CHAPTER 8 Friction To introduce the concept of

CHAPTER 8 Friction To introduce the concept of dry friction and show to analyze the equilibrium of rigid bodies subjected to this force

Dry 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.

For the body shown in the figure to be in equilibrium, the following must be true: F = P, N = W, and Wx = Ph.

Impending motion FS = s. N coefficient of static friction Motion Fk = k. N coefficient of kinetic friction

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.

PROBLEMS INVOLVING DRY FRICTION 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.

A block with weight W The inclination, s, is noted. Analysis of the block just before it begins to move gives (using Fs = s N): + Fy = N – W cos s = 0 + FX = S N – W sin s = 0 s = (W sin s ) / (W cos s ) = tan s

Pushing the uniform crate that has a weight W and sits on the rough surface. Static, Slip or tip over?

Verge of slipping Tip over

Number of unknowns? 3 Number of E-of-E 3 ∑Fx = 0 ∑Fy = 0 ∑M = 0

Example 8. 1 The uniform crate has a mass of 20 kg. If a force P = 80 N is applied on to the crate, determine if it remains in equilibrium. The coefficient of static friction is μ = 0. 3.

Solution Resultant normal force NC act a distance x from the crate’s center line in order to counteract the tipping effect caused by P 3 unknowns to be determined by 3 equations of equilibrium

SOLUTION SOLVING

Solution Since x is negative, the resultant force acts (slightly) to the left of the crate’s center line No tipping will occur since x ≤ 0. 4 m Maximum frictional force which can be developed at the surface of contact Fmax = μs. NC = 0. 3(236 N) = 70. 8 N Since F = 69. 3 N < 70. 8 N, the crate will not slip.

Example 8. 2 It is observed that when the bed of the dump truck is raised to an angle of θ = 25° the vending machines begin to slide off the bed. Determine the static of coefficient of friction between them and the surface of the truck

Solution Idealized model of a vending machine lying on the bed of the truck Dimensions measured and center of gravity located Assume machine weighs W

FBD Solution Dimension x used to locate position of the resultant normal force N 4 unknowns N, F, µs, x

4 unknowns N, F, µs, x Equations Of Equilibrium, Since slipping impends at θ=25°, Using the first two equations, we have

Solution Angle θ = 25°is referred as the angle of repose By comparison, θ = Φs θ is independent of the weight of the vending machine so knowing θ provides a method for finding coefficient of static friction θ = 25°, x = 0. 233 m (moment equation) Since 0. 233 m < 0. 3 m the vending machine will slip before it can tip.

Quiz 1 The horizontal force is P=80 N, Determine the normal and frictional forces acting on the crate of weight 300 N. The friction coefficients are μs= 0. 3 and μk=0. 2

Quiz 2 Determine the magnitude of force P needed to start towing the crate of mass M=40 kg. Also determine the location of the resultant normal force acting on the crate, measured from point A. μs= 0. 3

Quiz 3 Determine the friction force on the crate of mass M=40 Kg, and the resultant normal force and its position x, measured from point A, if the force is P=300 N. μs= 0. 3 and μk=0. 2

Quiz 4 30 o The uniform dresser has a weight of 360 N and rest on a tile floor which s = 0. 25 1) Determine the smallest magnitude of force F needed to move the dresser. [ F = 121. 45 N] 2) If the man has a weight of 600 N, determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip. [ s = 0. 195]

FBD 1 FBD 2

1) Determine the smallest magnitude of force F needed to move the dresser.

2) If the man has a weight of 600 N, determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip. Fm

Tutorial 2 Q 1 The crate has a mass M=350 N and is subjected to a towing force P acting at an angle 20° with the horizontal. If the coefficient of static friction is μ=0. 5, determine the magnitude of P to just start the crate moving down the plane.

Tutorial 2 Q 2 A packing crate of mass 40 kg is pulled by a rope as shown. The coefficient of static friction between the crate and the floor is 0. 35. If α =40°, determine (a) the magnitude of the force P required for impending motion of the crate (b) whether sliding or tipping is impending.