Ch. 4: Friction 203 4.0 Outline 203 Introduction 204 Types of Friction 205 Dry Friction 206
4.0 Outline Ch. 4: Friction 204 4.1 Introduction
In real situation, the forces of action and reaction between contacting surfaces have their components both in the tangential and normal directions to the contacting surface. Tangential forces are known as Friction forces. Whenever a tendency exists for one contacting surface to slide along another surface, the friction forces developed are always in a direction to oppose this tendency. In some systems, friction is undesirable since it normally spoils the required behavior. But in many situations, friction functions the systems. In real case where sliding motion between parts occurs, the friction forces result in a loss of energy. 4.1 Introduction Ch. 4: Friction 205 4.2 Types of Friction
a) Dry (Coulomb) friction when unlubricated surfaces are in contact under a condition of sliding or tendency to slide. Friction force tangent to the surfaces of contact is developed both during the interval leading up to impending slippage and while slippage takes place. Its direction always opposes the motion or impending motion which would occur if no friction were present. b) Fluid friction c) Internal friction
4.2 Types of Friction Ch. 4: Friction 206 4.3 Dry Friction Mechanism of friction
4.3 Dry Friction Ch. 4: Friction 207 Three regions of static motion transition a) No motion is the region up to the point of slippage or impending motion. Friction force is determined by the equations of equilibrium because the system is in equilibrium. When the motion is not
impending, FF< max b) Impending motion is the moment where the body is on the verge of slipping. Static friction force reaches the max value. For a given pair of mating surfaces,
FF=max = µ s N. c) Motion The body starts moving in the direction of the applied force. Here, friction force drops to a
lower value called kinetic friction FN= µk . It will drop further with higher velocity. 4.3 Dry Friction Ch. 4: Friction 208 Friction cone Friction coefficient reflects the roughness of a pair of mating surfaces. The smaller the coefficient value, the smoother the surfaces. Direction of resultant R is specified by tan α = F/N .
When the friction force reaches max value, tan φµ ss = .
When slippage occurs, tan φµ kk = .
The friction angle φφ sk , defines the limiting position of the total reaction force R. The friction cone of vertex angle 2 φφ sk , 2 represents the locus of possible positions for the reaction force R.
Friction force is independent of the apparent or projected area of contact.
4.3 Dry Friction Ch. 4: Friction 209 Friction cone
4.3 Dry Friction Ch. 4: Friction 210 Types of dry friction problems First step is to identify which of these categories applies.
1) Condition of impending motion is known to exist The body is in equilibrium and on the verge of slipping.
Friction force is the max static friction FN= µs 2) Relative motion is known to exist
Friction force is the kinetic friction FN= µk 3) Unknown status of the problem Assume static equilibrium and solve for the required friction force F. Then check and conclude the status.
4.3 Dry Friction Ch. 4: Friction 211 Possible outcomes
a) FN< µs friction force for the assumed equilibrium can be provided and so the body is in static equilibrium. b) FN = µ s max friction force is required for the static equilibrium condition and so motion impends. c) FN> µs surfaces cannot support more friction than
µs N. So the equilibrium assumption is invalid and motion occurs instead. Friction force
is the kinetic friction FN = µ k . Even with the correct kinetic friction substituted, equilibrium equations are still not hold accelerated motion
4.3 Dry Friction Ch. 4: Friction 212 P. 4/1 Determine the max angle θwhich the adjustable incline may have with the horizontal before the block of mass m begins to slip. The coefficient of static friction between the block and the
inclined surface is μs.
4.3 Dry Friction Ch. 4: Friction 213 P. 4/1
at the moment of slipping, friction is FN = µ s upward =−=θ ∑ Fy 0 N mgcos 0 =µθ −= ∑ Fxs 0 N mgsin 0 −1 µss=tan θθ or = tan µ
φs when the friction force reaches max value, tanφµss= −1 R by equilibrium, R = W and φθs = ∴=θµtan s
4.3 Dry Friction Ch. 4: Friction 214 P. 4/2 Determine the range of values which the mass
mo may have so that the 100 kg block shown in the figure will neither start moving up the plane nor slip down the plane. The coefficient of static friction for the contact surface is 0.30.
4.3 Dry Friction Ch. 4: Friction 215 P. 4/2
bounded mo values block start moving FN= µs =−== ∑ Fy 0 N 100gcos20 0, N 922 N
Case I: max mo, start moving up, friction downward = −−µ = = ∑ Fx 0 m os g N 100gsin20 0, mo 62.4 kg
Case II: min mo, start moving down, friction upward = +−µ = = ∑ Fx 0 m os g N 100gsin20 0, mo 6.0 kg
∴≤6.0 mo ≤ 62.4 kg and F ≤ Fmax = 277 N up/downward
4.3 Dry Friction Ch. 4: Friction 216 P. 4/3 Determine the magnitude and direction of the friction force acting on the 100 kg block shown if, first, P = 500 N and, second, P = 100 N. The coefficient of static friction is 0.20, and the coefficient of kinetic friction is 0.17. The force are applied with the block initially at rest.
4.3 Dry Friction Ch. 4: Friction 217 P. 4/3
don’t know if the block is impending or is moving assume static equilibrium P = 500 N: assume the block tends to move up friction downward =−−== ∑ Fy 0 N 500sin 20 100gcos20 0, N 1092.85 N max supportable friction =µs N = 218.6 N = −− = = <µ ∑ Fxs 0 500cos20 F 100gsin20 0, F 134.3 N N ∴the assumption is valid P = 100 N: assume the block tends to slide down friction upward =−−== ∑ Fy 0 N 100sin 20 100gcos20 0, N 956.04 N
max supportable friction =µs N = 191.21 N =+− ==>µ ∑ Fxs 0 F 100cos 20 100gsin20 0, F 241.55 N N ∴ the assumption is invalid, block is moving downward kinetic friction upward=µ N = 162.5 N k 4.3 Dry Friction Ch. 4: Friction 218 P. 4/4 The homogeneous rectangular block of mass m, width b, and height H is placed on the horizontal surface and subjected to a horizontal force P which moves the block along the surface with a constant velocity. The coefficient of kinetic friction between
the block and the surface is μk. Determine (a) the greatest value that h may have so that the block will slide without tipping over and (b) the location of a point C on the bottom face of the block through which the resultant of the friction and normal forces acts if h = H/2.
4.3 Dry Friction Ch. 4: Friction 219 P. 4/4
a) On the verge of tipping over, reaction acts at the corner A
When slippage occurs, tanθµ= k Block moves w/ const. velocity equilibrium Three-force member: reaction at A must pass through B
tanθµ=kk = b/2h, ∴= h b/( 2µ) b) When slippage occurs, tanθµ= k Block moves w/ const. velocity equilibrium Three-force member: reaction at C must pass through G
tanθµ=kk = x/( H/2) , ∴= xµ H/2
4.3 Dry Friction Ch. 4: Friction 220 P. 4/5 The three flat blocks are positioned on the 30°incline as shown, and a force P parallel to the incline is applied to the middle block. The upper block is prevented from moving by a wire which attaches it to the fixed support. The coefficient of static friction for each of the three pairs of mating surfaces is shown. Determine the maximum value which P may have before any slipping takes place.
4.3 Dry Friction Ch. 4: Friction 221 P. 4/5
4.3 Dry Friction Ch. 4: Friction 222 P. 4/5 =−== ∑ Fy1 0 N 30gcos30 0, N1 254.87 N
N21−− N 50gcos30 = 0, N2 = 679.66 N
N32−− N 40gcos30 = 0, N3 = 1019.5 N Since 30 kg-block cannot slide and 50 kg-block is pulled, 50 kg-block tends to move and only 2 cases are possible. Either 50 kg-block alone or 50&40 kg-blocks move together.
50 kg-block tends to move alone F1 & F2 max (either one alone will not slip)
F1=µµ s1 N = 76.46 N, F2= s2 N = 271.86 N
F3max=µ s N 3 = 458.8 N
block #3: F2−+ F 3 40gsin30 = 0, F3 = 468.06 N > F3max ∴block #3 cannot stay still -- the assumption is invalid
50&40 kg-blocks tend to move together F1 & F3 max (either one alone will not slip)
block #3: F2−+ F 3 40gsin30 = 0, F2 = 262.6 N < F2max ∴block #2 & #3 does not slip relative to each other block #2: P−−+ F F 50gsin30 = 0, P = 93.8 N 12 4.3 Dry Friction Ch. 4: Friction 223 P. 4/6 The light bar is used to support the 50 kg block in its vertical guides. If the coefficient of static friction is 0.30 at the upper and 0.40 at the lower end of the bar, find the friction force acting at each end for x = 75 mm. Also find the maximum value of x for which the bar will not slip.
4.3 Dry Friction Ch. 4: Friction 224 N P. 4/6 R Bar is a two-force member. Assume the system is in equilibrium. Hence the reaction forces at both ends F act along the axial direction. = −= = ∑ Fy 0 N 50g 0, N 490.5 N y F limitation of the reaction force on each end −−11 φµAB=tan =°= 21.8 , φµ tan =° 16.7
R N
−1 at x = 75 mm: θ=sin( 75/ 300) = 14.5 °<φφBA < R inside the static friction cone, system is in equilibrium and F= Ntanθ = 126.6 N max x before slipping when the bar angle = that of small friction cone
x/300= sinφB , x = 86.2 mm
4.3 Dry Friction Ch. 4: Friction 225 P. 4/7 Find the tension in the cable and force P that makes the 15 kg lower block (a) to start sliding downward (b) to start sliding upward
4.3 Dry Friction Ch. 4: Friction 226 P. 4/7 N1 = 8gcos20 = 73.75 N
N21−− N 15gcos20 = 0, N2 = 212 N
F1max = 0.3N1 = 22.12 N
F2max = 0.4N2 = 84.81 N
a) pulling down, 15 kg block impends to slide downward
T 8g P−−+ F1max F 2max 15gsin20 = 0, P = 56.6 N
F1max + 8gsin20 −= T 0, T = 49 N
F1
N1 15g N1 F1
P
F2 N2 4.3 Dry Friction Ch. 4: Friction 227 P. 4/7 N1 = 8gcos20 = 73.75 N
N21−− N 15gcos20 = 0, N2 = 212 N
F1max = 0.3N1 = 22.12 N
F2max = 0.4N2 = 84.81 N
b) pushing up, assume 15&8 kg blocks impends to slide upward together the cable slacks T=0 T 8g 8 kg block: 8gsin20−= F1 0, F 1 = 26.84 N > F1max ∴15 kg block impends to slide upward alone
−+P F1max + F 2max + 15gsin20 = 0, P = 157.3 N N1 F1 −−T F1max + 8gsin20 = 0, T = 4.72 N 15g N1
F1 P F2
N2 4.3 Dry Friction Ch. 4: Friction 228 P. 4/8 The uniform slender rod of mass m and length L is initially at rest in a centered horizontal position on the fixed circular surface of radius R = 0.6L. If a force P normal to the bar is gradually applied to its end until the bar begins to slip at the angle θ= 20°, determine the coefficient of static friction.
4.3 Dry Friction Ch. 4: Friction 229 P. 4/8
no slip until θ=20° distance on bar = length on curve
π α [θπ=a/r] a = 20 R= R/9 180
α
L/(2tan20) 20°
(L/2−π R/9) µα=tan = F/N = = 0.211 s L/( 2tan20)
4.3 Dry Friction Ch. 4: Friction 230 P. 4/9 The three identical rollers are stacked on a horizontal surface as shown. If the coefficient
of static friction μs is the same for all pairs of contacting surfaces, find the minimum value
of μs for which the rollers will not slip.
4.3 Dry Friction Ch. 4: Friction 231
P. 4/9 Lower roller tends rolling out at upper contact while tends to slide out at lower contact
condition: one or more contacts impend to slip
FBD: lower left roller (three-force member) = = R ∑ MO 0 F AB F mg α NA from the figure, N< N FA AB ∴< A FAmax F Bmax so F A reaches the limit value before FB ∴ O slipping does occur first at contact A ∴=FA F Amax and F B determined by equilibrium equation
FB B lower roller: three-force member N reaction force at A must pass through contact B B A r from geometry, 30° 15° F= FAmax , ∴=== tanαµs tan15 0.268 O
r B 4.3 Dry Friction Ch. 4: Friction 232 P. 4/10 The industrial truck is used to move the solid 1200 kg roll of paper up the 30°incline. If the coefficients of static and kinetic friction between the roll and the vertical barrier of the truck and between the roll and the incline are both 0.40, compute the required tractive force P between the tires of the truck and the horizontal surface.
4.3 Dry Friction Ch. 4: Friction 233 P. 4/10 To move the paper roll, 3 possibilities 1) A and B both slip 2) only B slips 3) only A slips after calculation, only case 3) is viable
slipping at A, F= 0.4N 1200g A = = ∑ MO 0 F BA 0.4N 0.4NA =−− == ∑ Fx 0 N AB F cos30 N B sin 30 0, NBA 1.307N F =−−− + = B ∑ Fy 0 0.4N A 1200g F BB sin30 N cos30 0 NA A O NA= 22.1 kN, N BB= 28.9 kN, F= 8853 N < 0.4N B B
NB
4.3 Dry Friction