic friction ?k tan-1 ( Fk / N)
tan-1 (?k N / N) tan-1 ?kBy comparison, ?s
? ?k
Angle of Friction
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Angle of Response

• To measure the coefficient of friction
  experimentally, a block is placed on a plane of
  different material than the block.
• The plane is inclined to the angle ?s
• The block is on the verge of moving
• Fs ?s N

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Angle of Response

• At equilibrium
• Fs W sin ?s
• N W cos ?s
• Since Fs ?s N
• W sin ?s ?s ( W cos ?s )
• ?s tan-1?s
• ?s is called the angle of response
• ?s tan ?s

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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

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Types of Friction Problems

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

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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.

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Example

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

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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
determined Check if FA ? 0.3 NA and FC ? 0.5
NC If inequalities are satisfied ? OK If
inequalities are not satisfied ? frame is not in
equilibrium
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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.

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

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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.

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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

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• 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.

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Example 2

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

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• 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

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• 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.

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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).

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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.